CONTENTS
S.NO.
TOPIC
PAGE NO.
01.
KTG & THERMODYNAMICS
01 - 73
02.
THERMAL PHYSICS
74 - 135
03.
FLUID MECHANICS
136 - 183
k
KTG AND
THERMODYNAMICS
Recap of Early Classes
W now direct our attention to the study of thermodynamics, which involves situations in which the
We
temperature of state (solid, liquid, gas) of a system changes due to energy transfers. As we shall see,
thermodynamics is very successful in explaining the bulk properties of matter and the correlation between these
properties and the mechanics of atoms and molecules.
Thermodynamics also addresses more practical questions. Have you ever wondered how a refrigerator
is able to cool its contents, or what types of transformations occur in a power plant or in the engine of your
automobile, or what happens to the kinetic energy of a moving object when the object comes to rest? The laws
of thermodynamics can be used to provide explanations for these and other phenomena.
1.0
2.0
KINETIC THEORY OF GASES
1.1
Ideal Gas Concept
1.2
Equation of State for Ideal Gas
1.3
Gas Laws
1.4
Expression for Pressure of an Ideal Gas
1.5
Degree of Freedom (F)
1.6
Maxwell's Law of Equipartition of Energy
THERMODYNAMICS
2.1
first Law of Thermodynamics
2.2
Application of First Law of Thermodynamics
2.3
Isometric or Isochoric Process
2.4
Isobaric Process
2.5
Isothermal Process
2.6
Adiabatic Process
2.7
Free Expansion
2.8
Carnot cycle
2.9
Heat Engine
EXERCISE-1
EXERCISE-2
EXERCISE-3
EXERCISE-4
EXERCISE-5
tinku
KTG & Thermodynamics
KTG & THERMODYNAMICS
1.0 KINETIC THEORY OF GASES
The properties of the gases are entirely different from those of solid and liquid. In case of gases, thermal
expansion is very large as compared to solids and liquids .To state the conditions of a gas, its volume, pressure
and temperature must be specified.
Intermolecular force
Solid > liquid > real gas > ideal gas (zero)
Potential energy
Solid < liquid < real gas < ideal gas (zero)
Internal energy, internal kinetic energy, internal potential energy
At a given temperature for solid, liquid and gas:
(i)
Internal kinetic energy : Same for all
(ii)
Internal potential Energy : Maximum for ideal gas (PE = 0) and Minimum for solids (PE = –ve)
(iii)
Internal Energy : Maximum for Ideal gas and Minimum for solid
At a given temperature for rared and compressed gas :
(i)
Internal kinetic energy
®
Same
(ii)
Internal potential energy ®
(PE)Rared > (PE)compressed
(iii)
Internal Energy
®
(U)Rared > (U)compressed
Temperature
Pressure
Volume
1.1
N.T.P.
(Normal temperature)
0° C = 273.15 K
1 atm = 1.01325 × 105 N/m2
= 1.01325 × 105 pascal
22.4 litre
S.T.P.
(Standard Temperature and Pressure)
0.01° C = 273.16K
1 atm
22.4 litre
Ideal Gas Concept
A gas which follows all gas laws and gas equation at every possible temperature and pressure is known as ideal
or perfect gas.
Volume of gas molecules is negligible as compared to volume of container so volume of
gas = volume of container (Except 0 K)
No intermoleculer force act between gas molecules.
Potential energy of ideal gas is zero so internal energy of ideal gas is perfectly translational K.E. of gas. It is
directly proportional to absolute temperature.
So, internal energy depends only and only on its temperature.
Etrans µ T
For a substance U = UKE + UPE
UKE : depends only on T, UPE : depends upon intermolecular forces (Always negative)
Specific heat of ideal gas is constant quantity and it does
not change with temperature
All real gases behaves as ideal gas at high temperature and
low pressure.
CV
ideal gas
temperature
CV
real gas
temperature
Volume expansion coefficient (a) and pressure expansion coefficient (b) is same for a ideal gas and value of
1
1
per °C a = b =
per °C
273
273
Gas molecule have point mass and negligible volume and velocity is very high (10 7 cm/s). That's why there is no
effect of gravity on them.
each is
1.2
Equation of State for Ideal Gas
PV = µRT
Illustration 1.
Solution
where m = number of moles of gas Þ PV =
éRù
é mN ù
M
RT = ê m N ú RT = ê N ú N T = NkT
ë
0û
ë 0û
Mw
By increasing temperature of gas by 5° C its pressure increases by 0.5% from its initial value at
constant volume then what is initial temperature of gas ?
DT
DP
5 ´ 100
´ 100 =
´ 100 = 0.5 Þ T =
= 1000K
Q At constant volume T µ P \
T
P
0.5
1
JEE-Physics
Illustration 2.
Calculate the value of universal gas constant at STP.
Solution
Universal gas constant is given by R = PV
T
One mole of all gases at S.T.P. occupy volume V = 22·4 litre = 22·4 × 10 –3 m3
P = 760 mm of Hg = 760 × 10–3 × 13·6 × 103 × 9.80 N m–2 T = 273 K
\R=
Illustration 3.
Solution
760 ´ 10-3 ´ 13.6 ´ 103 ´ 9.80 ´ 22.4 ´ 10-3
= 8·31 J mol–1 K–1
273
A closed container of volume 0.02 m3 contains a mixture of neon and argon gases at a temperature
of 27°C and pressure of 1 × 105 Nm2. The total mass of the mixture is 28 g. If the gram molecular
weights of neon and argon are 20 and 40 respectively, find the masses of the individual gases in the
container, assuming them to be ideal. Given : R = 8.314 J/mol/K.
Let m gram be the mass of neon. Then, the mass of argon is (28 – m)g.
Total number of moles of the mixture, m =
Now,
By (i) and (ii),
or
m 28 - m 28 + m
+
=
20
40
40
...(i)
PV 1 ´ 105 ´ 0.02
m= =
=0.8 ...(ii)
RT
8.314 ´ 300
28 + m
= 0.8 Þ 28 + m = 32 Þ m = 4 gram
40
mass of argon = (28 – 4)g = 24 g
Illustration 4.
Calculate the temperature of the Sun if density is 1.4 g cm –3, pressure is 1.4 × 109 atmosphere and
average molecular weight of gases in the Sun in 2 g/mole. [Given R = 8.4 J mol –1K–1]
Solution
PV = mRTÞ T =
PV
...(i)
mR
From equation (i) T =
M
rV
M
But m = M and r =
\ m=M
V
w
w
P V Mw P Mw 1.4 ´ 109 ´ 1.01 ´ 105 ´ 2 ´ 10 -3
=
=
= 2.4 × 107 K
rV R
rR
1.4 ´ 1000 ´ 8.4
Illustration 5.
At the top of a mountain a thermometer reads 7°C and barometer reads 70 cm of Hg. At the bottom
of the mountain they read 27°C and 76 cm of Hg respectively. Compare the density of the air at the
top with that at the bottom.
Solution
P
R
By gas equation PV = M RT Þ
=
rT M w
Mw
é
ù
M
M
and
= rú
êQ m =
M
V
ë
û
w
éPù
éPù
Now as MW and R are same for top and bottom ê rT ú = ê rT ú
ë û T ë ûB
rT PT TB 70 300 75
´
=
= 0.9868
So r = P ´ T =
76 280 76
B
B
T
Illustration 6.
Solution
During an experiment an ideal gas is found to obey an additional law VP 2 = constant. The gas is
initially at temperature T and volume V. What will be the temperature of the gas when it expands to
a volume 2V.
By gas equation PV = mRT and VP2 = constant on eliminating P
é A ù
ê
ú V = mRT Þ
ë Vû
2
V=
mR
T
A
\
éT ù
= ê 1ú Þ
V2 ë T2 û
V1
V
2V
=
T
T'
Þ T' =
( 2) T
KTG & Thermodynamics
1.3
Gas Laws
•
Boyle's Law
According to it for a given mass of an ideal gas at constant temperature, the volume of a gas is inversely
1
proportional to its pressure, i.e., V µ
if m and T = Constant
P
m=constant
T=constant
PV
PV
P
P
V
V
Illustration 7.
A sample of oxygen with volume of 500 cc at a pressure of 2 atm is compressed to a volume of
400 cc. What pressure is needed to do this if the temperature is kept constant ?
Solution
Temperature is constant, so P1 V1 = P2V2 \ P2 = P1
Illustration 8.
An air bubble doubles in radius on rising from bottom of a lake to its surface. If the atmosphere
pressuer is equal to that due to a column of 10 m of water, then what will be the depth of the lake.
(Assuming that surface tension is negligible) ?
Given that constant temperature, we use P1V1 = P2V2
P2 = (10) dg (for water column)
P1 = (10+h) dg (where h=depth of lake)
Solution
V1 =
V1
é 500 ù
= 2ê
ú = 2.5 atm
V2
ë 400 û
4p
3
æ 4p 3 ö
4p 3
2r ) = 8 ç
r
(
r , V2 =
= 8V1 Thus for P2V2 = P1V1,
è 3 ÷ø
3
3
We have 10 dg (8V1) = (10 + h) dg V1Þ 80 = 10 + h Þ h = 70 m
Illustration 9.
Solution
A vessel of volume 8.0 × 10–3 m3 contains an ideal gas at 300 K and 200 k Pa. The gas is allowed
to leak till the Pressure falls to 125 kPa. Calculate the amount of the gas leaked assuming that the
temperature remains constant.
As the gas leaks out, the volume and the temperature of the remaining gas do not change. The
PV
number of moles of the gas in the vessel in given by n =
.
RT
PV
The number of moles in the vessel before the leakage is n1 = 1 and that after the leakage is
RT
P2 V
n2 =
.
RT
The amount leaked is n1 – n2 =
•
(P1 - P2 )V (200 - 125) ´ 103 ´ 8.0 ´ 10 -3
=
= 0.24 mole
RT
8.3 ´ 300
Charle's Law
According to it for a given mass of an ideal gas at constant pressure, volume of a gas is directly proportional to
its absolute temperature, i.e. V µ T if m and P = Constant
V
V
T
V
T
T(k)
V
T
3
JEE-Physics
Illustration 10. 1500 ml of a gas at a room temperature of 23°C is inhaled by a person whose body temperature is
37°C, if the pressure and mass stay constant, what will be the volume of the gas in the lungs of the
person ?
Solution
T1 = 273 + 37 = 310 K; T2 = 273 + 23 = 296 K. Pressure and amount of the gas are kept
constant,
T2
V1 V2
293
So T = T \ V2 = V1 ´ T = 1500 ´
= 1417.74 ml
1
310
1
2
•
Gay–Lussac's Law
According to it, for a given mass of an ideal gas at constant volume, pressure of a gas is directly proportional to
its absolute temperature, i.e., P µ T if m and V = constant
P
P
T
T(k)
P
T
P
T
Illustration 11. A sample of O2 is at a pressure of 1 atm when the volume is 100 ml and its temperature is 27°C.
What will be the temperature of the gas if the pressure becomes 2 atm and volume remains
100 ml.
Solution
T1 = 273 + 27 = 300 K
P1 P2
For constant volume T = T
1
2
Þ T2 = T1 ´
•
2
P2
= 300 ´ = 600 K = 600 – 273 = 327°C
1
P1
Avogadro's Law
According to it, at same temperature and pressure of equal volumes of all gases contain equal number of molecules,
i.e., N1 = N2 if P,V and T are same.
•
Dalton's Partial Pressure Mixture Law
According to it, the pressure exerted by a gasesous mixture is equal to the sum of partial pressure of each
component gases present in the mixture, ie., P = P1 + P2 + ....
Illustration 12. The mass percentage in composition of dry air at sea level contains approximately 75.5% of N 2. If
the total atmospheric pressure is 1 atm then what will be the partial pressure of nitrogen ?
Solution
Mnitrogen
æ
ö
29
= 0.755 ´
The mole fraction of nitrogen m1 = ç
= 0.78
÷
Molecular
weight
è
ø
28
The partial pressure of nitrogen P1 = m1
RT m1 mRT æ m1 ö
= ç ÷ P = (0.78) × 1 = 0.78 atm
=
è mø
m V
V
The kinetic theory of gases
•
4
Rudolph Claussius (1822–88) and James Clark Maxwell (1831–75) developed the kinetic theory of gases
in order to explain gas laws in terms of the motion of the gas molecules. The theory is based on following
assumptions as regards to the motion of molecules and the nature of the gases.
KTG & Thermodynamics
Basic postulates of Kinetic theory of gases
•
Every gas consists of extremely small particles known as molecules. The molecules of a given gas are all
identical but are different than those another gas.
•
The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses.
•
The size is negligible in comparision to inter molecular distance (10 –9 m)
Assumptions regarding motion :
•
Molecules of a gas keep on moving randomly in all possible direction with all possible velocities.
•
The speed of gas molecules lie between zero and infinity (very high speed).
•
The number of molecules moving with most probable speed is maximum.
Assumptions regarding collision:
•
The gas molecules keep colliding among themselves as well as with the walls of containing vessel. These
collision are perfectly elastic. (ie., the total energy before collision = total energy after the collisions.)
Assumptions regarding force:
•
No attractive or repulsive force acts between gas molecules.
•
Gravitational attraction among the molecules is ineffective due to extremely small masses and very high
speed of molecules.
Assumptions regarding pressure:
•
Molecules constantly collide with the walls of container due to which their momentum changes. This
change in momentum is transferred to the walls of the container. Consequently pressure is exerted by gas
molecules on the walls of container.
Assumptions regarding density:
•
The density of gas is constant at all points of the container.
Properties/Assumptions of Ideal gas
•
The molecules of a gas are in a state of continuous random motion. They move with all possible velocities
in all possible directions. They obey Newton's law of motion.
r
•
Mean momentum = 0 ; Mean velocity = 0.< v > = 0 ; < v2 > ¹ 0 (Non zero); < v3 > = < v5 > = 0
•
The average distance travelled by a molecule between two successive collisions is called as mean free
path (lm) of the molecule.
•
The time during which a collision takes place is negligible as compared to time taken by the molecule to cover
the mean free path so NTP ratio of time of collision to free time of motion 10–8 : 1.
•
When a gas taken into a vessel it is uniformly distributed in entire volume of vessel such that its density,
moleculer density, motion of molecules etc. all are identical for all direction, therefore root mean velocity
v 2x = v 2y = v z2 ® equal Pressure exerted by the gas in all direction Px = Py = Pz = P ® equal
•
1.4
All those assumptions can be justified, if number of gas molecules are taken very large
i.e., 1023 molecules/cm3.
Expression for Pressure of an Ideal Gas
Consider an ideal gas enclosed in a cubical vessel of length l. Suppose there
are 'N' molecules in a gas which are moving with velocities
r r
r
v1 , v 2 ........v N .
If we consider any single molecule than its instantaneous velocity vr can be
r
expressed as v = v xˆi + v yˆj + v z kˆ
z
A
C
Due to random motion of the molecule vx = vy = vz
v = vx 3 = vy 3 = vz 3 =
2
x
2
y
v +v +v
2
z
0
y
B
l
D
x
l
Suppose a molecule of mass m is moving with a velocity vx towards the face ABCD. It strikes the face of the
cubical vessel and returns back to strike the opposite face.
5
JEE-Physics
Change in momentum of the molecule per collision Dp = – mvx – mvx = – 2 mvx
Momentum transferred to the wall of the vessel per molecule per collision Dp= 2 mvx
The distance travelled by the molecule in going to face ABCD and coming back is 2l.
So, the time between two successive collision is Dt =
Number of collision per sec per molecule is fc =
2l
vx
v rms
vm
molecule velocity
vx
=
, fc = l or fc = l
mean
free
path
2l
m
m
Hence momentum transferred in the wall per second by the molecule is = force on the wall
vx
mv 2x mv 2
=
=
2l
l
3l
force F = (2 mvx)
Pressure exerted by gas molecule P =
Pressure exerted by gas P =
v 2rms =
F
1 mv 2
=
A
3l´A
1 mv 2
=
V
å3
1 mv 2
3 V
[Q A ´ l = V ]
1 mv 2 N 1 mN å v 2
1 mN 2
´ =
v rms
=
3 V
V
N 3 V N
å3
3PV 3mRT
=
v =
M
µM w Þ rms
Average number of molecules for each wall =
v 2x = v 2y = v z2 =
Þ P=
3RT
1M 2
1
v rms = rv 2rms
,P=
Mw
3V
3
N
N
. No. of molecules along each axis =
(Nx = Ny = Nz)
6
3
2
v rms
Root mean square velocity along any axis for gas molecule is
3
(vrms)c= (vrms)y= (vrms)z=
v rms
3
All gas laws and gas equation can be obtained by expression of pressure of gas (except Joule’s law)
Illustration 13. The mass of a hydrogen molecule is 3.32 × 10 –27 kg. If 1023 molecules are colliding per second on
a stationary wall of area 2 cm2 at an angle of 45° to the normal to the wall and reflected elastically
with a speed 103 m/s. Find the pressure exerted on the wall will be (in N/m 2)
Solution
As the impact is elastic
A
r
r
–24
\ p1 = p2 = p = mv = 3.32 × 10 kg m/s
P1
The change in momentum along the normal
r
r
Dp = p 2 - p1 = 2p cos 45° = 2p
If f is the collision frequency then force applied on the wall
F=
O
B
Dp
= Dp ´ f = 2pf
Dt
\ Pressure P =
1.5
45°
45°
P2
F
2pf
=
=
A
A
2 ´ 3.32 ´ 10-24 ´ 1023
= 2.347 ´ 103 N / m2
2 ´ 10-4
Degree of Freedom (F)
The number of independent ways in which a molecule or an atom can exhibit motion or have energy is called it's
degrees of freedom.
The number of independent coordinates required to specify the dynamical state of a system is called it's degrees
of freedom.
6
KTG & Thermodynamics
f=1
For example
(a)
Block has one degree of freedom, because it is confined to move in a straight line
and has only one transistional degree of freedom.
f=1
y
(b)
The projectile has two degrees of freedom becomes it is confined to move
in a plane and so it has two translational degrees of freedom.
f=2
x
(c)
The sphere has two degrees of freedom one rotational and another translational.
Similarly a particle free to move in space will have three translational degrees of
freedom.
Note : In pure rolling sphere has one degree of freedom as KE =
f=2
1
7
K2
mv2 (1+ 2 ) =
mv2
2
10
R
The degrees of freedom are of three types :
(a)
Translational Degree of freedom : Maximum three degree of freedom are there corresponding to
translational motion.
(b) Rotational Degree of freedom : The number of degrees of freedom in this case depends on the
structure of the molecule.
(c)
Vibrational Degree of freedom : It is exhibited at high temperatures.
Degree of freedom for different gases according to atomicity of gas at low temperature
Atomicity of gas
Translational
Rotational
Total
Monoatomic
Ex. Ar, Ne, Ideal gas etc
3
0
3
y
x
z
Diatomic
Ex. O2, Cl2, N2 etc.
3
2
5
Triatomic (linear)
Ex. CO 2, C2H2
3
2
5
Triatomic (Non–linear)
or Polyatomic
Ex. H2O, NH3, CH4
3
3
6
O=C=O
At high temperatures a diatomic molecule has 7 degrees of freedom. (3 translational, 2 rotational and 2 vibrational)
Illustration 14. Calculate the total number of degrees of freedom possessed by the molecules in one cm 3 of H2 gas
at NTP.
Solution
22400 cm3 of every gas constains 6.02 × 1023 molecules.
6.02 ´ 1023
= 0.26875 × 1020
22400
Number of degrees of freedom of a H2 gas molecule = 5
\ Total number of degrees of freedom of 0.26875 × 1020 × 5 = 1.34375 × 1020.
\ Number of molecules in 1 cm3 of H2 gas =
1.6
Maxwell's Law of Equipartition of Energy
The total kinetic energy of a gas molecules is equally distributed among its all degree of freedom and the energy
1
associated with each degree of freedom at absolute temperature T is kT
2
7
JEE-Physics
For one molecule of gas
Energy related with each degree of freedom =
Energy related with all degree of freedom =
1
kT
2
f
1
3
v2
kT Q v x2 = v 2y = v z2 = rms Þ mv 2rms = kT
2
2
2
3
So energy related with one degree of freedom =
1 v 2rms 3 kT 1
m
=
= kT
2
3
2 3
2
Illustration 15. A cubical box of side 1 meter contains helium gas (atomic weight 4) at a pressure of 100 N/m2.
During an observation time of 1 second, an atom travelling with the root–mean–square speed
parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without
any collision with other atoms. Take R =
Solution
25
J/mol–K and k = 1.38 × 10–23 J/K.
3
(a) Evaluate the temperature of the gas.
(b) Evaluate the average kinetic energy per atom.
(c) Evaluate the total mass of helium gas in the box.
Volume of the box = 1m3, Pressure of the gas = 100 N/m2. Let T be the temperature of the gas
(a) Time between two consecutive collisions with one wall =
1
sec
500
2I
This time should be equal to v , where l is the side of the cube.
rms
2lv rms =
\
1
Þ vrms = 1000 m/s
500
l
(1000)2 M (10)6 ( 3 ´ 10-3 )
3RT
=
= 160K
= 1000 Þ T =
M
æ 25 ö
3R
3ç ÷
è 3ø
(b) Average kinetic energy per atom =
(c) From PV = nRT =
Substituting the values, m =
3
3
kT =
[(1.38 × 10–23) = 160] J = 3.312 ×–21 J
2
2
m
PVM
RT, Mass of helium gas in the box m=
M
RT
(100) (1) ( 4 ´ 10-3 )
æ 25 ö (
çè ÷ø 160)
3
= 3.0 × 10–4 kg
Different K.E. of gas (Internal Energy)
Translatory kinetic energy (ET) ET =
•
Kinetic energy of volume V is =
1
3
Mv 2rms = PV
2
2
1
Mv 2rms Note : Total internal energy of ideal gas is kinetic
2
Energy per unit volume or energy density (EV)
•
EV =
8
1 2
1 éMù 2
2 é1 2 ù
Total energy E
3
=
; E V = ê ú v rms = r v rms Q P = ê rv rms ú \ E V = P
2
3 ë2
2ëVû
Volume
V
2
û
KTG & Thermodynamics
•
Molar K.E. or Mean Molar K.E. (E)
E=
•
1
M w v 2rms for N0 molecules or Mw (gram)
2
E=
3
3
RT = N 0kT
2
2
Molecular kinetic energy or mean molecular K.E. ( E )
E=
E
3 RT 3
1
=
= kT
Mw v 2rms , E =
N0 2 N 0 2
2
l
l
Except 0 K, at any temperature T , E > Em > E
At a common temperature, for all ideal gas
l
E and E are same while Em is different and depends upon nature of gas (M w or m)
For thermal equilibrium of gases, temperature of each gas is same and this temperature called as temperature
of mixture (Tm) which can be find out on basis of conservation of energy (All gases are of same
atomicity). Tm =
l
å NT = N T
åN
1 1
+ N2 T2 + ......... + N n Tn
N1 + N2 ......N n
1 mole gas : Mean kinetic energy =
3
RT
2
1 molecule of gases : Mean kinetic energy =
;
Total kinetic energy =
f
RT
2
f
3
kT ; Total kinetic energy = kT
2
2
¦ ® Degree of freedom
Illustration 16. Two ideal gases at temperature T1 and T2 are mixed. There is no loss of energy. If the masses of
molecules of the two gases are m 1 and m2 and number of their molecules are n1 and n2 respectively.
Find the temperature of the mixture.
3
Solution
Total energy of molecules of first gas = n1kT1 ,
2
3
Total energy of molecules of second gas = n2 kT2
2
3
Let temperature of mixture be T then total energy of molecules of mixture = k (n 1 + n 2 )T
2
\
n T + n2 T2
3
3
(n1 + n2 )kT = k(n1 T1 + n2 T2 ) Þ T = 1 1
(n1 + n2 )
2
2
Illustration 17. The first excited state of hydrogen atom is 10.2 eV above its ground state. What temperature is
needed to excite hydrogen atoms to first excited level.
3
Solution
K.E. of the hydrogen atom kT = 10.2 eV = 10.2 ´ 1.6 ´ 10 -19 J
2
(
Þ T=
)
2 10.2 ´ 1.6 ´ 10 -19
´
= 7.88 ´ 104 K
3
1.38 ´ 10 -23
Explanation of Gas Laws from Kinetic Theory
•
Boyle's Law
According to this law, the product of the pressure and the volume of a given
mass of gas at constant temperature is constant. From the kinetic
theory of gases, the pressure of a given mass of an ideal gas is given by
P
T =con stant
V
9
JEE-Physics
P=
1 mN 2
v rms . mN is the mass of the gas which is constant. If the tempera3 V
PV
ture remains constant, the mean–square–velocity of the molecules. ( v 2rms )
T =constant
also remains constant. Thus, from the above equation, we have
PV = constant.
This is Boyle's law. V µ
•
1
(T = constant) or PiVi = PfVf
P
P or V
Charle's Law
According to this law, the volume of a given mass of gas at constant pressure is directly proportional to the absolute temperature of the gas. From
kinetic theory, we have
V
P =con stan t
2 Næ1
ö
V=
mv 2rms ÷
ç
ø
3 Pè2
V=
T (in K )
2 Næ3 ö N
kT = kT
3 P çè 2 ÷ø P
1
3 ù
é
2
êQ 2 mv rms = 2 kT ú
ë
û
V
T
If the pressure P is constant, then for a given mass of the gas,
we have
VµT
This is Charle's Law.
P =co nsta n t
V or T
•
Gay Lussac's law of Pressure law
For a given mass of a gas the pressure of a gas at constant volume (called
Isochoric process) is directly proportional to its absolute temperature,
P=
or
1 mN 2
v rms Q v2rmsµ T \
3 V
P
= constant
T
or
PµT
(V = constant)
Pi Pf
=
Ti Vf
T (in K )
P–T graph in an isochoric process is a straight line passing through
origin or
P
P
v/s P or T graph is a straight line parallel to P or T axis.
T
P
T
P or T
•
Avogadro's Law
Equal volumes of 'all gases' under the same conditions of temperature and pressure contain equal number of
molecules.
At same pressure equal volumes V of different gases contain respectively N 1 and N2 molecules of masses m1
and m2. PV =
1
m1 N1 v 2rms1
3
and PV =
1
2
m 2 N 2 v 2rms2 Þ m1 N1 v 2rms = m 2 N 2 v rms
1
2
3
Now, if the gases are at the same temperature, their average kinetic energies of translation per molecule are
equal. That is
•
1
1
2
m1 v rms
= m2 v 2rms2 Þ N1 = N2.
1
2
2
Dalton's Law of Partial Pressures
The total pressure exerted by a mixture of non–reacting gases occupying a vessel is equal to the sum of the
individual pressures which each gas would exert if it alone occupied the whole vessel.
10
KTG & Thermodynamics
Let as consider a mixture of gases occupying a volume V. Suppose the first gas contains N 1 molecules, each of
mass m1 having mean–square–speed v 2rms , the second gas contains N2 molecules each of mass m2 and
1
mean–square–speed v 2rms , and so on. Let P1, P2,........ respectively the partial pressures of the gases. Each gas
2
fills the whole volume V. According to kinetic theory, we have
P1 V =
Additing, we get
1
1
m 1N1 v 2rms1 , P2 V = m 2 N 2 v 2rms2 , and so on.
3
3
(P1 + P2 + .......)V =
1
(m1N1v 2rms1 + m 2N 2 v 2rms2 + ......)
3
......(i)
1
1
1
2
2
2
Now, the whole mixture is at the same temperature. \ m1 v rms1 = m2 v rms2 ...... = mv rms
2
2
2
Substituting this result in eqn. (i) we have (P1 + P2 + ......)V =
1
2
(N1 + N 2 + ......)mv rms
3
The mixture has a total number of molecules (N 1 + N2 + .......). Hence the pressure P exerted by the mixture
is given by PV =
1
(N1 + N2 + ......)
3
That P = P1 + P2 + ........
mv 2rms
This is Dalton's law of partial pressures.
Different Speeds of Gas Molecules
•
Average velocity
Because molecules are in random motion in all possible direction in all possible velocity. Therefore, the average
r
r
r
v + v 2 +............ v N
r
velocity of the gas in molecules in container is zero. < v > = 1
=0
N
RMS speed of molecules v
=
rms
3P
=
r
3RT
=
Mw
3kT
kT
= 1.73
m
m
r
Mean speed of molecules :By maxwell’s velocity distribution law vM or < |v|> = v
mean
r
< |v|> = v
mean
=
r
r
r
|v 1 |+|v 2 |+..........|v n |
8P
=
=
N
pr
8RT
=
pM w
8kT
kT
= 1.59
pm
m
Most probable speed of molecules (vmp)
At a given temperature, the speed to which maximum number of molecules belongs is called as most probable
speed (vmp)
vmp =
2P
=
r
2RT
Mw =
2kT
= 1.41
m
kT
m
Maxwell's law of distribution of velocities
n u m b e r of m ole cu le s ( N )
N max
(T 1 )
T 1 =5 00 K
v m p =m ost p rob a ble speed
v m a x =m a xim um sp eed of m o lecule
N max
(T 2 )
N max
(T 3 )
T 2 =1 000K
T 3 =2 00 0K
v m p (T
1
)
v m p (T
2
)
v m p (T
velocity of m o lecule
3
)
v
11
JEE-Physics
•
At any given temperature graph drawn in between molecular velocity and number of molecules is known as
velocity distribution curve.
•
The velocities of molecules of a gas are in between zero and infinity (0 –
•
With the increase in the temperature, the most probable velocity and maximum molecule velocity both increases.
•
The number of molecules within certain velocity range is constant although the velocity of molecule changes
continuously at particular temperature.
•
The area enclosed between the (N – v) curve and the velocity axis presents the total number of molecules.
¥)
On the basis of velocity distribution Maxwell established gives the law of equipartition of energy for gases of any
temperature.
Velocity of sound in gas medium (vs )
vsound =
gP
=
r
• At any temperature
gRT
=
Mw
gkT
m
vrms > vMean > vMP > vsound
v rms
• For a gas at any temperature (T) v
=
sound
v rms
3
=
, v
g
MP
(always)
3
2
• A temperature is not possible at which above order can be changed
vrms ¹ vMean ¹ vMP ¹ vsound
(always )
Illustration 18. The velocities of ten particles in ms–1 are 0, 2, 3, 4, 4, 4, 5, 5, 6, 9. Calculate
(i) average speed and
(ii) rms speed (iii) most probable speed.
Solution
(i)
average speed,
(ii) rms speed,vrms
vav=
0+2+ 3+ 4 + 4 + 4 +5 + 5 + 6+ 9
42
=
= 4·2 ms–1
10
10
é (0)2 + (2)2 + (3)2 + (4)2 + (4)2 + (4)2 + (5)2 + (5)2 + (6) 2 + (9) 2 ù
ú
= êë
10
û
12
1/ 2
é 228 ù
= ê
ú
ë 10 û
= 4.77 ms–1
(iii) most probable speed vmp = 4 m/s
Illustration 19. At what temperature, will the root mean square velocity of hydrogen be double of its value at
S.T.P., pressure remaining constant ?
Solution
Let v1 be the r.m.s. velocity at S.T.P. and v 2 be the r.m.s. velocity at unknown temperature T2.
\
v12
T
= 1 or T = T
2
2
1
T2
v2
2
é v2 ù
2
ê ú = 273 × (2) = 273 × 4 = 1092 K = (1092 – 273) = 819°C
v
ë 1û
Illustration 20. Calculate rms velocity of oxygen molecule at 27°C
Solution
Temperature, T = 27° C
Þ
273 + 27 = 300 K,
Molecular weight of oxygen = 32 × 10–3 kg and R = 8·31 J mol–1 K–1
rms velocity is vrms =
12
3RT
=
M
3 ´ 8 × 31 ´ 300
= 483·5 ms–1
32 ´ 10-3
KTG & Thermodynamics
Illustration 21. Calculate the kinetic energy of a gram moelcule of argon at 127°C.
Solution
Temperature, T = 127°C = 273 + 127 = 400 K, R = 8.31 J/mol K
K.E. per gram molecule of argon =
1.
3
3
RT=
× 8.31 × 400 = 4986 J
2
2
At pressure P and absolute temperature T a mass M of an ideal gas fills a closed container of volume V. An
V
additional mass 2M of the same gas is added into the container and the volume is then reduced to
and the
3
T
temperature to . The pressure of the gas will now be:
3
(A)
P
3
(B) P
(C) 3 P
(D) 9 P
2.
Number of collisions of molecules of a gas on the wall of a container per m2 will :
(A) Increase if temperature and volume both are doubled.
(B) Increase if temperature and volume both are halved.
(C) Increase if pressure and temperature both are doubled.
(D) Increase if pressure and temperature both are halved.
3.
The pressure of one mole of an ideal gas varies according to the law P = P0 - aV 2 , where P0 and a are positive
constants. The highest temperature that the gas may attain is:–
(A)
4.
2P0
3R
æ P0 ö
çè ÷ø
3a
1/ 2
(B)
3P0
2R
æ P0 ö
çè ÷ø
3a
1/ 2
(C)
P0
R
æ P0 ö
çè ÷ø
3a
1/ 2
(D)
P0
R
æ P0 ö
çè ÷ø
3a
1/ 2
Figure shows a cylindrical tube of cross-sectional area A fitted with two frictionless
pistons. The pistons are connected to each other by a metallic wire. Initially, the
temperature of the gas is T0 and its pressure is p0 which equals the atmospheric
pressure. (a) What is the tension in the wire? (b) What will be the tension if the
temperature is increased to 2T0?
2.0 THERMODYNAMICS
Branch of physics which deals with the inter–conversion between heat energy and any other form of energy is
known as thermodynamics. In this branch of physics we deals with the processes involving heat, work and
internal energy. In this branch of science the conversion of heat into mechanical work and vice versa is studied.
•
Thermodynamical System
The system which can be represented in of pressure (P), volume (V) and temperature (T), is known thermodynamic
system. A specified portion of matter consisting of one or more substances on which the effects of variables
such as temperature, volume and pressure are to be studied, is called a system. e.g. A gas enclosed in a cylinder
fitted with a piston is a system.
•
Surroundings
Anything outside the system, which exchanges energy with the system and which tends to change the properties
of the system is called its surroundings.
•
Heterogeneous System
A system which is not uniform throughout is said to be heterogeneous. e.g. A system consisting of two or more
immiscible liquids.
13
JEE-Physics
•
Homogeneous System
A system is said to be homogeneous if it is completely uniform throughout.
e.g. Pure solid or liquid.
•
Isolated System
A system in which there can be no exchange of matter and energy with the surroundings is said to be an
isolated system.
•
Universe
The system and its surroundings are together known as the universe.
•
Thermodynamic variables of the system
(i) Composition (m)
(ii) Temperature (T)
(iii) Volume (V)
(iv) Pressure (P)
•
Thermodynamic state
The state of a system can be described completely by composition, temperature, volume and pressure.
If a system is homogeneous and has definite mass and composition, then the state of the system can be
described by the remaining three variables namely temperature, pressure and volume. These variables are
interrelated by equation PV = µRT. The thermodynamic state of the system is its condition as identified by two
independent thermodynamic variables (P, V or P, T or V, T).
•
Zeroth law of thermodynamics
If objects A and B are separately in thermal equilibrium with a third object C (say thermometer), then objects
A and B are in thermal equilibrium with each other. Zeroth law of thermodynamics introduce the concept of
temperature. Two objects (or systems) are said to be in thermal equilibrium if their temperatures are the same.
In measuring the temperature of a body, it is important that the thermometer be in the thermal equilibrium with
the body whose temperature is to be measured.
•
Thermal equilibrium
Thermal equilibrium is a situation in which two objects in thermal contact cease to exchange energy by the
process of heat. Heat is the transfer of energy from one object to another object as a result of a difference in
temperature between them.
•
Internal Energy
Internal energy of a system is the energy possessed by the system due to molecular motion and molecular
configuration. The energy due to molecular motion is called internal kinetic energy (Uk) and that due to
molecular configuration is called internal potential energy (Up). dU = dUk + dUp
If there no intermolecular forces, then dUp = 0 and dU = dUk = m cv dT
cv = Specific heat at constant volume and dT = Infinitesimal change in temperature
m = Mass of system
M = Molecular weight
m
C v dT
M
Internal energy in the absence of inter–molecular forces is simply the function of temperature and state only, it
is independent of path followed.
DU = Uf – Ui
Ui = Internal energies in initial state and U f = Internal energies in final state
Molar heat capacity Cv = Mcv
For µ–moles of ideal gas dU = µC v dT =
•
Thermodynamic Processes
In the thermodynamic process pressure, volume, temperature and entropy of the system change with time.
Thermodynamic process is said to take place if change occurs in the state of a thermodynamic system.
•
Sign convention used for the study of thermodynamic processes
Heat gained by a system
Positive
Heat lost by a system
Negative
The work done by a system
Positive
Work done on the system
Negative
Increase in the internal energy of system
Positive
Decrease in the internal energy of system
Negative
•
Indicator Diagram or P–V Diagram
In the equation of state of a gas PV = µRT
Two thermodynamic variables are sufficient to describe the behavior of a thermodynamic system.
14
KTG & Thermodynamics
•
•
•
•
If any two of the three variables P, V and T are known then the third can be calculated.
P–V diagram is a graph between the volume V and the pressure P of the system.
The volume is plotted against X–axis while the pressure is plotted against Y–axis.
The point A represents the initial stage of the system. Initial pressure of the system is P i and initial volume of
the system Vi .
(Pi , Vi)
The point B represents the final state of the system. Pf and Vf
P A
are the final pressure and final volume respectively of the system. The
points between A and B represent the intermediate states of the system.
B
With the help of the indicator diagram we calculate the amount of work
(P f, V f)
done by the gas or on the gas during expansion or compression.
Cyclic process
V
Cyclic process is that thermodynamic process in which the system returns to its initial stage after undergoing a
series of changes.
Non–cyclic process
Non–cyclic process is that process in which the system does not return to its initial stage.
Quasi–static or equilibrium process
Quasi–static is a thermodynamic process which proceeds extremely slowly such that at every instant of time,
the temperature and pressure are the same in all parts of the system.
Reversible and Irreversible processes
A reversible process is one in which the changes in heat and work of direct process from initial to a final state
are exactly retraced in opposite sense in the reverse process and the system and surroundings are left in their
initial states. The reversibility is an ideal concept and can not be realized in practice.
The process which is not reversible is the irreversible process. In nature the processes are irreversible.
Work done by Thermodynamic System
•
•
•
dx
One of the simple Illustration of a thermodynamic system
area of
(P, V)
piston is A
is a gas in a cylinder with a movable piston.
F = P× A
If the gas expands against the piston
Gas exerts a force on the piston and displace it
through a distance and does work on the piston.
If the piston compresses the gas
When piston moved inward, work is done on the gas.
The work associated with volume changes
(Pi , Vi)
P A
If pressure of gas on the piston = P.
(P f, Vf)
Then the force on the piston due to gas is F = PA
B
When the piston is pushed outward an infinitesimal distance dx,
P
the work done by the gas is dW = F × dx = PA dx
dV
The change in volume of the gas is dV = Adx, \ dW = PdV
V
For a finite change in volume from V i to Vf, this equation is then integrated between V i to Vf to find the
Vf
Vf
i
i
net work done W = òV dW = òV PdV
Hence the work done by a gas is equal to the area under P–V graph.
Following different cases are possible.
(i)
Volume is constant
P
initial
V
initial
final
or
final
V
V = constant and WAB = 0
P
15
JEE-Physics
(ii)
Volume is increasing
P
P
Initial
A
Initial
A
Final
B
Final
B
or
V
V
V is increasing
(iii)
WAB > 0
WAB = Shaded area
Volume is decreasing
P
Final
P
B
Initial
A
Final
B
Initial
A
or
V
V is decreasing
(iv)
WAB < 0
V
WAB = – Shaded area
Cyclic process
P
P
V
V
Wclockwise cycle = + Shaded area
Wanticlockwise cycle = – Shaded area
Work Done in Clockwise Cycle
P
P
P
B
B
=
A
A
V
B
+
V
V
WAB I Positive
Wcyclic = WAB I (positive) + WAB II (negative) =
16
A
WAB II Negative
f
a o th
are d pa
se
clo
KTG & Thermodynamics
2.1
First Law of Thermodynamics
Let a gas in a cylinder with a moveable piston changes from an initial
equilibrium state to a final equilibrium state.
System change its state through path 'a' :
The heat absorbed by the system in this process = dQa
The work done by the system = dWa
P initial
a
b
c
final
V
Again for path 'b' :
Heat absorbed by the system = dQb , Work done by the system = dWb.
It is experimental fact that the dQa – dWa = dQb – dWb
Both dQ and dW depend on the thermodynamic path taken between two equilibrium states, but difference
(dQ – dW) does not depends on path in between two definite states of the system.
So, there is a function (internal energy) of the thermodynamic coordinates (P, V and T) whose final value (U f)
minus its initial value (Ui) equals the change dQ – dW in the process.
dU = dQ – dW.
This is the first law of thermodynamics.
Heat supplied to the system and work done by the system are path dependent so they are denoted by dQ and
dW respectively. Change in internal energy DU = Uf – Ui does not depends on path it depends only on initial
and final positions of the system. So, it is denoted by dU (or DU)
First Law of Thermodynamics
If some quantity of heat is supplied to a system capable of doing external work, then the quantity of heat
absorbed by the system is equal to the sum of the increase in the internal energy of the system and the external
work done by the system. dQ = dU + dW or Q = W + DU
* This law is applicable to every process in nature
* The first law of thermodynamics introduces the concept of internal energy.
* The first law of thermodynamics is based on the law of conservation of energy.
* dQ, dU and dW must be expressed in the same units (either in units of work or in units of heat).
* This law is applicable to all the three phases of matter, i.e., solid, liquid and gas.
* dU is a characteristic of the state of a system, it may be any type of internal energy–translational kinetic
energy, rotational kinetic energy, binding energy etc.
Illustration 22. The pressure in monoatomic gas increases linearly from 4 × 10 5 Nm–2 to 8 × 10+5 Nm–2 when its
volume increases from 0.2m3 to 0.5 m3. Calculate.
(ii) Increase in the internal energy,
(iii) Amount of heat supplied,
(iv) Molar heat capacity of the gas R = 8.31 J mol–1 K–1
P1 = 4 × 105 Nm–2
(i)
P2 = 8 × 10+5 Nm–2, V1 = 0.2 m3, V2 = 0.5 m3
Work done by the gas = Area under P–V graph (Area ABCDEA)
1
1
= (AE +BD) × AC = (4 × 105 + 8 × 105) × (0.5–0.2)
2
2
=
(ii)
1
× 12 × 105 × 0.3 = 1.8 × 105 J
2
Increase in internal energy DU = CV (T2 – T1) =
P
8
(10 5 × N/m 2)
Solution
(i) Work done by the gas,
4
0
B
C
A
E
0.2
3
(m )
D
0.5 V
CV
CV
R(T2–T1) =
(P2V2 – P1V1)
R
R
For monoatomic gas
CV =
3
3
R \ DU =
[(8 × 105 × 0.5) – (4 × 105 × 0.2)]
2
2
=
3
2
[4 × 105 – 0.8 × 105 ] = 4.8 × 105 J
17
JEE-Physics
(iii) Q = DU + W = 4.8 × 105 + 1.8 × 105 = 6.6 × 105 J
(iv) C =
QR
Q
QR
6.6 ´ 105 ´ 8.31
=
=
=
= 17.14 J/mole K
h(P2 V2 - P1 V1 )
hDT hRDT
1 ´ 3.2 ´ 105
Illustration 23. When a system is taken from state a to state b, in figure along the
P
path a ® c ® b, 60 J of heat flow into the system, and 30 J of
c
b
work is done :
(i) How much heat flows into the system along the path
a
d
a ® d ® b if the work is 10 J.
(ii) When the system is returned from b to a along the curved
V
path, the work done by the system is –20 J. Does the system
absorb or liberate heat, and how much?
(iii) If, Ua = 0 and Ud = 22 J, find the heat absorbed in the process
a ® d and d ® b.
Solution
For the path a, c, b, DU = Q – W = 60 – 30 = 30 J or Ub – Ua = 30 J
(i)
Along the path a, d, b,Q = DU + W = 30 + 10 = 40 J
(ii) Along the curved path b, a, Q = (U a – Ub) + W = (–30) + (–20) = –50 J, heat flows out the
system
(iii) Qad = 32 J; Qdb = 8 J
2.2
Application of First Law of Thermodynamics
•
Melting Process
When a substance melts, the change in volume (dV) is very small and can, therefore, be neglected. The
temperature of a substance remains unchanged during melting process.
Let us consider the melting of a mass m of the solid. Let L be the latent heat of fusion i.e., the heat required
L to change a unit mass of a solid to liquid phase at constant temperature.
Heat absorbed during melting process, dQ = mL
By the first law of thermodynamics , dQ = DU + dW Þ mL = DU [Q dW=PdV=P × 0=0]
So, the internal energy increases by mL during the melting process.
•
Boiling Process
When a liquid is heated, it changes into vapour at constant temperature (called boiling point) and pressure.
When water is heated at normal atmospheric pressure, it boils at 100°C. The temperature remains unchanged
during the boiling process.
Let us consider the vaporisation of liquid of mass m. Let V l and Vv be the volumes of the liquid and vapours
respectively.
The work done in expanding at constant temperature and pressure P, dW = PDV = P (Vv – Vl)
Let the latent heat of vaporisation = L \ Heat absorbed during boiling process, dQ = mL
Let Ul and Uv be the internal energies of the liquid and vapours respectively then DU = Uv – Ul
According to the first law of thermodynamics, dQ = DU + dW \ mL = (Uv – Ul) + P(Vv – Vl)
Illustration 24. 1 kg of water at 373 K is converted into steam at same temperature. Volume of 1 cm 3 of water
becomes 1671 cm 3 on boiling. What is the change in the internal energy of the system if the
latent heat of vaporisation of water is 5.4 × 10 5 cal kg –1?
Solution
Volume of 1 kg of water = 1000 cm³ = 10–3 m³,
Volume of 1 kg of steam = 10³ × 1671 cm³ = 1.671 m³
Change in volume, DV = (1.671 – 10–3) m³ = 1.670 m³ ,
Pressure, P = 1 atm. = 1.01 × 105 N m–2
In expansion work done, W = PDV = 1.01 × 105 × 1.67 J =
1.691 ´ 105
cal = 4.026 × 104 cal
4.2
But DU = Q – W (first law of thermodynamics)
or DU = (5.4 × 105 – 0.4026 × 105) cal = 4.9974 × 105 cal
18
KTG & Thermodynamics
2.3
Isometric or Isochoric Process
•
Isochoric process is a thermodynamic process that takes place at constant volume of the system, but pressure
and temperature varies for change in state of the system.
Equation of state P = constant × T (P and T are variable, V is constant)
Work done In this process volume remains constant DV = 0 or dV = 0 Þ W =
Form of first Law Q = DU
It means whole of the heat supplied is utilized for change in internal
energy of the system. Q = DU = µ Cv DT
Slope of the P–V curve
dP
=¥
dV
ò
Vf
Vi
PdV = 0
V
P
dP
=¥
dV
dV
=0
dP
V
P
Specific heat at constant volume (CV)
•
•
•
The quantity of heat required to raise the temperature of 1 gram mole gas through 1 °C at constant volume is
equal to the specific heat at constant volume.
A gas enclosed in a cylinder having rigid walls and a fixed piston. When heat is added to the gas, there would
be no change in the volume of the gas.
When a substance melts, the change in volume is negligibly small. So, this may be regarded as a nearly
isochoric process.
Heating process in pressure cooker is an Illustration of isometric process.
Illustration 25. An ideal gas has a specific heat at constant pressure CP =
5R
. The gas is kept in a closed vessel of volume
2
0.0083 m3 at a temperature of 300K and a pressure of 1.6 × 106 Nm–2. An amount of 2.49 × 104 J of
heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas.
Solution
5R
3R
-R=
, DV = 0, T1 = 300 K, V = 0.0083 m3, P1 = 1.6 × 106 Nm–2
2
2
From first law of thermodynamics Q = DU + PDV Þ DU = Q = 2.49 × 104 J
CV = CP – R =
From gas equation n =
PV 1.6 ´ 106 ´ 0.0083 16
=
=
RT
8.3 ´ 300
3
\ DU = nCVDT Þ DT =
DU 2.49 ´ 104 ´ 6
=
= 375 K
nC v
3 ´ 8.3 ´ 16
Final temperature = 300 + 375 = 675K
P2 T2
According to pressure law P µ T Þ P = T
1
1
T2
1.6 ´ 106 ´ 675
= 3.6 × 106 Nm–2
Þ P2 = T ´ P1 =
300
1
Illustration 26. 5 moles of oxygen is heated at constant volume from 10°C to 20°C. What will be change in the
internal energy of the gas? The gram molecular specific heat of oxygen at constant pressure.
CP = 8 cal/mole and R = 8.36 joule/mole °C.
Solution
\ CV = CP – R = 8 – 2 = 6 cal/mole °C
\ Heat absorbed by 5 moles of oxygen at constant volume
Q = nCV D T = 5 × 6 (20 – 10) = 30 × 10 = 300 cal
At constant volume D V = 0.
\ DW = 0
\ From first law of thermodynamics Q = DU + W Þ 300 = DU + 0 Þ DU = 300 cal
19
JEE-Physics
2.4
Isobaric Process
•
•
Isobaric process is a thermodynamic process that takes place at constant pressure, but volume and temperature
varies for change in state of the system.
Equation of state V = constant × T or V µ T
Work done
In this process pressure remains constant
DP = 0
Vf
Work done W = òV PdV = P(Vf - Vi )
i
Form of first Law Q = DU + P (Vf – Vi)
•
µ Cp dT = µ Cv dT + P(Vf – Vi)
It is clear that heat supplied to the system is utilized for :
(i) Increasing internal energy and
(ii) Work done against the surrounding atmosphere.
Slope of the PV curve :
•
P
æ dP ö
=0
çè
÷
dV ø isobaric
V
dP
=0
dV
dV
=¥
dP
V
P
Specific heat at constant pressure (CP)
The quantity of heat required to raise the temperature of 1 gram mole gas through 1°C at constant
pressure is equal to the specific heat. Heating of water at atmospheric pressure. • Melting of solids and
boiling of liquids at atmospheric pressure.
•
Illustration 27. At normal pressure and 0°C temperature the volume of 1 kg of ice is reduced by 91 cm 3 on melting.
Latent heat of melting of ice is 3.4 × 105 J/kg. Calculate the change in the internal energy when
2kg of ice melts at normal pressure and 0°C. (P=1.01 × 10 5 Nm–2)
Solution
Heat energy absorbed by 2 kg of ice for melting Q = mL = 2 × 3.4 × 10 5 = 6.8 × 105J
Change in volume of 2 kg of ice = 2 × 91 = 182 cm 3 = 182 × 10–6 m3
\
W = PDV = 1.01 × 105 × (–182 × 10-6) = –18.4 J
Since, work is done on ice so work W is taken –ve. Now from first law of thermodynamics
Q = DU + W
Þ DU = Q – W
= 6.8 × 105 – (–18.4)
= (6.8 × 105 + 18.4)J
Illustration 28. What amount of heat must be supplied to 2.0 × 10 –2 kg of nitrogen (at room temperature) to raise
the temperature by 45°C at constant pressure. Molecular mass of N 2 = 28, R = 8.3 J mol–1 K–1.
Solution
7
m
2 ´ 10-2
5
=
= & CP = R
Here m = 2 × 10 kg, Þ n =
3
M 28 ´ 10
7
2
–2
\ Q = nCP DT =
2.5
5 7
´ ´ 8.3 ´ 45 = 933.75 J
7 2
Isothermal Process
In this process pressure and volume of system change but temperature remains constant.
In an isothermal process, the exchange of heat between the system and the surroundings is allowed.
Isothermal process is carried out by either supplying heat to the substance or by extracting heat from it.
A process has to be extremely slow to be isothermal.
Equation of state
P V = constant (µRT) (T is constant)
Work Done
Consider µ moles of an ideal gas, enclosed in a cylinder, at absolute temperature T, fitted with a frictionless
piston. Suppose that gas undergoes an isothermal expansion from the initial state (P1, V1) to the final state
(P2, V2).
20
KTG & Thermodynamics
\
Work done : W =
ò
V2
V1
V2 dV
µRT
V2
dV = µRT ò
= µRT [ log e V ]V1
V1 V
V
é V2 ù
= µRT [loge V2 – loge V1] = µRT log e ê V ú
êë 1 úû
Þ
é P1 ù
W = 2.303µRT log10 ê ú [Q P1V1 = P2V2]
ë P2 û
Form of First Law
There is no change in temperature and internal energy of the system depends on temperature only
So
é V2 ù
DU = 0, Q = 2.303 µRT log10 ê ú
ë V1 û
It is clear that Whole of the heat energy supplied to the system is utilized by the system in doing external work.
There is no change in the internal energy of the system.
Slope of the isothermal curve
P
Pi
For an isothermal process, PV = constant
Differentiating,
PdV + VdP = 0
Þ
VdP = – PdV Þ
dP
P
=dV
V
isotherm
P V = constant
hyperbola
Pf
P
é dP ù
=Slope of isothermal curve, ê ú
V
ë dV û isothermal
Vi
Vf V
For a given system :
• The product of the pressure and volume of a given mass of a perfect gas remains constant in an isothermal
process.
•
Boyle's law is obeyed in an isothermal process.
•
A graph between pressure and volume of a given mass of a gas at constant temperature is known as
isotherm or isothermal of the gas.
•
Two isotherms for a given gas at two different temperatures
P
T2 > T1
T1 and T2 are shown in figure.
•
The curves drawn for the same gas at different temperatures are mutually
parallel and do not cut each other.
•
If two isotherms intersect each other at a single point we get same value of
P and V at intersection point. PV = µ RT1 for temperature T1 and PV = µ
RT2 for temperature T2.
It means T1 = T2 which is not possible.
T2
T1
V
at const P V µ T so T 2 > T 1
•
An ideal gas enclosed in a conducting cylinder fitted with a conducting
piston. Let the gas be allowed to expand very–very slowly.
•
This shall cause a very slow cooling of the gas, but heat will be conducted into the cylinder from the
surrounding. Hence the temperature of the gas remains constant. If the gas is compressed very–very
slowly, heat will be produced, but this heat will be conducted to the surroundings and the temperature of
the gas shall remain constant.
•
The temperature of a substance remains constant during melting. So, the melting process is an isothermal
process.
•
Boiling is an isothermal process, when a liquid boils, its temperature remains constant.
•
If sudden changes are executed in a vessel of infinite conductivity then they will be isothermal.
21
JEE-Physics
Illustration 29. Two moles of a gas at 127°C expand isothermally until its volume is doubled. Calculate the amount
of work done.
Solution
V2
n = 2, T = 127 + 273 = 400K, V =2
1
V2
W = 2.3026 nRT log10 V = 2.3026 × 2 × 8.3 × 400 × log10 2
1
From formula
= 2.3026 × 2 × 8.3 × 400 × 0.3010 » 4.6 × 103J
Illustration 30. Figure shows a process ABCA performed on an ideal gas.
Find the net heat given to the system during the process.
Solution
Since the process is cyclic, hence the change in internal energy is
zero.
The heat given to the system is then equal to the work done by it.
The work done in part AB is W1 = 0 (the volume remains constant).
The part BC represents an isothermal process so that the work done
V2
by the gas during this part is W2 = nRT2 ln V
1
During the part CA
VµT
V
V2
V1
C
A
T1
B
T2 T
nRT
is constant
V
The work done by the gas during the part CA is W 3 = P(V1 – V2) = nRT1 – nRT2 = – nR (T2 – T1).
The net work done by the gas in the process ABCA is
So, V/T is constant and hence, P =
é
ù
V2
- (T2 - T1 )ú
W = W1 + W2 + W3 = nR ê T2 ln
V1
ë
û
The same amount of heat is given to the gas.
2.6
Adiabatic Process
It is that thermodynamic process in which pressure, volume and temperature of the system do change but there
is no exchange of heat between the system and the surroundings.
A sudden and quick process will be adiabatic since there is no sufficient time available for exchange of heat so
process adiabatic.
Equation of state : PV = µRT
g
Equation for adiabatic process PV = constant
Work done
Let initial state of system is (P1, V1, T1) and after adiabatic change final state of system is (P 2, V2, T2) then we
can write P1 V1g = P2 V2g = K (here K is const.)
So
W=
ò
V2
V1
PdV = K ò
V2
V1
æ V - g +1 ö
V dV = K ç - g + 1÷
è
ø
V2
=
-g
V1
K
g
( - + 1)
é V2 - g +1 - V1 - g +1 ù
ë
û
(Q
K = P1 V1g = P2 V2g
)
1
1 é
g
g
g
P1 V1 V1- .V1 - P2 V2g V2- .V2 ùû =
ÞW = g
[P1V1– P2V2]
( - 1) ë
( g - 1)
µR
( T1 - T2 ) (Q PV = µRT)
Þ W= g
( - 1)
Form of first law : dU = – dW
It means the work done by an ideal gas during adiabatic expansion (or compression) is on the cost of change
in internal energy proportional to the fall (or rise) in the temperature of the gas.
22
KTG & Thermodynamics
If the gas expands adiabatically, work is done by the gas. So, W adia is positive.
The gas cools during adiabatic expansion and T1 > T2.
If the gas is compressed adiabatically, work is done on the gas. So, W adia is negative.
The gas heats up during adiabatic compression and T1 < T2.
(Adiabatic
expansion of gas)
Slope of the adiabatic curve
For an adiabatic process, PVg = constant
Differentiating,
Pg Vg – 1 dV + Vg dP = 0
P
g -1
g
Þ V dP = – g PV
g–1
gPV
dP
=dV Þ
dV
Vg
mon
o
atom
ic
g= 1.33
di at
po
om
ic
ly a
to
g= 1.4
m
ic g= 1.67
P
æ Pö
= -g
= g ç- ÷
è Vø
V
gP
é dP ù
=Slope of adiabatic curve, ê dV ú
V
ë û adiabatic
V
(expansion of gas)
Slope of adiabatic is greater than the slope of isotherm
P
slope of adiabatic changes
é dP ù
é Pù
é dP ù
ê dV ú = g ê - V ú = g ê dV ú Þ slope of isothermal changes =g
ë
û adia
ë
û
ë
û iso
isot h
e
a di
a ba
Since g is always greater than one so an adiabatic is steeper than an isotherm
rm a
l
ti c
V
Illustrations of adiabatic process
•
A gas enclosed in a thermally insulated cylinder fitted with a non–conducting piston. If the gas is compressed
suddenly by moving the piston downwards, some heat is produced. This heat cannot escape the cylinder.
Consequently, there will be an increase in the temperature of the gas.
•
If a gas is suddenly expanded by moving the piston outwards, there will be a decrease in the temperature of the
gas.
•
Bursting of a cycle tube.
•
Propagation of sound waves in a gas.
•
In diesel engines burning of diesel without spark plug is done due to adiabatic compression of diesel vapour and
air mixture
Illustration 31. Why it is cooler at the top of a mountain than at sea level?
Solution
Pressure decreases with height. Therefore if hot air rises, it suffers adiabatic expansion.
From first law of thermodynamics DQ = DU + DW Þ DU = – DW [Q DQ = 0]
This causes a decrease in internal energy and hence a fall of temperature.
Illustration 32. 2m3 volume of a gas at a pressure of 4 × 105 Nm–2 is compressed adiabatically so that its volume
becomes 0.5m3. Find the new pressure. Compare this with the pressure that would result if the
compression was isothermal. Calculate work done in each process. g = 1.4
Solution
V1 = 2m3, P1 = 4 × 105 Nm–2,
V2 = 0.5m3
1.4
In adiabatic process
In isothermal process P1V1 = P2V2 Þ P2 =
Now work done in adiabatic process
W=
é 2 ù
ú
ë 0.5 û
P1 V1g = P2 V2g Þ P2 = 4 × 105 ê
=4×105 (4)1.4 = 2.8 × 106 Nm–2
P1 V1 4 ´ 105 ´ 2
=
= 1.6 × 106 Nm–2.
V2
0.5
P2 V2 - P1 V1
(2.8 ´ 106 ´ 0.5) - (4 ´ 105 ´ 2)
=
= 1.48 × 106 J.
g -1
1.4 - 1
Work done in isothermal process W = 2.3026RT log
V2
V1
= 2.3026P1V1 log
V2
V1
é 0.5 ù
æ 1ö
= 2.3026 × 4 × 105 × 2 × log ê
= 2.3026 × 4 × 105 × 2log çè ÷ø = –1.1 × 106 J
ú
4
ë 2.0 û
23
JEE-Physics
Illustration 33. Two samples of a gas initially at same temperature and pressure are compressed from a volume V
to
V
. One sample is compressed isothermally and the other adiabatically. In which sample is the
2
pressure greater?
Solution
Let initial volume, V1 = V and pressure, P1 = P, final volume, V2 =
P2 = ?
For isothermal compression P2V2 = P1V1 or P2 =
V
and final pressure,
2
P1 V1 PV
=
= 2P
V
V2
2
g
g
é V1 ù
é V ù
g
For adiabatic compression P2' = P1 ê ú Þ P2 ' = P ê
ú =2 P
V
V
/
2
ë
û
ë 2û
g
g
g
Þ P ´ = 2 P > 1 \ 2 > 2 and P ' > P
2
2
2
Pressure during adiabatic compression is greater than the pressure during isothermal compression.
Illustration 34. Calculate the work done when 1 mole of a perfect gas is compressed adiabatically. The initial
pressure and volume of the gas are 105 N/m2 and 6 litre respectively. The final volume of the gas is
2 liters. Molar specific heat of the gas at constant volume is
Solution
g
g
For an adiabatic change PV = constant P1V1 = P2V2
As molar specific heat of gas at constant volume C v =
g
3
2
3R
2
[(3)5/3 = 6.19]
R
5
R
3
5
CP 2
5
CP = CV + R = R + R = R Þ g =
=
=
2
2
CV 3
3
R
2
g
5/3
é V1 ù
é6 ù
5
\ P2 = ê ú P1 = ê ú ´ 10 = (3)5/3 × 105 = 6.19 × 105 N/m2
ë2 û
ë V2 û
Work done W =
1
1-g
[P2 V2 - P1 V1 ] =
1
1 - (5 / 3)
5
[6.19 ´ 10 ´ 2 ´ 10
-3
- 10
-5
-3
´ 6 ´ 10 ]
é 2 ´ 102 ´ 3
ù
= -ê
(6.19 - 3) ú = – 3 × 102 × 3.19 = – 957 joules
2
ë
û
– ive sign shows external work done on the gas
Illustration 35. A motor tyre pumped to a pressure of 3 atm. It suddenly bursts. Calculate the fall in temperature
due to adiabatic expansion. The temperature of air before expansion is 27°C. Given g=1.4.
g
Solution
g
We know that T2 P2
1.4
é T2 ù
Þ ê
ú
ë 300 û
é1 ù
=ê ú
ë3û
1–g
g
= T1 P1
1–g
é T2 ù
é P1 ù
Þ ê ú =ê ú
ë T1 û
ë P2 û
0.4
ÞT2 = 219.2 K
Þ T1 – T2 = (300 – 219.2) K = 80.8 K
24
1- g
1.4
é T2 ù
Þ ê
ú
ë 300 û
1-1.4
é3ù
= ê ú
ë1 û
KTG & Thermodynamics
•
When a gas expands its volume increases, then final pressure is less for adiabatic expansion. But, when a gas
compresses its volume decreases, then the final pressure is more in case of adiabatic compression.
P
is o
ad iab at ic
th
er
m
P
P fin al
DP
al
iso b aric
W IB > W IT > W A D
P IB > P IT > P A D
P
DP A D > DP IT > DP IB
IB
DP
AD
IT
IT
iso b aric
is o t h e r
AD
m al
W IB > W IT > W A D
P A D > P IT > P IB
P fin al
DP A D > DP IT > DP IB
IB
ad iab at ic
V
V
2V
V
V/2
V
V
First Law of Thermodynamics Applied to Different Processes
Process
Q
DU
W
Cyclic
W
0
Area of the closed curve
Isochoric
DU
µCvDT (µ mole of gas)
0
Isothermal
W
0
éV ù
éP ù
mRT log e ê f ú = mRT log e ê i ú
V
ë iû
ë Pf û
Adiabatic
0
–W
µR(Tf - Ti )
1-g
Isobaric
µCPDT
P (Vf – Vi) = µR(Tf – Ti)
mCV DT
Illustration 36. Plot P – V , V – T graph corresponding to the P–T graph for an ideal
gas shown in figure. Explain your answers.
Solution
D
A
V
C
B
T
(V - T curve)
(P - V curve)
For process AB T = constant so P µ
1
V
For process BC
For process CD T = constant so V µ
1
P
For process DA P = constant so V µ T
P = constant so V µ T
25
JEE-Physics
1.
Figure shows a cyclic process abca for one mole of an ideal gas. If ab is isothermal process, then which of the
following is the P–T diagram for the cyclic process ?
(A)
(B)
(C)
(D)
2.
When a gas filled in a closed vessel is heated through 1ºC, its pressure increases by 0.4%. The initial temperature
of the gas was :
(A) 25ºC
(B) 250ºC
(C) 250 K
(D) 2500 K
3.
Figure shows a cyclic process. Which of the following is its PV conversion ?
(A)
(B)
(C)
(D)
4.
When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which
increases the internal energy of the gas is
(A) 2 / 5
(B) 3 / 5
(C) 3 / 7
(D) 5 / 7
6.
A monatomic & a diatomic gas both at the N.T.P. having same no. of moles are compressed adiabatically to
half of initial volume :
(A) Increase in temp. will be more for the diatomic gas
(B) Increase in temp. will be more for the monatomic gas
(C) Increase in temp. will be same for both the gases.
(D) Increase in pressure will be more for the monatomic gas.
7.
n moles of a gas filled in a container at temperature T is in thermodynamic
equilibrium initially. If the gas is compressed slowly and isothermally to
half its initial volume the work done by the atmosphere on the piston is:
(A)
26
nR T
2
(B) -
nR T
2
1ö
æ
(C) n R T ç l n 2 - ÷
2ø
è
(D) - n R T l n 2
KTG & Thermodynamics
2.7
Free Expansion
Take a thermally insulated bottle with ideal gas at some temperature T1 and, by means of a pipe with a
stopcock, connect this to another insulated bottle which is evacuated. If we suddenly open the stopcock, the
gas will rush from the first bottle into the second until the pressures are equalized.
free expansion of a gas
Experimentally, we find that this process of free expansion does not change the temperature of the gas – when
the gas attains equilibrium and stops flowing, the final temperature of both botles are equal to the initial
temperature T1. This process is called a free expansion.
The change in the internal energy of the gas can be calculated by applying the first law of thermodynamics to
the free–expansion process.
The process is adiabatic because of the insulation, So Q = 0.
No part of the surroundings moves so the system does no work on its surroundings.
•
•
For ideal gas
(dW)ext. = Work done against external atmosphere
= P dV = 0 (because P = 0)
(dW)int. = Work done against internal molucular forces = 0
dQ = dU + dW Þ 0 = dU + 0
The internal energy does not change dU = 0 So U and T are constant.
The initial and final states of this gas have the same internal energy.
Which implies that the internal energy of an ideal gas does not depend on the volume at all.
The free–expansion process has led us to the following conclusion :
The internal energy U(T) of an ideal gas depends only on the temperature.
For real gas
In free expansion of real gases, measurements show that the temperature changes slightly in a free expansion.
Which implies that the internal energy of a real gas depends on the volume also.
dQ
= 0 (dW)ext. = 0 (Q P = 0)
(dW)int. ¹ 0 (Intermolucular forces are present in real gases)
dQ
= dU + dW Þ 0 = dU + (dW)int. Þ dU = – (dW)int.
Þ U decreases. So T decreases.
Relation between Degree of Freedom and Specific Heat of Gas
Energy related with each degree of freedom =
1
f
kT , Energy related with all degree of freedom = kT
2
2
Internal energy of one mole of ideal gas (total K.E.) U =
f
RT for Isometric process (volume constant)
2
dW = 0
By first law of thermodynamics
dQ = dW + dU
Þ
CV dT = dU
CV =
dU f
R
= R=
dT 2
g -1
gR
éf
ù
. C P = C V + R = ê + 1ú R =
2
g
-1
ë
û
CV =
R
g -1
gR
g -1
,
CP =
and
g =1+
Þ CV =
CP
2
and g = C = 1 + f
V
2
f
General expression for C (C P or CV) in the process PVx = constant C =
For isobaric process P = constant so
x=0
dU
dT
\ C = CP =
R
R
+
g -1 1 - x
R
+ R = CV + R
g -1
27
JEE-Physics
For isothermal process, PV = constant so
g
For adiabatic process PV = constant so
\ C=¥
\ C=0
x=1
x=g
Values of f, U, CV, CP and g for different gases are shown in table below.
Atomicity of gas
f
Monoatomic
3
Diatomic
5
Triatomic and Triatomic linear
(at high temperature)
7
Poly atomic Triangular Non-linear
•
•
•
•
6
Cv
CP
g
3
R
2
5
R
2
7
R
2
6
R = 3R
2
5
R
2
7
R
2
9
R
2
8
R = 4R
2
5
3
7
5
9
7
4
3
= 1.67
= 1.4
= 1.28
= 1.33
1<g<2
If atomicity of gases is same U, CP, CV and g is same for gas mixture.
If in a gas mixture gases are of different atomicity, then for gas mixture g changes according to following
condition. Diatomic g1 < mixture < g2 mono atomic where g1 < g2
If 'f' is the degree of freedom per molecule for a gas, then
Total energy of each molecule =
fkT
2
Total energy per mole of gas = N 0
f
f
kT = RT
2
2
•
According to kinetic theory of gases, the molecule are not interacting with each other. So potential energy
is zero and internal energy of gas molecules is only their kinetic energy.
•
For 'm' mole of a gas : Internal energy at temperature T is U =
•
Change in internal energy is given by dU =
mfRT
= mC V T
2
mfR
(dT) = mC V dT
2
This change is process independent.
CP is greater than CV
If a gas is heated at constant volume, the gas does no work against external pressure. In this case, the whole of
the heat energy supplied to the gas is spend in raising the temperature of the gas.
If a gas is heated at constant pressure, its volume increases. In this case, heat energy is required for the following
two purpose :
(i)
To increase the volume of the gas against external pressure.
(ii)
To increase the temperature of 1 mole of gas through 1 K.
Thus, more heat energy is required to raise the temperature of 1 mole of gas through 1 K when it is heated at
constant pressure than when it is heated at constant volume. \ CP > CV
The difference between CP and CV is equal to thermal equivalent of the work done by the gas in expanding
against external pressure.
Mayor's formula : CP – CV = R
Q
At constant pressure dQ = µCPdT, dU = µCVdT & dW = PdV = µRdT
Now from first law of thermodynamics dQ = dW + dU
Þ
28
µCPdT = µRdT + µCVdT Þ CP = R + CV
Þ CP – CV = R
KTG & Thermodynamics
Illustration 37. Calculate the difference between two specific heats of 1 g of helium gas at NTP. Molecular weight of
helium = 4 and J = 4.186 × 107 erg cal–1.
Solution
Gas constant for 1 g of helium
r=
R
PV
76 ´ 13.6 ´ 981 ´ 22400
=
=
= 2.08 × 107 erg g–1 K–1
Mw T ´ Mw
273 ´ 4
CP – CV =
r
2.08 ´ 107
=
= 0.5 cal g–1 K–1
J 4.186 ´ 107
Illustration 38. Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant
pressure is 6.85 cal mol–1 K–1 and density of hydrogen = 0.0899 g cm –3. One mole of gas = 2.016 g,
J = 4.2 × 107 erg cal–1 and 1 atmosphere = 10 6 dyne cm–2.
Solution
Since the density of hydrogen is 0.0899 g cm–3 therefore volume occupied by 0.0899 g of hydrogen
at NTP is 1000 cm3. So, volume of 1 mole (2.016 g) of gas, V =
1000
´ 2.016 cm3
0.0899
R
PV
106 ´ 1000 ´ 2.016
=
=
= 1.96 cal mol–1 K–1
J T ´ J 0.0899 ´ 273 ´ 4.2 ´ 107
\ CV = CP – 1.96 = (6.85 – 1.96) = 4.89 cal mol–1 K–1
CP – CV =
Illustration 39. The specific heat of argon at constant volume is 0.075 kcal/kg K. Calculate its atomic weight,
[R = 2cal/mol K]
Solution
As argon is monoatomic, its molar specific heat at constant volume will be
3
3
C V = R = ´ 2 = 3 cal/mol K, CV = Mw cV and
cV = 0.075 cal/g K
2
2
So 3 = Mw × 0.075
Þ Mw =
3
= 40 gram/mole
0.075
Effeciency of a cycle (h) :
total Mechanical work done by the gas in the whole process
h=
Heat absorbed by the gas (only + ve)
=
area under the cycle in P-V curve
Heat injected into the system
Illustration 40. n moles of a diatomic gas has undergone a cyclic process ABC as
shwon in figure. Temperature at a is T0. Find
(i) Volume at C.
(ii) Maximum temprature.
(iii) Total heat given to gas.
(iv) Is heat rejected by the gas, if yes how much heat is rejected.
(v) Find out the efficiency.
Solution
(i)
2P0 P0
Since triangle OAV0 and OCV are simillar therefore V = V Þ V = 2V0
0
(ii) Since process AB is isochoric hence
Since process BC is isobaric therefore
PA PB
=
TA TB Þ TB = 2T0
TB
T
= C Þ T = 2T = 4 T
C
B
0
VB VC
(iii) Since process is cyclic therefore Q = W = area under the cycle =
1
PV .
2 0 0
29
JEE-Physics
(iv) Since U and W both are negative in process CA
\ Q is negative in process CA and heat is rejected in process CA
QCA = WCA + UCA
= -
1
5
[P + 2P0] V0 – nR (Tc – Ta )
2 0
2
= -
æ 4P0 V0 P0 V0 ö
5
1
[P0 + 2P0] V0 – nR çè
nR
nR ÷ø
2
2
= –9P0V0 = Heat injected.
(v) h = efficiency of the cycle =
work done by the gas P0 V0 / 2
=
heat injected
Qinjected × 100
é5
ù é5
ù 19
PV.
where Qinj = QAB + QBC = ê nR(2T0 - T0 ) ú + ê nR(2T0 ) + 2P0 (2V0 - V0 ) ú =
ë2
û ë2
û 2 0 0
Therefore h =
100
%
19
P
T1
st a
Q1
on
nd
P2
D
on
B
s in k
sta
P4
so u
r ce
on
nd
Carnot devised an ideal engine which is based on a
reversible cycle of four operations in succession :
(i) Isothermal expansion,
A®B
(ii) Adiabatic expansion,
B®C
(iii) Isothermal compression
C®D
(iv) Adiabatic compression.
D®A
A
on
2.8 CARNOT CYCLE
P1
Q2 T2
P3
V1
V4
C
V2
V3
V
Main parts of Carnot's Engine are
•
Source of heat
It is a hot body of very large heat capacity kept at a constant high temperature T 1. Its upper surface is perfectly
conducting so that working substance can take heat from it.
•
Mechanical arrangements and working substance
It is a hollow cylinder whose walls are perfectly non–conducting and its base is perfectly conducting fitted with
non–conducting piston. This piston move without any friction. Ideal gas enclosed in cylinder as a working
substance.
•
Heat sink
It is a cold body at low temperature T2. It is a body of large heat capacity and its upper surface is highly
conducting so that working substance can reject heat to it.
•
Stand
It is made by perfectly insulating material. When cylinder placed on it working substance can expended or
compressed adiabatic.
•
Working
A set of reversible processes through them working substance is taken back to initial condition to get maximum
work from this type of ideal engine.
Processes of Carnot's cycle can be denoted by an indicator diagram.
•
Isothermal expansion A ® B
Initially the cylinder is taken to be in thermal equilibrium with the high temperature
T1, this is initial state of working substance denoted by point A (P 1, V1, T1). After
that the piston is allowed to move outward slowly. With the movement of the piston.
The process is very slow so that it is isothermal. Heat from reservoir flows through
the base of cylinder into the gas so temperature of the gas remains T 1. Gas expand
and receive heat Q1 from source and gets state B(P2, V2, T1)
This heat input Q1 to the gas from path A to B is utilized for doing work W 1.
30
Q
source
KTG & Thermodynamics
By path A to B the heat input to the gas = the work done against the external
pressure. W1 = Q1 =
•
ò
V2
V1
PdV =
ò
V2
V1
µRT1
V
dV = µRT1 ln 2
V
V1
Adiabatic expansion B ® C
Now cylinder is put in contact with a non–conducting stand and piston is allowed to
move outward, because no heat can enter in or leave out so the expansion of gas is
adiabatic. The temperature falls to T2K and gas describes the adiabatic from B to
point C (P3, V3, T2). During this expansion more work is done (W2) at the expense of
the internal energy.
Work done in adiabatic path BC is W2 =
•
µR
( T - T2 )
g -1 1
Q=0
Isothermal compression C ® D
Now the gas cylinder is placed in contact with sink at temperature T2. The piston is
moved slowly inward so that heat produced during compression passes to the sink.
The gas is isothermally compressed from C to point D (P4, V4, T2). The heat rejected
Q2 to the cold reservoir (sink) at T2 occurs over this path. Amount of work done on
gas W3 = amount of heat rejected to the sink
æV ö
æV ö
Q2 = W3 = mRT2ln ç 4 ÷ Þ Q2 = µRT2ln ç 4 ÷
è V3 ø
è V3 ø
•
Q2
sink
Adiabatic compression D ® A
The cylinder is removed from the sink and is put in contact with insulating stand now
piston moves inward. Heat is not allowed to go out and it increases the internal
energy of the system. Now work is done on the gas during adiabatic compression
from state D to initial point A (P1, V1, T1). No heat exchanges occur over the
µR
(T2 - T1 )
adiabatic path. Work done on the system W4 = g
-1
This cycle of operations is called a Carnot cycle.
In first two steps work is done by engine W 1 and W2 are positive
In last two steps work is done on gas
W3 and W4 are negative
The work done in complete cycle W = the area of the closed part of the P–V cycle.
W = W1 + W2 + W3 + W4
Q=0
V2
V4
V2
µR
V4
µR
\ W = µRT1ln V + g - 1 (T1 - T2 ) + µRT2 ln V + g - 1 (T2 - T1 ) = µRT1ln V + µRT2 ln V
1
3
1
3
Efficiency of Carnot Engine, h =
W
Q1
Þh=
µRT1 l n
V2
V1
+ µ RT2 ln
µRT1 l n
V4
V3
V2
V1
B to C and D to A are adiabatic paths
V
V
2
3
so T1V2(g – 1) = T2 V3(g – 1) and T1V1(g – 1) = T2 V4(g – 1) Þ V = V
1
4
Þ h=
h=
T1 - T2
Q1 - Q2
Q2
T2
Þ h =1- Q = 1- T
=
T1
Q1
1
1
T1 - T2
´ 100%
T1
Þ h=
Q1 Q2
=
T1
T2
Q1 - Q2
´ 100%
Q1
31
JEE-Physics
The efficiency for the Carnot engine is the best that can be obtained for any heat engine. The efficiency of a
Carnot engine is never 100% because it is 100% only if temperature of sink T 2 = 0 which is impossible.
Carnot Theorem
No irreversible engine (I) can have efficiency greater than Carnot reversible engine (R) working between same hot
T2
Q2
and cold reservoirs. hR > hI Þ 1 - T > 1 - Q
1
1
2.9
source
hot reservoir
T1
Heat Engine
Heat engine is a device which converts heat into work.
Three parts of a heat engine:
Q1
(i)
Source of high temperature reservoir at temperature T1
working
W
(ii)
Sink of low temperature reservoir at temperature T2
substance
(iii)
Working substance.
Q2
In a cycle of heat engine the working substance extracts heat Q 1 from source, does
sink
some work W and rejects remaining heat Q2 to the sink.
cold reservoir
T1 - T2 Q1 - Q2
work done (W)
Efficiency of heat engine h =
Þ
=
Q1
T1
heat taken from source (Q1 )
T2
Types of Heat Engine
•
External combustion engine
Steam engine is the Illustration of external combustion engine. Its efficiency is 10 to 20% (small).
•
Internal combustion engine
The Illustrations are petrol engine and diesel engine.
There are four strokes in internal combustion engine.
(i)
Intake stroke or Charging stroke
(ii)
Compression stroke
(iii)
Working (or power) stroke of Expansion stroke
(iv)
Exhaust stroke
The useful work is done in third stroke called work stroke or power stroke.
The efficiency of internal combustion engine is 40 to 60%.
Illustration 41. A carnot engine working between 400 K and 800 K has a work output of 1200 J per cycle. What is
the amount of heat energy supplied to the engine from source per cycle?
Solution
W = 1200 J, T1 = 800 K, T2 = 400 K
\h=1–
400 1200
1200
T2
W
=
Þ 1 - 800 = Q Þ 0.5 = Q
T1 Q1
1
1
Heat energy supplied by source Q1 =
1200
= 2400 joule per cycle
0.5
Illustration 42. The temperatures T1 and T2 of the two heat reservoirs in an ideal carnot engine are 1500°C and
500°C respectively. Which of the following : increasing T1 by 100°C or decreasing T2 by 100°C would
result in a greater improvement in the efficiency of the engine?
Solution
T1 = 1500°C = 1500 + 273 = 1773 K and T2 = 500°C = 500 + 273 = 773 K.
The efficiency of a carnot's engine h = 1 -
T2
T1
When the temperature of the source is increased by 100°C, keeping T2 unchanged, the new
temperature of the source is T´1 = 1500 + 100 = 1600°C = 1873 K.
The efficiency becomes h´= 1 32
T2
773
=1= 0.59
T1 '
1873
KTG & Thermodynamics
On the other hand, if the temperature of the sink is decreased by 100°C, keeping T 1 unchanged, the
new temperature of the sink is T´2 = 500 – 100 = 400°C = 673 K. The efficiency now becomes
h´´= 1 -
T´2
673
=1= 0.62
T1
1773
Since h´´ is greater than h´, decreasing the temperature of the sink by 100°C results in a greater
efficiency than increasing the temperature of the source by 100°C.
Illustration 43. A heat engine operates between a cold reservoir at temperature T 2 = 300 K and a hot reservoir at
temperature T1. It takes 200 J of heat from the hot reservoir and delivers 120 J of heat to the cold
reservoir in a cycle. What could be the minimum temperature of hot reservoir?
Solution
Work done by the engine in a cycle is W = 200 – 120 = 80 J. h =
W
80
=
= 0.4
Q 200
T2
300
300
300
From carnot's Theorem 0.4 £ 1 - T = 1 - T Þ T £ 0.6 Þ T1 ³
Þ T1 ³ 500
0.6
1
1
1
Refrigerator
It is inverse of heat engine. It extracts heat (Q 2) from a cold reservoir,
same external work W is done on it and rejects heat (Q 1) to hot reservoir.
The coefficient of performance of a refrigerator.
Q2
Q2
1
=
=
Heat extracted from cold reservoir =
Q
W Q1 - Q2
b=
1
-1
Work done on refrigerator
Q2
Hot reservoir
T1
Q1 = Q2 + W
W
Q2
Cold reservoir
T2
Q1 T1
T2
Q
1
1
=
For Carnot reversible refrigerator Q = T \ b = 2
Þ b=
W é Q1
ù é T1
ù
2
2
T1 - T2
ê Q - 1ú ê T - 1ú
ë 2
û ë 2
û
Illustration 44. A carnot engine works as a refrigrator between 250 K and 300 K. If it receives 750 cal of heat from
the reservoir at the lower temperature. Calculate the amount of heat rejected at the higher temperature.
Solution
T1 = 300 K
T2 = 250 K Q2 = 750
Q1 = ?
Q1
Q2
=
T1
T2
Q1 =
300
´ 750 = 900 cal
250
Illustration 45. The temperature inside and outside of refrigerator are 260 K and 315 K respectively. Assuming that
the refrigerator cycle is reversible, calculate the heat delivered to surroundings for every joule of work
done.
Solution
T2 = 260 K, T1 = 315 K
; W = 1 joule
Q2
T2
Coefficient of performance of Carnot refrigerator b = W = T - T
1
2
\
Q2
260
260
260
=
=
= 6.19 J
Þ Q2 =
1
315 - 260
42
42
Illustration 46. A refrigerator takes heat from water at 0°C and transfer it to room at 27°C. If 100 kg of water is
converted in ice at 0°C then calculate the work done. (Latent heat of ice is 3.4 × 10 5 J/kg)
Solution
T2
273
273
Coefficient of performance (COP) = T - T = 300 - 273 = 27
1
2
W=
Q2
mL
100 ´ 3.4 ´ 105 100 ´ 3.4 ´ 105 ´ 27
=
=
=
= 3.36 × 107 J
COP COP
273 / 27
273
33
JEE-Physics
1.
A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant
pressure until volume is doubled. The average molar heat capacity for whole process is:
(A)
2.
13R
6
(B)
19R
6
(C)
23R
6
(D)
17R
6
An ideal gas undergoes an expansion from a state with temperature T1 and volume V1 to V2 through three
different polytropic processes A, B and C as shown in the P-V diagram. If ½DEA½, ½DEB½ and ½DEC½ be the
magnitude of changes in internal energy along the three paths respectively, then :
(A) ½DEA½ < ½DEB½ < ½DEC½ if temperature in every process decreases
(B) ½DEA½ > ½DEB½ > ½DEC½ if temperature in every process decreases
(C) ½DEA½ > ½DEB½ > ½DEC½ if temperature in every process increases
(D) ½DEB½ < ½DEA½ < ½DEC½ if temperature in every process increases
Comprehension (3 - 5)
An ideal gas initially at pressure p0 undergoes a free expansion (expansion against vaccum under adiabatic
conditions) until its volume is 3 times its initial volume. The gas is next adiabatically compressed back to its
original volume. The pressure after compression is 32/3 p0.
3.
The pressure of the gas after the free expansion is :
(A)
4.
p0
3
The gas
(A) is monoatomic.
(C) is polyatomic.
5.
1/ 3
(B) p 0
(C) p0
(D) 3p0
(B) is diatomic.
(D) type is not possible to decide from the given information.
One mole of an ideal gas is taken from state A to state B by three different processes, (a) ACB (b) ADB (c)
AEB as shown in the P - V diagram. The heat absorbed by the gas is :
(A) greater in process (b) then in (a)
(C) the same in (a) and (c)
6.
(B) the least in process (b)
(D) less in (c) then in (b)
During an experiment, an ideal gas is found to obey a condition
P2
= constant [r = density of the gas]. The gas
r
is initially at temperature T, pressure P and density r. The gas expands such that density changes to
(A) The pressure of the gas changes to
2 P.
(B) The temperature of the gas changes to 2 T..
(C) The graph of the above process on the P-T diagram is parabola.
(D) The graph of the above process on the P-T diagram is hyperbola.
34
r
2
KTG & Thermodynamics
SOME WORKED OUT ILLUSTRATIONS
Illustration 1.
Certain perfect gas is found to obey PVn = constant during adiabatic process.
The volume expansion coefficient at temperature T is
(A)
1
(B) (
1 - n) T
1-n
T
n
T
(C)
(D)
1
nT
Ans. (B)
Solution
æ 1 ö
ç
÷
è
ø
PVn = constant & PV = mRT Þ V µ T 1- n Þ
Þ volume expansion coefficient =
DV æ 1 ö DT
=ç
è 1 - n ÷ø T
V
1
DV
=
(
VD T
1 - n) T
Illustration 2.
5n, n and 5n moles of a monoatomic, diatomic and non-linear polyatomic gases (which do not react chemically
with each other) are mixed at room temperature. The equivalent degree of freedom for the mixture is(A)
25
7
(B)
48
11
(C)
52
11
(D)
50
11
Ans. (D)
Solution
f1n1 + f2 n2 + f3 n 3 (5n)(3) + (n)(5) + (5n)(6) 50
=
=
n1 + n2 + n3
5n + n + 5n
11
Illustration 3.
Figure shows the adiabatic curve on log–log scale performed on a
ideal gas. The gas must be :–
(A) Monoatomic
(B) Diatomic
(C) A mixture of monoatomic and diatomic
(D) A mixture of diatomic and polyatomic
Ans. (C)
Solution
For adiabatic process TVg–1 = constant
7
6
5
4
logT
feq =
3
2
1
0 1
2
3
æ 5ö
3
Þ log T + (g–1) log V = constant Þ slope = – (g–1) = – ç ÷ Þ g =
è 10 ø
2
4 5 6
log V
7
8
9 10 11
5
7
7
3 5
;
For diatomic gas g =
As < g = <
3
5
5
2 3
Hence, the gas must be a mixture of monoatomic & diatomic gas.
For monoatomic gas g =
Illustration 4.
Figure demonstrates a polytropic process (i.e. PV n = constant) for an ideal gas. The work done by the gas will
be in process AB is
(A)
15
PV
2 0 0
(B)
14
PV
3 0 0
(C) 8P0V0
(D) Insufficient information
Ans. (B)
Solution
For a polytropic process PVn = constant Þ 16P0V0n = P0(2V0)n Þ n=4
Work done =
P1 V1 - P2 V2 16P0 V0 - P0 (2V0 ) 14
=
=
PV
n -1
4 -1
3 0 0
35
JEE-Physics
Illustration 5.
+
–
A vessel contains 14 g (7 moles) of hydrogen and 96 g (3 moles) of oxygen at
STP. Chemical reaction is induced by passing electric spark in the vessel till one
of the gases is consumed. The temperature is brought back to it's starting
H
value 273 K. The pressure in the vessel is(A) 0.1 atm
(B) 0.2 atm
O
(C) 0.3 atm
(D) 0.4 atm
Spark
Ans. (A)
Solution
When electric spark is passed, hydrogen reacts with oxygen to form water (H 2O). Each gram of hydrogen reacts
with eight grams of oxygen. Thus 96 g of oxygen will be totally consumed together with 12 g of hydrogen.
2
The gas left in thevessel will be 2g of hydrogen i.e. number of moles m = = 1 .
2
2
2
P2 m2
P2
1
Using PV = mRT Þ P µ m Þ P = m Þ 1 = 10 Þ P2 = 0.1atm
1
1
3
Vf(m )
Illustration 6.
Suppose 0.5 mole of an ideal gas undergoes an isothermal expansion as
0.3
energy is added to it as heat Q. Graph shows the final volume V f versus Q.
The temperature of the gas is (use ln 9 = 2 and R= 25/3 J/mol-K)
0.2
(A) 293 K
(B) 360 K
0.1
(C) 386 K
(D) 412 K
Ans. (B)
0
Solution
Vf
Q
1500
1500
Q = W = nRT ln V ; T = nRlnV / V = 0.5 ´ 25 / 3 ´ ln3 = 0.5 ´ 25 / 3 ´ 1 = 360K
f
i
i
500
1000 1500
Q(J)
Illustration 7.
The respective speeds of five molecules are 2,1.5,1.6,1.6 and 1.2 km/s. The most probable speed in
km/s will be
(A) 2
(B) 1.58
(C) 1.6
(D) 1.31
Ans. (C)
Solution
Since maximum number of molecules travel with speed 1.6 km/s so vmp = 1.6 km/s
Illustration 8.
When water is boiled at 2 atm pressure, the latent heat of vaporization is 2.2 × 10 6 J/kg and the boiling point
is 120°C. At 2 atm pressure, 1 kg of water has volume of 10 –3 m3 and 1 kg of steam has a volume of 0.824 m 3.
The increase in internal energy of 1 kg of water when it is converted into steam at 2 atm pressure and 120°C is
[1 atm pressure = 1.013 × 105 N/m2]
(A) 2.033 J
(B) 2.033 × 106 J
(C) 0.167 × 106 J
(D) 2.267 × 106 J
Ans. (B)
Solution
Total heat given to correct convert water into steam at 120 °C is Q = mL = 1 × 2.2 × 10 6 = 2.2 × 106 J
The work done by the system against the surrounding is
PDV = 2 × 1.013 × 105 (0.824 × 0.001) = 0.167 × 106 J \ DU = Q – W = 2.033 × 106 J
1
Illustration 9.
2
T
Given T–P curve for three processes. If initial and final pressure are same for
3
all processes then work done in process 1, 2 and 3 is W1, W2 & W3 respectively.
Correct order is
(A) W1 < W2 > W3
(B) W1 > W2 > W3
(C) W1 < W2 < W3
(D) W1 = W2 = W3
Ans. (B)
P
Solution
Here T µ Pn where for : graph-1, n >1 Þ W > 0 ; graph-2, n =1 Þ W = 0; graph-3, n <1 Þ W < 0
36
KTG & Thermodynamics
Illustration 10.
Figure shows the variation of the internal energy U with density r of one
mole of an ideal monatomic gas for thermodynamic cycle ABCA. Here
process AB is a part of rectangular hyperbola :(A) process AB is isothermal & net work in cycle is done by gas.
(B) process AB is isobaric & net work in cycle is done by gas.
(C) process AB is isobaric & net work in cycle is done on the gas.
(D) process AB is adiabatic & net work in cycle is done by gas.
Ans. (B)
Solution
For the process AB : Ur = constant (hyperbola)
U=
U(J)
A
C
B
r kg3
m
3
3
RT (monoatomic ideal gas); rRT = constant
2
2
æ 1ö
Comparing it with ideal gas equation P = çè ÷ø rRT Þ P is constant.
M
P
P-V graph for the cycle is
C
. Thus work done in cycle is positive
A
B
X
Illustration 11.
One mole of an ideal gas undergoes a process whose molar heat capacity is 4R and in which work done by gas
CP
for small change in temperature is given by the relation dW = 2RdT, then the ratio C is
V
(A) 7/5
(B) 5/3
(C) 3/2
Ans. (C)
Solution
For small change dQ = dU + dW
nCdT = nCV dT + 2nRdT \ C = CV + 2R; 4R = CV + 2R
Also CV =
(D) 2
\ CV = 2R
R
R
3
\
= 2R Þ 2g – 2 = 1 Þg =
g -1
g -1
2
Illustration 12.
æ 8 ö
3
A gas undergoes an adiabatic process in which pressure becomes ç
times and volume become
of
è 3 3 ÷ø
4
initial volume. If initial absolute temperature was T, the final temperature is
(A)
32T
(B)
9 3
2T
(C) T3/2
3
(D)
3T
2
Ans. (B)
Solution
g
For adiabatic process, P1V1 =P2V2
g
æ 8
öæ3 ö
P0 ç V0 ÷
Þ P0 V0 g = ç
è 3 3 ÷ø è 4 ø
3
Also, T1 V1g -1 = T2 V2g -1 Þ TV0g -1 = T2 æç V0 ö÷
è4 ø
Solving (i) & (ii), T2 =
g
....(i)
g -1
....(ii)
2T
3
37
JEE-Physics
Illustration 13.
n moles of an ideal triatomic linear gas undergoes a process in which the temperature changes with volume
as T = k1V2 where k1 is a constant. Choose correct alternative(s).
(A) At normal temperature Cv =
5
R
2
(B) At any temperature Cp – Cv = R
(C) At normal temperature molar heat capacity C = 3R
(D) At any temperature molar heat capacity C = 3R
Ans. (ABC)
Solution
At normal temperature Cv=
æf
ö f
f
5
R=
R ; At any temperature Cp – Cv = çè + 1÷ø - R = R
2
2
2
2
from process T = k1V2 & ideal gas equation PV = nRT we have PV–1 = constant Þ x =–1
Þ C = CV +
R
R
R
= CV +
= CV +
1- x
1+1
2
At normal temperature C =
5
R
R + = 3R
2
2
Illustration 14.
An ideal gas expands in such a way that PV 2 = constant throughout the process. Select correct alternative
(A) This expansion is not possible without heating
(B) This expansion is not possible without cooling
(C) Internal energy remains constant in this expansion
(D) Internal energy increases in this expansion
Ans. (B)
Solution
PV = nRT & PV2 = constant Þ V µ
1
Þ gas can expand only if it cools
T
As tempeature decreases during expansion so internal energy will decrease.
Illustration 15.
Which of the following processes must violate the first law of thermodynamics (Q = W + DEint)?
(A) W > 0, Q < 0 and DEint > 0
(B) W > 0, Q < 0 and DEint < 0
(C) W < 0, Q > 0 and DEint < 0
(D) W > 0, Q > 0 and DEint < 0
Ans. (AC)
Solution
For (A) : W > 0 & DEint > 0 Þ Q > 0
For (B) : W > 0 & DEint < 0 Þ Q > 0 or Q < 0
For (C) : W < 0 & DEint < 0 Þ Q < 0
For (D) : W > 0 & DEint < 0 Þ Q > 0 or Q < 0
Illustration 16.
In a process on a closed system of ideal gas, the initial pressure and volume is equal to the final pressure and
volume
(A) initial internal energy must be equal to the final internal energy
(B) the work done on the system is zero
(C) the work done by the system is zero
(D) the initial temperature must be equal to final temperature
Ans. (AD)
Solution
Here n = constant so if P2 = P1 and V2 = V1 then T2 = T1
Also work done by the system may be zero or may not be zero.
38
KTG & Thermodynamics
V
Illustration 17.
The graph below shows V-P curve for three processes.
Choose the correct statement(s)
(A) Work done is maximum in process 1.
(B) Temperature must increase in process 2 & 3.
(C) Heat must be supplied in process 1.
(D) If final volume of gas in process 1, 2 and 3 are same then temperature must be same.
3
2 1
P
Ans. (AC)
Solution
Area under P-V curve and volume axis represent work.
Internal energy in 1 increases (expansion at constant pressure)
In process 2 and 3 internal energy may decrease depending on the process.
Since final pressure of 1, 2 & 3 are different, final temperature must also be different.
Illustration 18.
During an experiment, an ideal gas is found to obey a condition
P2
= constant [r=density of the gas]. The gas
r
is initially at temperature T, pressure P and density r. The gas expands such that density changes to
r
2
(A) The pressure of the gas changes to Ö2P
(B) The temperature of the gas changes to Ö2T.
(C) The graph of the above process on the P-T diagram is parabola.
(D) The graph of the above process on the P-T diagram is hyperbola.
Ans. (B,D)
Solution
PV = nRT
P µ rT .....(i)
Tµ
1
r
...(ii)
Tµ
1 ...(iii)
r
Illustration 19 to 21
One mole of a monoatomic ideal gas occupies two chambers of a cylinder partitioned by means of a movable
piston. The walls of the cylinder as well as the piston are thermal insulators. Initially equal amounts of gas fill
both the chambers at (P0, V0, T0). A coil is burnt in the left chamber which absorbs heat and expands, pushing
the piston to the right. The gas on the right chamber is compressed until to pressure becomes 32 P 0.
P0,V 0,T 0
19.
The final volume of left chamber is
(A)
20.
V0
8
(B)
15
V
8 0
(C)
7
V
8 0
(D)
9
V
8 0
(C)
13
PV
2 0 0
(D)
17
PV
2 0 0
(D)
131
RT0
4
The work done on the gas in the right chamber is
(A)
21.
P0,V 0,T 0
9
PV
2 0 0
(B) –
9
PV
2 0 0
The change in internal energy of the gas in the left chamber is
(A)
186
RT0
4
(B)
177
RT0
4
(C)
59
RT0
2
39
JEE-Physics
Solution
19. Ans. (B)
Since the compression on the right is adiabatic so P0V0g = PRVRg
15
7
V
V0 =
V
Þ P0 V05/3 = 32P0 VR5/3 Þ VR = 0 Þ VL = V0 +
8 0
8
8
20. Ans. (A)
P V - P0 V0 4P0 V0 - P0 V0 9
=
= P0 V0
Work done on the gas = R R
g -1
5 / 3 -1
2
21. Ans. (B)
For mechanical equilibrium PL = PR = 32 P0
æ 15V0 ö
So PLVL = (32P0) çè
÷ = 60P0V0 = 60nRT0 = nRTL Þ TL = 60T0
8 ø
The change in the internal energy of the gas in the left chamber
DU = nCV DT =
1 3
177
RT0
´ R ´ 59T0 =
2 2
4
P
Illustration 22 to 24.
One mole of an ideal monoatomic gas undergoes a cyclic process as shown in
figure. Temperature at point 1 = 300 K and process 2-3 is isothermal.
22. Net work done by gas in complete cycle is
(A) ( 9ln3 + 12) P0 V0
(B) ( 9ln3 + 4 ) P0 V0
(C) ( 9ln3 - 4 ) P0 V0
(D) ( 9ln3 - 8 ) P0 V0
Heat capacity of process 1 ® 2 is
23.
R
2
24. The efficiency of cycle is
æ 9ln3 + 4 ö
(A) çè
÷
9ln3 + 12 ø
Solution
22. Ans. (C)
(A)
W = W12 + W23 + W31 =
23.
Ans. (D)
(B)
3R
2
æ 9ln3 - 4 ö
(B) çè
÷
9ln3 + 12 ø
(C)
5R
2
æ 9ln3 - 4 ö
(C) çè
÷
9ln3 + 16 ø
2
3P0
P0
3
1
V0
3V 0
v
(D)2R
æ 9ln3 + 12 ö
(D) çè
÷
9ln3 + 16 ø
1
( 4P0 )(2V0 ) + nRTln (3) - P0 (8V0 ) = -4P0 V0 + 9P0 V0ln3 = (9 ln3 - 4 ) P0 V0
2
æ 3R
ö
nç
.DT÷ + DW
PV
è 2
ø
3R 4P0 V0 3R
3R R
DQ
=
=
+
=
+ 0 0 =
+ = 2R
C=
nDT
nDT
2
nDT
2 n ( 2T0 )
2
2
DQ123 = n (2R ) ( DT12 ) + ( W23 ) = n (2R ) ( 8T0 ) + nR ( 9T0 ) ln (3) = nRT0 (16 + 9 ln3 )
Ans. (C) , h =
24.
Illustration 25.
W
æ 9ln3 - 4 ö
=ç
÷
DQ123 è 9 ln3 + 16 ø
7
is expanded according to the law P=2V. The initial volume
5
25
units)
of the gas is equal to V0= 1 unit. As a result of expansion the volume increases 4 times. (Take R =
3
Column - I
Column - II
(A) Work done by the gas
(p) 25 units
(B) Increment in internal energy of the gas
(q) 45 units
(C) Heat supplied to the gas
(r) 75 units
(D) Molar heat capacity of the gas in the process
(s) 15 units
(t) 55 units
Ans. (A) s; (B) r ; (C) q ; (D) p
An ideal gas whose adiabatic exponent equals to g =
40
KTG & Thermodynamics
Solution
W = ò PdV =
4 V0
ò
2VdV = ( V 2 ) V0 = 15V02 = 15 units
4 V0
V0
From PV = nRT, 2V2 = nRT Þ 2 ( V22 - V12 ) = nR ( DT ) Þ nRDT = 30V02
30V02 30 (1)
nR
30
( 5) = 75 units
DT =
=
=
7
g -1
g -1
2
-1
5
2
DU = nCV DT =
Q = W + DU = 15 + 30 = 45 units
Molar heat capacity : C = C V +
R
5
R
5
R
25
= 25 units
= R+
= R + = 3R = 3 ´
1- x 2
1 - ( -1) 2
2
3
Illustration 26.
The figure given below show different process for a given amount for an ideal gas. W is work done by the
system and DQ is heat absorbed by the system.
P
P
(i)
(ii)
1/V
(A)
(B)
(C)
(D)
P
(iii)
adiabatic
V
Column-I
DQ > 0
W<0
DQ < 0
W>0
P
P
(iv)
(v)
V
(p)
(q)
(r)
(s)
(t)
V
V
Column-II
In figure (i)
In figure (ii)
In figure (iii)
In figure (iv)
In figure (v)
Ans. (A) (p,r,t) ; (B) s ; (C) s; (D) p,q,r,t
Solution
figure (i) P µ
1
ÞPV = constant
V
Isothermal (T = constant), so DU = 0
1
is decreasing; So V is increasing hence, DW > 0
V
\\\\\\\\\\\\
\\\\\\\\\\\\
DQ = DU +DW = DW > 0
Illustration 27.
A cylinder of cross-section area A has two pistons of negligible mass separated by distances l loaded with spring
of negligible mass. An ideal gas at temperature T 1 is in the cylinder where the springs are relaxed. When the gas
is heated by some means its temperature becomes T2 and the springs get
l
compressed by
each. If P0 is atmospheric pressure and spring constant k
2
2P0 A
=
, then find the ratio of T2 and T1.
l
Ans. 4
Solution
kx
T
PV
P1 V1 P2 V2
+ P0 = 2P0 Þ 2 = 2 2 = 4
where V1 = lA and V2 = 2lA and P1 = P0 and P2 =
=
A
T
P1 V1
T1
T2
1
Illustration 28.
A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel
containing one mole of He gas (molar mass 4) at a temperature 2T has a pressure of xP. Find the value of x.
Ans. 2
41
JEE-Physics
Solution
PV = nRT, V is constant
Illustration 29.
One mole of a gas is taken from state A to state B as shown in figure.
Work done by the gas is a × 10b J. Find the value of a + b.
25
(Given : T1=320 K, R =
)
3
Ans. 7
Solution
2P0
P
B
T1
P0
1
3
P0 + 2P0 ][ 2V0 - V0 ] = P0 V0
[
2
2
and P0V0 = R × 320
3
3 25
´ R ´ 320 = ´
´ 320 = 4000J = 4 ´ 103 J Þa + b = 4 + 3 = 7
So work done
2
2 3
Illustration 30.
A container having base area A0. Contains mercury upto a height l0. At its bottom
A
a thin tube of length 4l0 and cross-section area A (A<<A0) having lower end
closed is attached. Initially the length of mercury in tube is 3l0. In remaining part 2
mole of a gas at temperature T is closed as shown in figure. Determine the work
done (in joule) by gas if all mercury is displaced from tube by heating slowly the gas
in the rear end of the tube by means of a heater. (Given : density of mercury = r,
atmospheric pressure P0 = 2l0rg, CV of gas = 3/2 R, A= (3/r)m2, l0 = (1/9) m, all
units in S.I.)
Ans. 5
Solution
If x is length of mercury in tube then pressure of gas
A
2V0
V0
V
Work done =
P' = P0 + rgl0 + rgx = 3rgl0 + rgx Þ W = -
0
ò ( 3rgl
3l 0
0
container
P0
0
Hg
A
gas
+ rgx ) Adx = 13.5 rgAl20 = 5
Illustration 31.
The relation between internal energy U, pressure P and volume V of a gas in an adiabatic process is: U=a+bPV
where a = b = 3. Calculate the greatest integer of the ratio of specific heats [g].
Ans. 1
Solution
For an adiabatic process, dQ=dU+PdV =0
dV
dP
+b
=0
Þ d[a+bPV] + PdV=0 Þ bPdV + bV dP + PdV =0 Þ (b+1)PdV+bV dP=0 Þ (b + 1)
V
P
b+1 4
b +1
= = 1.33
Þ (b+1)logV + b logP= constant; V b+1Pb =constant Þ PV b = constant \ g =
b
3
Illustration 32.
One mole of an ideal monatomic gas undergoes the process P = aT1/2, where a is constant. If molar heat
capacity of the gas is bR when R = gas constant then find the value of b.
Ans. 2
ANSWERS
BEGINNER'S BOX-1
1. (C)
2. (B,C)
3. (A)
4. (a) zero
(b) p0 A
3. (C)
4. (D)
6. (B,D)
7. (A)
3. (A)
4. (A)
5. (D)
6. (B,D)
BEGINNER'S BOX-2
1. (D)
2. (C)
BEGINNER'S BOX-3
1. (B)
42
2. (A,C)
KTG & Thermodynamics
1.
Pressure versus temperature graphs of an ideal gas are as shown in figure. Choose the wrong statement:–
P
P
(i)
P
(ii)
(iii)
T
2.
T
(A) Density of gas is increasing in graph (i)
(B) Density of gas decreasing in graph (ii)
(C) Density of gas is constant in graph (iii)
(D) None of the above
In a process the density of a gas remains constant. If the temperature is doubled, then the change in the
pressure will be :–
(A) 100 % increase
3.
T
(B) 200 % increase
(C) 50 % decrease
(D) 25 % decrease
The expansion of unit mass of a perfect gas at constant pressure is shown in the diagram. Here :–
a
O
4.
b
(A) a = volume, b = °C temperature
(B) a = volume, b = K temperature
(C) a = °C temperature, b = volume
(D) a = K temperature, b = volume
Air is filled at 60° C in a vessel of open mouth. The vessel is heated to a temperature T so that
1 th
part of air
4
escapes. The value of T is :–
(A) 80° C
5.
(B) 444° C
One mole of an ideal gas undergoes a process P =
(C) 333° C
(D) 171° C
P0
Here P0 and V0 are constants. Change in
1 + (V0 / V)2
temperature of the gas when volume is changed from V =V0 to V = 2V0 is :–
(A) 6.
(B)
11P0 V0
10R
(C) -
5P0 V0
4R
(D) P0 V0
A gas has volume V and pressure P. The total translational kinetic energy of all the molecules of the gas is:–
(A)
3
PV only if the gas is monoatomic.
2
(C) >
7.
2P0 V0
5R
3
PV if the gas is diatomic.
2
(B)
3
PV only if the gas is diatomic.
2
(D)
3
PV in all cases.
2
CP
A mixture of n1 moles of monoatomic gas and n2 moles of diatomic gas has C = g = 1.5 :–
V
(A) n1 = n2
(B) 2n1 = n2
(C) n1 = 2n2
(D) 2n1 = 3n2
43
JEE-Physics
8.
1023 molecules of a gas strike a target of area 1 m 2 at angle 45° to normal and rebound elastically with speed
1 kms–1. The impulse normal to wall per molecule is :– [Given : mass of molecule = 3.32 × 10–27kg]
(A) 4.7 × 10–24 kg ms–1
9.
(B) 7.4 × 10–24 kg ms–1
(C) 3.32× 10–24 kg ms–1
(D) 2.33 kg ms–1
V
From the following V–T diagram we can conclude:–
(A) P1 = P2
(B) P1>P2
(C) P1 < P2
(D) Can't say anything
P1
T1
10.
P2
T2
T
A gas is expanded from volume V0 to 2V0 under three different processes. Process 1 is isobaric process,
process 2 is isothermal and process 3 is adiabatic. Let DU1, DU2 and DU3, be the change in internal energy of
the gas is these processes.
P
P0
1
2
3
V0
2V0
V
Then :–
(A) DU1 > DU2 > DU3
11.
(B) DU1 < DU2 < DU3
(C) DU2 < DU1 < DU3
(D) DU2 < DU3 < DU1
For an ideal gas PT11 = constant then volume expansion coefficient is equal to :–
(A)
12.
11
T
(B)
1
T
(C)
12
T
(D)
2
T
The internal energy of a gas is given by U = 5 + 2PV. It expands from V 0 to 2V0 against a constant pressure
P0. The heat absorbed by the gas in the process is :–
(A) –3P0V0
13.
(B) 3P0V0
(C) 2P0V0
(D) P0V0
When water is heated from 0°C to 4°C and CP and CV are its specific heats at constant pressure and constant
volume respectively, then :–
(A) CP > CV
(B) CP < CV
(C) CP = CV
(D) CP – CV = R
The molar specific heat of the process V µ T4 for CH4 gas at room temperature is:–
14.
(A) 4R
15.
(B) 7R
(C) 3R
(D) 8R
5n, n and 5n moles of a monoatomic, diatomic and non-linear polyatomic gases (which do not react chemically
with each other) are mixed at room temperature. The equivalent degree of freedom for the mixture is :–
(A)
25
7
(B)
48
11
(C)
52
11
(D)
50
11
The internal energy of a gas in an adiabatic process is given by U = a + bPV, find g :–
16.
(A)
17.
a +1
a
(B)
b +1
b
(C)
b +1
a
(D)
a
b +1
The ratio of translational and rotational kinetic energies at 100K temperature is 3 : 2. Then the internal energy
of one mole gas at that temperature is [R = 8.3 J/mol-K]
(A) 1175 J
44
(B) 1037.5 J
(C) 2075 J
(D) 4150 J
KTG & Thermodynamics
18.
Which of the following will have maximum total kinetic energy at temperature 300 K.
(A) 1 kg, H2
19.
(B) 1 kg, He
(C)
1
1
kg H 2 + kg He
2
2
(D)
1
3
kg H 2 + kg He
4
4
P-T curve for a cyclic process is as shown
P
T
P-V graph for this process will be :
P
P
(A)
P
(B)
(C)
V
20.
(D)
V
V
V
A gas has molar heat capacity C = 24.9 J mol–1 K–1 in the process P2T = constant. Then (R = 8.3 J/mol k)
(A) Gas is monoatomic
21.
P
(B) Gas is diatomic
(C) Gas is triatomic
(D) Atomicity of gas is 4
P-T diagram is shown below then choose the corresponding V-T diagram
P
C
A
D
B
T
A
V B
D
V B
D
V
C
V
(A)
B
C
(B)
A
T
22.
C
(C)
A
T
(D)
D
T
D
B
C
A
T
Some of the thermodynamic parameters are state variables while some are process variables. Some grouping
to the parameters are given. Choose the correct one.
(A) State variables : Temperature, No. of moles
Process variables : internal energy, work done by the gas.
(B) State variables ; Volume, Temperature
Process variables : Internal energy, work done by the gas.
(C) State variables : work done by the gas, heat rejected by the gas
Process variables : Temperature, volume.
(D) State variables : internal energy, volume
Process variables : work done by the gas, heat absorbed by the gas.
45
JEE-Physics
23.
The state of an ideal gas is changed through an isothermal process at temperature T0 as shown in figure. The
work done by gas in going from state B to C is double the work done by gas in going from state A to B. If the
pressure in the state B is
(A)
P0
2
P
(C) 0
6
24.
P0
then the pressure of the gas in state C is.
2
(B)
P0
4
P
P0
A
B
P0
2
P
(D) 0
8
C
V0
T0
V
In the P–V diagram shown. the gas does 5 J of work in isothermal process a b and 4J in adiabatic process b
c. What will be the change in internal energy of the gas in straight path c to a ?
a
b
P
c
V
(A) 9 J
46
(B) 1 J
(C) 4 J
(D) 5 J
KTG & Thermodynamics
1.
The figure shows two paths for the change of state of a gas from A to B. The ratio of molar heat capacities in
path 1 and path 2 is :–
P
2
B
A
1
V
(A) > 1
2.
(B) < 1
(C) 1
(D) Data insufficient
During the melting of a slab of ice at 273 K at atmospheric pressure :–
(A) Positive work is done by the ice–water system on the atmosphere.
(B) Positive work is done on the ice–water system by the atmosphere.
(C) The internal energy of ice–water system increases.
(D) The internal energy of the ice–water system decreases.
3.
A partition divides a container having insulated walls into two compartments I and II.The same gas fills the
two compartments whose initial parameters are given. The partition is a conducting wall which can move
freely without friction. Which of the following statements is/are correct, with reference to the final equilibrium
position
(A) The pressure in the two compartments are equal.
(B) Volume of compartment I is
3V
5
P,V,T
I
(C) Volume of compartment II is
12V
5
(D) Final pressure in compartment I is
4.
2P,2V,T
II
5P
3
Pressure versus temperature graph of an ideal gas is shown in figure.Density of the gas at point A is r 0.
Density at B will be :–
P
3P 0
P0
B
A
T0
(A)
3
r0
4
(B)
3
r0
2
T
2T0
(C)
4
r0
3
(D) 2r0
47
JEE-Physics
5.
A closed vessel contains a mixture of two diatomic gases A and B. Molar mass of A is 16 times that of B and
mass of gas A contained in the vessel is 2 times that of B. Which of the following statements are true?
(A) Average kinetic energy per molecule of A is equal to that of B
(B) Root mean square value of translational velocity of B is four times that of A
(C) Pressure exerted by B is eight times of that exerted by A
(D) Number of molecules of B in the cylinder is eight times that of A
6.
The internal energy of a system remains constant when it undergoes :–
(A) a cyclic process
(B) an isothermal process
(C) an adiabatic process
(D) any process in which the heat given out by the system is equal to the work done on the system
7.
CP is always greater than CV due to the fact that :–
(A) No work is being done on heating the gas at constant volume.
(B) When a gas absorbs heat at constant pressure its volume must change so as to do some external work.
(C) The internal energy is a function of temperature only for an ideal gas.
(D) For the same rise of temperature, the internal energy of a gas changes by a smaller amount at constant
volume than at constant pressure.
8.
The specific heats of a gas are CP = 0.2 cal/g °C & CV = 0.15 cal/g °C. [ Take R=2 cal/mole0 C ]
(A) The molar mass of the gas is 40 g
(B) The molar mass of the gas cannot be determined from the data given
(C) The number of degrees of freedom of the gas molecules is 6
(D) The number of degrees of freedom of the gas molecules is 8
9.
A gas expands such that its initial and final temperatures are equal. Also, the process followed by the gas traces
a straight line on the P–V diagram :–
(A) The temperature of the gas remains constant throughout.
(B) The temperature of the gas first increases and then decreases.
(C) The temperature of the gas first decreases and then increases.
(D) The straight line has a negative slope.
10.
Three identical adiabatic containers have helium, neon and oxygen gases at the same pressure. The gases are
compressed to half their original volume. Then:–
(A) The final temperature of the gas in each container is same
(B) The final pressure of the gas in each container is same
(C) The final temperature of both helium and neon is same
(D) The final pressure of both helium and neon is same
Match the column
11.
Column–I
48
Column–II
(A) Isobaric process
(p) No heat exchange
(B) Isothermal process
(q) Constant pressure
(C) Isoentropy process
(r) Constant internal energy
(D) Isochoric process
(s) Work done is zero
KTG & Thermodynamics
12.
An ideal gas whose adiabatic exponent equals to g =
7
is expanded according to the law P=2V. The initial volume
5
of the gas is equal to V0= 1unit . As a result of expansion the volume increases 4 times. (Take R =
Column – I
13.
25
units)
3
Column – II
(A) Work done by the gas
(p) 25 units
(B) Increment in internal energy of the gas
(q) 90 units
(C) Heat supplied to the gas
(r) 75 units
(D) Molar heat capacity of the gas in the process
(s) 15 units
For a ideal monoatomic gas match the following graphs for constant mass in different processes (r = Density
of gas)
Column I
Column II
P
P
B
B
(A)
A
(p)
C
C
A
V
T
r
P
B
(B)
A
A
(q)
C
C
B
T
T
r
V
B
(C)
A
A
(r)
C
B
C
T
T
r
r
B
(D)
A
A
(s)
C
T
C
B
T
(t) None
Comprehension Based Questions
Comprehension-1.
Molar heat capacity of an ideal gas in the process PVx = constant, is given by : C=
R
R
+
. An ideal
g -1 1- x
5R
occupies a volume V1 at a pressure P1. The gas undergoes a process in which the
2
pressure is proportional to the volume. At the end of the process the rms speed of the gas molecules has
doubled from its initial value.
diatomic gas with CV =
49
JEE-Physics
14.
The molar heat capacity of the gas in the given process is :–
(A) 3 R
15.
(B) 3.5 R
(C) 4 R
(D) 2.5 R
(C) 9 P1V1
(D) 10 P1V1
Heat supplied to the gas in the given process is :
(A) 7 P1V1
(B) 8 P1V1
Comprehension-2
Five moles of helium are mixed with two moles of hydrogen to form a mixture. Take molar mass of helium
M1 = 4g and that of hydrogen M2 = 2g
16.
The equivalent molar mass of the mixture is
(A) 6g
17.
(B)
13g
7
(C)
18g
7
(D) None
The equivalent degree of freedom f of the mixture is
(A) 3.57
(B) 1.14
(C) 4.4
(D) None
(C) 1.56
(D) None
The equivalent value of g is
18.
(A) 1.59
19.
(B) 1.53
If the internal energy of He sample is 100J and that of the hydrogen sample is 200J, then the internal energy
of the mixture is
(A) 900 J
(B) 128.5 J
(C) 171.4J
(D) 300J
Comprehension-3
Refrigerator is an apparatus which takes heat from a cold body, work is done on it and the work done together
with the heat absorbed is rejected to the source. An ideal refrigerator can be regarded as Carnot's ideal heat
engine working in the reverse direction. The coefficient of performance of refrigerator is defined as
b=
Q2
T2
Heat extracted from cold reservoir Q2
=
=Q -Q = T -T
work done on workingsubs tance
W
1
2
1
2
source
T1
Q1
working
substance
Q2
sink
T2
W
A Carnot's refrigerator takes heat from water at 0°C and discards it to a room temperature at 27°C. 1kg of
water at 0°C is to be changed into ice at 0°C. (Lice = 80 kcal/kg)
20.
How many calories of heat are discarded to the room ?
(A) 72.8 kcal
21.
(B) 87.9 kcal
(C) 80 kcal
(D) 7.9 kcal
What is the work done by the refrigerator in this process (1 cal = 4.2 joule)
(A) 7.9 kJ
22.
(B) 33.18 kJ
(C) 43.18 kJ
(D) 23.18 kJ
What is the coefficient of performance of the machine ?
(A) 11.1
50
(B) 10.1
(C) 9.1
(D) 8.1
KTG & Thermodynamics
1.
For a gas
R
= 0.4 . For this gas calculate the following (i) Atomicity and degree of freedom (ii) Value of C V
CP
and g (iii) Mean gram – molecular kinetic energy at 300 K temperature
2.
Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, calculate
P
the following quantities in this process.
A
2 atm
(i) The net change in the heat energy.
B
(ii) The net work done
(iii) The net change in internal energy
3.
1 atm
T
400K
Examine the following plots and predict whether in (i) P1 < P2 and T1> T2, in (ii) T1=T2<T3, in(iii) V1 > V2,
in(iv) P1 > P2 or P2>P1
P
PV
2
2P
3
2
(i)
(ii)
1
1
P
V
V
V
2
2
(iii)
2V
V
P
(iv)
1
1
T
T
4.
C
D
300K
In the given figure an ideal gas changes its state from A to state C by two paths ABC and AC.
P
15
-2
P(N/m)
B
C
10
5
A
V
0
2
4
6
(i) Find the path along which work done is the least.
(ii) The internal energy of gas at A is 10J and amount of heat supplied to change its state to C through the
path AC is 200J. Calculate the internal energy at C.
(iii) The internal energy of gas at state B is 20J. Find the amount of heat supplied to the gas from A to B.
5.
The pressure in monoatomic gas increases linearly from 4× 105 N/m2 to 8× 105 N/m2 when its volume
increases from 0.2 m3 to 0.5 m3 . Calculate the following –
(i) Work done by the gas
(ii) Increase in internal energy
(iii) Amount of heat supplied
(iv) Molar specific heat of the gas
51
JEE-Physics
6.
At 27°C two moles of an ideal monoatomic gas occupy a volume V. Then gas is adiabatically expanded until
its volume becomes 2V. Calculate : (i) The final temperature of the gas (ii) Change in its internal energy (iii)
The work done by the gas during this process
7.
Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and
volume of the gas are 105 N/m2 and 6L respectively. The final volume of the gas is 2L, molar specific heat of
the gas at constant volume is
8.
3R
.
2
An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat
involved in these steps are Q1 = 5960 J, Q2 = –5585 J, Q3 = –2980 J and Q4 = 3645 J respectively. The
corresponding quantities of work involved are W1 = 2200 J, W2 = –825 J, W3 = –1100 J and W4 respectively.
(i) Find the value of W4. (ii) What is the efficiency of the cycle ?
9.
A gas has molar heat capacity C = 37.35 J mole –1K–1 in the process PT = constant. Find the number of degree
of freedom of molecules in the gas.
10.
P–V graph for an ideal gas undergoing polytropic process PV m = constant is shown here. Find the value of m.
P
(Pa)
5
2 ×10
37C0
5
4 ×10
52
V (m )
3
KTG & Thermodynamics
1.
One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is
adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct
statement
[JEE(Main) 2014]
(A) The change in internal energy in whole cyclic process is 250 R.
(B) The change in internal energy in the process CA is 700 R
(C) The change in internal energy in the process AB is –350 R
(D) The change in internal energy in the process BC is –500 R
2.
An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury
level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46
cm. what will be length of the air column above mercury in the tube now ? (Atomspheric pressure = 76 cm of
Hg)
[JEE(Main) 2014]
(A) 16 cm
3.
(B) 22 cm
(C) 38 cm
(D) 6 cm
A solid body of constant heat capacity 1 J°/C is being heated by keeping it in contact with reservoirs in two
ways JEE(Main) 2015]
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.
In both the cases body is brought from initial temperature 100°C to final temperature 200°C. Entropy change
of the body in the two cases respectively is (A) ln2, 2ln2
4.
(B) 2ln2, 8ln2
(C) ln2, 4ln2
(D) ln2, ln2
Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered
1æUö
as an ideal gas of photons with internal energy per unit volume u = U µ T 4 and pressure p = ç ÷ . If the
3è V ø
V
shell now undergoes an adiabatic expansion the relation between T and R is [JEE(Main) 2015]
(A) T µ
5.
1
R
(B) T µ
1
R3
(C) T µ e–R
(D) T µ e–3R
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion,
the average time of collision between molecules increases as Vq, where V is the volume of the gas. The value
Cp ö
æ
of q is :- ç g = C ÷
v ø
è
(A)
6.
g +1
2
[JEE(Main) 2015]
(B)
g -1
2
(C)
3g + 5
6
(D)
3g - 5
6
An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant.
If during this process the relation of pressure P and volume V is given by PV n = constant, then n is given by
(Here CP and CV are molar specific heat at constant pressure and constant volume, respectively) :[JEE(Main) 2016]
C - CV
(A) n = C - C
P
(B) n =
CP
CV
(C) n =
C – CP
C – CV
(D) n =
CP – C
C - CV
53
JEE-Physics
'n' moles of an ideal gas undergoes a process A ® B as shown in the figure. The maximum temperature of the
gas during the process will be :
[JEE(Main) 2016]
7.
P
A
2P0
B
P0
V0
(A)
8.
9P0 V0
nR
(B)
9 P0 V0
4nR
V
2V0
(C)
3P0 V0
2nR
(D)
9P0 V0
2nR
The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to sunshine. The
atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the room
before and after heating, then nf – ni will be :[JEE(Main) 2017]
(A) 2.5 × 1025
9.
(B) –2.5 × 1025
(C) –1.61 × 1023
(D) 1.38 × 1023
Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that
Cp – Cv = a for hydrogen gas
Cp – Cv = b for nitrogen gas
[JEE(Main) 2017]
The correct relation between a and b is :
(A) a = 14 b
10.
(B) a = 28 b
(C) a =
1
b
14
(D) a = b
The mass of a hydrogen molecule is 3.32 × 10–27kg. If 1023 hydrogen molecules strike, per second, a fixed
wall of area 2 cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the
pressure on the wall is nearly :
[JEE(Main) 2018]
(A) 4.70 × 103 N/m2
11.
(B) 2.35 × 102 N/m2
(C) 4.70 × 102 N/m2
(D) 2.35 × 103 N/m2
Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a
volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.
12.
(A) (a) 195 K (b) –2.7 kJ
(B) (a) 189 K (b) –2.7 kJ
(C) (a) 195 K (b) 2.7 kJ
(D) (a) 189 K (b) 2.7 kJ
[JEE(Main) 2018]
A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27°C. Amount of heat transferred to the
gas, so that rms velocity of molecules is doubled, is about :
[JEE(Main) 2019]
[Take R = 8.3 J/ K mole]
(A) 10 kJ
13.
(B) 0.9 kJ
(C) 6 kJ
(D) 14 kJ
A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass = 40 u) is kept
é Vrms (helium) ù
at 300 K in a container. The ratio of their rms speeds ê V (arg on) ú
ë rms
û
(A) 2.24
14.
(B) 0.45
(C) 0.32
[JEE(Main) 2019]
(D) 3.16
Two kg of a monoatomic gas is at a pressure of 4 × 104 N/m2 . The density of the gas is 8 kg /m3. What is the
order of energy of the gas due to its thermal motion ?
[JEE(Main) 2019]
(A) 103 J
54
(B) 105 J
(C) 106 J
(D) 104 J
KTG & Thermodynamics
15.
A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible
thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the
piston is l1, and that below the piston is l2 , such that l1 > l2. Each part of the cylinder contains n moles of an
ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by :
[JEE(Main) 2019]
(R is universal gas constant and g is the acceleration due to gravity)
(A)
16.
nRT é 1 1 ù
ê + ú
g ë l2 l1 û
nRT é l1 - l2 ù
ê
ú
g ë l1 l2 û
RT é 2l1 + l2 ù
(C) g ê l l ú
ë 1 2 û
RT é l1 - 3l2 ù
(D) g ê l l ú
ë 1 2 û
An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between
two successive collisions is 6 × 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the
mean time between two successive collisions will be close to :
[JEE(Main) 2019]
(A) 4 × 10–8 s
17.
(B)
(B) 3 × 10–6 s
(C) 2 × 10–7 s
(D) 0.5 × 10–8 s
An ideal gas occupies a volume of 2m3 at a pressure of 3 × 106 Pa. The energy of the gas is:
[JEE(Main) 2019]
(A) 3 × 102
18.
(C) 6 × 104 J
(D) 9 × 106 J
Two Carrnot engines A and B are operated in series. The first one, A, receives heat at T 1(= 600 K) and rejects
to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in turn,
rejects to a heat reservoir at T3(= 400 K). Calculate the temperature T2 if the work outputs of the two engines
are equal :
[JEE(Main) 2019]
(A) 400 K
19.
(B) 108 J
(B) 600 K
(C) 500 K
(D) 300 K
A gas can be taken from A to B via two different processes ACB and ADB.
When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If path ADB
is used work done by the system is 10 J. The heat Flow into the system in path ADB is :[JEE(Main) 2019]
(A) 80 J
20.
(C) 100 J
(D) 40 J
A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having
coefficient of linear expansion a/ °C. It is observed that an external compressive force F, is applied on each of
its ends, prevents any change in the length of the rod, when its temperature rises by DT K. Young’s modulus,
Y, for this metal is :
[JEE(Main) 2019]
(A)
21.
(B) 20 J
F
2AaDT
(B)
F
Aa(DT - 273)
(C)
F
AaDT
(D)
2F
AaDT
Half mole of an ideal monoatomic gas is heated at constant pressure of latm from 20 °C to 90°C. Work done
by gas is close to : ( Gas constant R = 8.31 J /mol.K)
[JEE(Main) 2019]
(A) 73 J
(B) 291 J
(C) 581 J
(D) 146 J
55
JEE-Physics
22.
Three Carnot engines operate in series between a heat source at a temperature T x and a heat sink at temperature T4 (see figure). There are two other reservoirs at temperature T 2, and T3, as shown, with T2 > T2 > T3 >
T4 . The three engines are equally efficient if:
[JEE(Main) 2019]
(
)
(
)
2
(A) T2 = T1 T4
3
(C) T2 = T1 T4
23.
1/ 3
1/ 4
; T3 = T1 T42
(
)
(
)
; T3 = T1 T43
1/ 3
1/ 4
(
)
; T3 = T12 T4
(
)
(
)
; T3 = T12 T4
(
)
2
(B) T2 = T1 T4
(D) T2 = T1 T4
1/ 3
1/ 2
1/ 3
1/ 3
Two rods A and B of identical dimensions are at temperature 30°C. If A is heated upto 180°C and B upto T°C,
then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then
the value of T is :
[JEE(Main) 2019]
(A) 270°C
(B) 230°C
(C) 250°C
(D) 200°C
24 . A thermometer graduated according to a linear scale reads a value x 0 when in contact with boiling water, and
x0/3 when in contact with ice.
What is the temperature of an object in 0 °C, if this thermometer in the contact with the object reads x0/2 ?
[JEE(Main) 2019]
(A) 35
25.
(B) 25
(C) 60
(D) 40
In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the
relation VT = K, where K is a constant. In this process the temperature of the gas is incresed by AT. The
amount of heat absorbed by gas is (R is gas constant) :
[JEE(Main) 2019]
(A)
26.
1
RDT
2
(B)
3
RDT
2
(C)
1
KRDT
2
(D)
2K
DT
3
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature,. The relation between temperature and volume of this process is TVx = constant, then x is :
[JEE(Main) 2019]
(A)
27.
5
3
(B)
2
5
(C)
2
3
(D)
3
5
The gas mixture constists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only
translational and rotational modes, the total inernal energy of the system is :
[JEE(Main) 2019]
(A) 12 RT
28.
(B) 20 RT
(C) 15 RT
For the given cyclic process CAB as shown for a gas, the work done is :
(A) 1 J
56
(B) 5 J
(C) 10 J
(D) 4 RT
[JEE(Main) 2019]
(D) 30 J
KTG & Thermodynamics
29.
The given diagram shows four processes i.e., isochoric, isobaric, isothermal and adiabatic. The correct
assignment of the processes, in the same order is given by :[JEE(Main) 2019]
a
P
b
d
c
V
(A) d a c b
30.
(B) a d c b
(C) a d b c
(D) d a b c
The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity
from the earth, is closest to :
[JEE(Main) 2019]
[Boltzmann Constant kB = 1.38 × 10–23 J/K
Avogadro Number NA = 6.02 × 1026 /kg
Radius of Earth : 6.4 × 106 m
Gravitational acceleration on Earth = 10ms–2]
(A) 650 K
31.
(B) 3 × 105 K
(C) 104 K
(D) 800 K
An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in
its gaseous phase is v , m is its mass and kB is Boltzmann constant, then its temperature will be :
[JEE(Main) 2019]
(A)
32.
mv 2
6k B
(B)
mv 2
5k B
(C)
mv 2
3k B
(D)
mv 2
7k B
Following figure shows two processes A and B for a gas. If DQA and DQB are the amount of heat absorbed by
the system in two cases, and DUA and DUB are changes in internal energies, respectively, then :
f
P
A
B
i
V
33.
(A) DQA = DQB ; DUA = DUB
(B) DQA > DQB ; DUA = DUB
(C) DQA > DQB ; DUA > DUB
(D) DQA < DQB ; DUA < DUB
For a given gas at 1 atm pressure, rms speed of the molecule is 200 m/s at 127°C. At 2 atm pressure and at
227°C, the rms speed of the molecules will be :
[JEE(Main) 2019]
(A) 80 m/s
34.
[JEE(Main) 2019]
(B) 100 5 m/s
(C) 80 5 m/s
(D) 100 m/s
A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. The
spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the
change in the temperature of water when the vibrations stop completely ? (Assume that the water container and
spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K)
[JEE(Main) 2019]
(A)
10–3K
(B)
10–4
K
(C)
10–1
K
(D) 10–5K
57
JEE-Physics
35.
The specific heats, CP and CV of a gas of diatomic molecules, A, are given (in units of J mol –1 K–1) by 29 and
22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are
treated as ideal gases, then :[JEE(Main) 2019]
36.
(A) A has one vibrational mode and B has two
(B) Both A and B have a vibrational mode each
(C) A is rigid but B has a vibrational mode
(D) A has a vibrational mode but B has none
A cylinder with fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the
temperature of the gas by 20°C is : [Given that R = 8.31 J mol–1 K–1]
[JEE(Main) 2019]
(A) 748 J
37.
(B) 374 J
(C) 350 J
(D) 700 J
A 25 × 10–3 m3 volume cylinder is filled with 1 mol of O2 gas at room temperature (300K). The molecular
diameter of O2, and its root mean square speed, are found to be 0.3 nm, and 200 m/s, respectively. What is
the average collision rate (per second) for an O2 molecule ?
[JEE(Main) 2019]
(A) ~1011
38.
(B) ~1013
(C) ~1010
(D) ~1012
n moles of an ideal gas with constant volume heat capcity CV undergo an isobaric expansion by certain
volume. The ratio of the work done in the process, to the heat supplied is :
[JEE(Main) 2019]
(A)
39.
4nR
C V - nR
(B)
nR
C V - nR
(C)
nR
C V + nR
(D)
4nR
C V + nR
When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by
DT. The heat required to produce the same change in temperature, at a constant pressure is :
[JEE(Main) 2019]
(A)
40.
7
Q
5
(B)
3
Q
2
(C)
5
Q
3
(D)
2
Q
3
One mole of an ideal gas passes through a process where pressure and volume obey the relation
2
é
1æ V ö ù
P = Po ê1 - ç 0 ÷ ú . Here Po and Vo are constants. Calculate the change in the temperature of the gas if its
2è V ø ú
êë
û
[JEE(Main) 2019]
volume changes from Vo to 2Vo.
(A)
41.
1 Po Vo
2 R
(B)
3 Po Vo
4 R
(C)
5 Po Vo
4 R
(D)
1 Po Vo
4 R
A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the
internal energy of the gas along the path ca is –180J. The gas absorbs 250 J of heat along the path ab and
60 J along the path bc. The work done by the gas along the path abc is :
[JEE(Main) 2019]
c
P
a
b
V
(A) 100 J
58
(B) 120 J
(C) 140 J
(D) 130 J
KTG & Thermodynamics
42.
Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the
molar specific heat of mixture at constant volume ? (R = 8.3 J/mol K)
[JEE(Main) 2019]
(A) 21.6 J/mol K
43.
(B) 19.7 J/mol K
(C) 17.4 J/mol K
(D) 15.7 J/mol K
4
The number density of molecules of a gas depends on their distance r from the origin as, n(r) = n0 e -ar . Then
[JEE(Main) 2019]
the total number of molecules is proportional to :
(A) n0a1/4
44.
(C) n0a–3/4
(D)
n0 a1/ 2
A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62ºC, its efficiency
is doubled. The temperatures of the source and the sink are, respectively
[JEE(Main) 2019]
(A) 124ºC, 62ºC
45.
(B) n0a–3
(B) 37ºC, 99ºC
(C) 62ºC, 124ºC
(D) 99ºC, 37ºC
A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be
the heat energy absorbed by the gas, in this process ?
[JEE(Main) 2019]
(A) 35 J
(B) 40 J
(C) 25 J
(D) 30 J
59
JEE-Physics
MCQ's with One Correct Answer
1.
A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the
average rotational kinetic energy per O2 molecule to per N2 molecule is :–
[IIT-JEE 1998]
2.
(A) 1 : 1
(B) 1 : 2
(C) 2 : 1
(D) depends on the moment of inertia of the two molecules
Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature
and the same volume V. The mass of the gas in A is mA and that in B is mB. The gas in each cylinder is now
allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found
to be DP and 1.5 DP respectively. Then :–
[IIT-JEE 1998]
(A) 4mA = 9mB
3.
(B) 2mA = 3mB
(C) 3mA = 2mB
(D) 9mA = 4mB
Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston
of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder.
If the rise in temperature of the gas in A is 30 K, then the rise in temp. of the gas in B is:– [IIT-JEE 1998]
(A) 30 K
4.
(B) 18 K
(C) 50 K
(D) 42 K
[IIT-JEE 1999]
The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300 K is :–
æ 2ö
çè ÷ø
7
(A)
5.
æ 1ö
çè ÷ø
7
(B)
3
5
(C)
6
5
(D)
A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational
modes, the total internal energy of the system is :–
[IIT-JEE 1999]
(A) 4RT
6.
(B) 15RT
(C) 9RT
(D) 11RT
A monoatomic ideal gas, initially at temperature T1, is enclosed in a cylinder fitted with a frictionless piston.
The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2
are the lengths of the gas column before and after expansion respectively, then T1 is given by
T2
[IIT-JEE 2000]
æ L1 ö
(A) ç ÷
è L2 ø
7.
2/3
æ L1 ö
(B) ç ÷
è L2 ø
æ L2 ö
(C) ç ÷
è L1 ø
æ L2 ö
(D) ç ÷
è L1 ø
2/3
Starting with the same initial conditions, an ideal gas expands from volume V 1 to V2 in three different ways,
the work done by the gas W1 if the process is purely isothermal, W2 if purely isobaric and W3 if purely
adiabatic, then
[IIT-JEE 2000]
(A) W2 > W1 > W3
(B) W2 > W3 > W1
(C) W1 > W2 > W3
(D) W1 > W3 > W2
An ideal gas is initially at temperature T and volume V. Its volume is increased by DV due to an increase in
DV
temperature DT, pressure remaining constant. The quantity d =
varies with temperature as:
VDT
[IIT-JEE 2000]
8.
d
d
(B)
(A)
T
60
T+ D T
d
d
(C)
T
T+ D T
(D)
T
T+ D T
T
T+ D T
KTG & Thermodynamics
9.
Two monoatomic ideal gases 1 and 2 of molecular masses m1 and m2 respectively are enclosed in separate
containers kept at the same temperature. The ratio of the speed of sound in gas 1 to the gas 2 is given by.
[IIT-JEE 2000]
m1
m2
(A)
10.
11.
m1
(C) m
2
m2
m1
(B)
m2
(D) m
1
[IIT-JEE 2001]
In a given process of an ideal gas, dW = 0 and dQ < 0. Then for the gas :–
(A) the temperature will decrease
(B) the volume will increase
(C) the pressure will remain constant
(D) the temperature will increase
P–V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond
respectively to :
[IIT-JEE 2001]
P
2
1
V
(A) He and O2
12.
(B) O2 and He
(C) He and Ar
(D) O2 and N2
An ideal gas is taken through the cycle A ® B ® C ® A, as shown in the figure. If the net heat supplied to the
gas in the cycle is 5 J, the work done by the gas in the process C ® A is :–
[IIT-JEE 2002]
2
3
V (m )
C
B
A
1
10
2
P (N / m )
(A) –5J
13.
(B) –10J
(C) –15J
Which of the following graphs correctly represent the variation of b = –
(D) –20J
dV / dP
with P for an ideal gas at
V
[IIT-JEE 2002]
constant temperature ?
b
(A)
(B)
(D)
P
P
P
An ideal gas expands isothermally from a volume V1 and V2 and then compressed to original volume V1
adiabatically. Initial pressure is P1 and final pressure is P3. The total work done is W. Then :– [IIT-JEE 2004]
(A) P3 > P1, W > 0
15.
b
(C)
P
14.
b
b
(B) P3 < P1, W < 0
(C) P3 > P1, W < 0
(D) P3 = P1, W = 0
An ideal gas is expanding such that PT2 = constant. The coefficient of volume expansion of the gas is:–
(A)
1
T
(B)
2
T
(C)
3
T
(D)
4
T
[IIT-JEE 2008]
61
JEE-Physics
16.
Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial
pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is :[IIT-JEE 2013]
(A) 1 : 4
17.
(B) 1 : 2
(C) 6 : 9
(D) 8 : 9
One mole of a monatomic ideal gas is taken along two cyclic processes E®F®G®E and E®F®H®E as
shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.
P
F
32P 0
P0
G
H
E
V0
V
Match the paths in List I with the magnitudes of the work done in the List II and select the correct answer using
the codes given blow the lists.
[IIT-JEE 2013]
P.
List I
List II
G®E
1. 160 P0V0 ln2
Q. G ® H
2. 36 P0V0
R. F ® H
3. 24 P0V0
S. F ® G
4. 31 P0V0
Codes :
18.
P
Q
R
S
(A)
4
3
2
1
(C)
3
1
2
4
P
Q
R
S
(B)
4
3
1
2
(D)
1
3
2
4
An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the
figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state.
The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this process the
piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct
statement(s) is(are) :[IIT-JEE 2015]
(A) If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is
1
P1V1
4
(B) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3 P1V1
(C) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is
7
P1 V1
3
(D) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is
62
17
P1 V1
6
KTG & Thermodynamics
19.
A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure
æ 1 ö
5
–3
3
÷ ´ 10 Pa and Vf = 8 ×10 m in
è 32 ø
Pi = 105 Pa and volume Vi = 10–3 m3 changes to a final state at Pf = ç
an adiabatic quasi-static process, such that P3V5 = constant. Consider another thermodynamic process that
brings the system from the same initial state to the same final state in two steps: an isobaric expansion at P i
followed by an isochoric (isovolumetric) process at volumes Vf. The amount of heat supplied to the system in
the two step process is approximately
[IIT-JEE 2016]
(A) 112 J
20.
(B) 294 J
(C) 588 J
One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the
PV-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is
adiabatic. Match the processes mentioned in List-I with the corresponding statements in List–II.
P
II
3P0
IV
V0
P.
III
I
P0
21.
(D) 813 J
3V0
List–I
List–II
In process I
1. Work done by the gas is zero
V
[IIT-JEE 2018]
Q. In process II
2. Temperature of the gas remains unchanged
R. In process III
3. No heat is exchanged between the gas and its surroundings
S. In process IV
4. Work done by the gas is 6 P0V0
(A) P ® 4 ; Q ® 3 ; R ® 1 ; S ® 2
(B) P ® 1 ; Q ® 3 ; R ® 2 ; S ® 4
(C) P ® 3 ; Q ® 4 ; R ® 1 ; S ® 2
(D) P ® 3 ; Q ® 4 ; R ® 2 ; S ® 1
Answer the following by appropriately matching the lists based on the information given in the
paragraph.
In a thermodynamics process on an ideal monatomic gas, the infinitesimal heat absorbed by the
gas is given by TDX, where T is temperature of the system and DX is the infinitesimal change in a thermodynamic
quantity X of the system. For a mole of monatomic ideal gas X =
constant, V is volume of gas, TA and VA are constants.
3
R ln
2
æ T ö
æ V ö
ç
÷ + R ln ç
÷ . Here, R is gas
T
è Aø
è VA ø
The List-I below gives some quantities involved in a process and List-II gives some possible values of these
quantities.
List-I
List-II
(III) Heat absorbed by the system in process 1 ® 2 ® 3
1
RT0 ln 2
3
1
(Q) RT0
3
(R) RT0
(IV) Heat absorbed by the system in process 1 ® 2
(S)
(I) Work done by the system in process 1 ® 2 ® 3
(II) Change in internal energy in process 1 ® 2 ® 3
(P)
4
RT0
3
1
(T) RT0 (3 + ln 2)
3
5
(U) RT0
6
63
JEE-Physics
If the process carried out on one mole of monatomic ideal gas is as shown in figure in the PV-diagram with
P0V0 =
1
RT0 , the correct match is,
3
[IIT-JEE 2019]
P
3P0
2
P0
3
1
(A) I ® Q, II ® R, III ® P, IV ® U
V
2V0
(B) I ® S, II ® R, III ® Q, IV ® T
(C) I ® Q, II ® R, III ® S, IV ® U
(D) I ® Q, II ® S, III ® R, IV ® U
V0
22.
2
Answer the following by appropriately matching the lists based on the information given in the
paragraph.
In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by
the gas is given by TDX, where T is temperature of the system and DX is the infinitesimal change in a
thermodynamic quantity X of the system. For a mole of monatomic ideal gas X =
æ T ö
æ V ö
3
R ln ç T ÷ + R ln ç V ÷
2
è Aø
è Aø
Here, R is gas constant, V is volume of gas, TA and VA are constants.
The List-I below gives some quantities involved in a process and List-II gives some possible values of these
quantities.
List-I
List-II
(I) Work done by the system in process 1 ® 2 ® 3
(P)
1
RT0 ln 2
3
(II) Change in internal energy in process 1 ® 2 ® 3
(Q)
1
RT0
3
(III) Heat absorbed by the system in process 1 ® 2 ® 3
(R) RT0
(IV) Heat absorbed by the system in process 1 ® 2
(S)
4
RT0
3
(T)
1
RT0 (3 + ln 2)
3
(U)
5
RT0
6
1
3
[IIT-JEE 2019]
If the process on one mole of monatomic ideal gas is an shown is as shown in the TV-diagram with P 0V0 =
RT0 , the correct match is
T
T0
T0
3
64
3
1
2
V0
2V0
V
(A) I ® S, II ® T, III ® Q, IV ® U
(B) I ® P, II ® R, III ® T, IV ® S
(C) I ® P, II ® T, III ® Q, IV ® T
(D) I ® P, II ® R, III ® T, IV ® P
KTG & Thermodynamics
MCQs with one or more than one correct answer
1.
Let n , nrms and np respectively denote the mean speed, root mean square speed and most probable speed of
the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m. Then :–
2 nrms
(A) No molecule can have energy greater than
(B) No molecule can have speed less than
[IIT-JEE 1998]
np
2
(C) np < n < nrms
(D) The average kinetic energy of a molecule is
2.
3
mn2p
4
[IIT-JEE 1998]
During the melting of a slab of ice at 273 K at atmospheric pressure :–
(A) Positive work is done by the ice–water system on the atmosphere
(B) Positive work is done on the ice–water system by the atmosphere
(C) the internal energy of the ice–water increases
(D) The internal energy of the ice–water system decreases
3.
CV and CP denote the molar specific heat capacities of a gas at constant volume and constant pressure,
respectively. Then
[IIT-JEE 2009]
(A) CP – CV is larger for a diatomic ideal gas than for a monoatomic ideal gas
(B) CP + CV is larger for a diatomic ideal gas than for a monoatomic ideal gas
(C) CP/CV is larger for a diatomic ideal gas than for a monatomic ideal gas
(D) CP . CV is larger for a diatomic ideal gas than for a monoatomic ideal gas
4.
The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA.
3
The part ABC is a semi-circle and CDA is half of an ellipse. Then,
2
(A) the process during the path A ® B is isothermal
(B) heat flows out of the gas during the path B ® C ® D
P
D
B
1
0
(C) work done during the path A ® B ® C is zero
A
C
1
3
V
[IIT-JEE 2009]
(D) positive work is done by the gas in the cycle ABCDA
5.
2
A container of fixed volume has a mixutre of one mole of hydrogen and one mole of helium in equilibrium at
temperature T. Assuming the gases are ideal, the correct statement(s) is (are) :[IIT-JEE 2015]
(A) The average energy per mole of the gas mixture is 2RT.
(B) The ratio of speed of sound in the gas mixture to that in helium gas is
6/5.
(C) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2.
(D) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1 / 2 .
6.
One mole of a monatomic ideal gas undergoes a cyclic process as shown in the figure (whre V is the volume
and T is the temperature). Which of the statements below is (are) true ?
[IIT-JEE 2018]
T
II
III
I
IV
(A) Process I is an isochoric process
V
(B) In process II, gas absorbs heat
(C) In process IV, gas releases heat
(D) Processes I and II are not isobaric
65
JEE-Physics
7.
One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus
temperature (V-T) diagram. The correct statement(s) is/are :
[IIT-JEE 2019]
[R is the gas constant]
V
3
2V0
V0
4
2
1
T0/2
T0
3T0/2 2T0
T
(A) Work done in this thermodynamic cycle (1®2®3®4®1) is |W| =
1
RT0
2
Q1® 2
5
=
(B) The ratio of heat transfer during processes 1®2 and 2®3 is Q
3
2 ®3
(C) The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
Q1® 2
1
=
(D) The ratio of heat transfer during processes 1®2 and 3®4 is Q
2
3 ®4
8.
A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at
pressure P0, volume V0 and temperature T0. If the gas mixture is adiabatically compressed to a volume V 0/4,
then the correct statement(s) is/are,
[IIT-JEE 2019]
(Give 2
1.2
= 2.3 ; 2
3.2
= 9.2; R is gas constant)
(A) The final pressure of the gas mixture after compression is in between 9P0 and 10P0
(B) The average kinetic energy of the gas mixture after compression is in between 18RT 0 and 19RT0
(C) The work |W| done during the process is 13RT0
(D) Adiabatic constant of the gas mixture is 1.6
Match the Column
1.
[IIT-JEE 2006]
For the given process :–
P
30
J
M
20
10
66
K
L
10
20
3
v(m )
Column I
Column II
(A) Process J ® K
(p) W > 0
(B) Process K ® L
(q) W < 0
(C) Process L ® M
(r) Q > 0
(D) Process M ® J
(s) Q < 0
KTG & Thermodynamics
2.
Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II
describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate
bubbles of the 4 × 4 matrix given in the ORS.
[IIT-JEE 2008]
Column I
Column II
(A) An insulated container has two chambers separated
(p) The temperature of the gas decreases
by a valve. Chamber I contains an ideal gas and the
Chamber II has vacuum. The valve is opened.
I
II
ideal gas
vacuum
(B) An ideal monoatomic gas expands to twice its original
volume such that its pressure P µ
1
, where V is
V2
the volume of the gas.
(C) An ideal monoatomic gas expands to twice its
original volume such that its pressure P µ
(q) The temperature of the gas
increases or remains constant
(r) The gas loses heat
1 ,
V4 /3
where V is its volume
(D) An ideal monoatomic gas expands such that its
pressure P and volume V follows the behaviour
shown in the graph
(s) The gas gains heat
P
V1
2V 1 V
Answer Q.3, Q.4 and Q.5 by appropriately matching the information given in the three columns of the
following table.
An ideal gas is undergoing a cyclic thermodynamics process in different ways as shown in the corresponding
P–V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the
corresponding work done on the system. The equations and plots in the table have standard notations as used
in thermodynamics processes. Here g is the ratio of heat capacities at constant pressure and constant volume.
The number of moles in the gas is n.
[IIT-JEE 2017]
Column-1
Column-2
Column-3
P
(I) W1® 2 =
1
(P2 V2 – P1V1 )
γ–1
(i) Isothermal
1
2
(P)
V
P
1
(II) W1®2 = –PV2 + PV1
(ii) Isochoric
(Q)
2
P
(III) W1®2 = 0
(iii) Isobaric
1
V
2
(R)
V
67
JEE-Physics
P
V2
(IV) W1®2 = –nRT ln V
1
(iv) Adiabatic
(S)
1
2
V
Which of the following options is the only correct representation of a process in which DU = DQ – PDV?
3.
4.
(A) (II) (iv) (R)
(B) (II) (iii) (P)
(C) (II) (iii) (S)
(D) (III) (iii) (P)
Which one of the following options is the correct combination ?
5.
(A) (III) (ii) (S)
(B) (II) (iv) (R)
(C) (II) (iv) (P)
(D) (IV) (ii) (S)
Which one of the following options correctly represents a thermodynamics process that is used as a correction
in the determination of the speed of sound in an ideal gas ?
(A) (III) (iv) (R)
(B) (I) (ii) (Q)
(C) (IV) (ii) (R)
(D) (I) (iv) (Q)
Assertion & Reason
1.
Statement–I : The total translation kinetic energy of all the molecules of a given mass of an ideal gas is
1.5 times the product of its pressure and its volume.
[IIT-JEE 2007]
Because :
Statement–II : The molecules of a gas collide with each other and the velocities of the molecules change due
to the collision.
(A) statement–I is true, statement–II is true; statement–II is a correct explanation for statement–I
(B) statement–I is true, statement–II is true, statement–II is NOT a correct explanation for statement–I
(C) statement–I is true, statement–II is false
(D) statement–I is false, statement–II is true
Comprehension Type Questions
Comprehension-1
[IIT-JEE 2007]
A fixed thermally conducting cylinder has a radius R and height L 0. The cylinder is open at its bottom and has
a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure.
The atmospheric pressure is P0.
2R
L
L0
P isto n
1.
The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder
between its top and the piston will then be :–
(A) P0
68
(B)
P0
2
(C)
P0 Mg
+
2 pR 2
(D)
P0 Mg
2 pR 2
KTG & Thermodynamics
2.
While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to
a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is :–
æ 2P0 pR 2 - Mg ö
(B) ç
÷ (2L)
pR 2 P0
è
ø
æ 2P pR 2 ö
(2L)
(A) ç 2 0
è pR P + Mg ÷ø
0
3.
æ P pR 2 + Mg ö
(2L)
(C) ç 0
è pR2 P0 ÷ø
æ P pR 2 ö
(2L)
(D) ç 2 0
è pR P0 - Mg ÷ø
The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below
the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the
cylinder as shown in the figure. The density of the water is r. In equilibrium, the height H of the water column
in the cylinder satisfies :–
(A) rg(L0 – H)2 + P0(L0 – H) + L0P0 = 0
(B) rg(L0 – H)2 – P0(L0 – H) – L0P0 = 0
L0
(C) rg(L0 – H)2 + P0(L0 – H) – L0P0 = 0
H
(D) rg(L0 – H)2 – P0(L0 – H) + L0P0 = 0
Comprehension-2
[IIT-JEE 2008]
5ö
æ
A small spherical monoatomic ideal gas bubble ç g = ÷ is trapped inside a liquid of density rl (see figure).
è
3ø
Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The
temperature of the gas when the bubble is at the bottom is T 0, the height of the liquid is H and the atmospheric
pressure is P0 (Neglect surface tension).
P0
Liquid
H
y
1.
As the bubble moves upwards, besides the buoyancy force the following forces are acting on it.
(A) Only the force of gravity
(B) The force due to gravity and the force due to the pressure of the liquid
(C) The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the
liquid
(D) The force due to gravity and the force due to viscosity of the liquid.
2.
When the gas bubble is at a height y from the bottom, its temperature is :–
æ P0 + rl gH ö
(A) T0 ç
è P0 + rl gy ÷ø
3.
2/ 5
æ P0 + rl g(H - y) ö
(B) T0 ç
è P0 + rl gH ÷ø
2/ 5
æ P0 + rl gH ö
(C) T0 ç
è P0 + rl gy ÷ø
3/5
æ P0 + rl g(H - y) ö
(D) T0 ç
è P0 + rl gH ÷ø
The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant)
(A) plnRgT0
(C) plnRgT0
(P0 + rl gH)2 / 5
(P0 + rl gy)
7/5
(P0 + rl gH)3 / 5
(P0 + rl gy)
8/5
rl nRgT0
(B) (P + r gH)2 / 5 [P + r g(H - y)]3 / 5
0
l
0
l
(D)
rl nRgT0
3/ 5
(P0 + rl gH)
[P0 + rl g(H
- y)]2 / 5
69
3/5
JEE-Physics
Comprehension-3
[IIT-JEE 2014]
In the figure a container is shown to have a movable (without friction) piston on top. The container
and the piston are all made of perfectly insulating material allowing no heat transfer between
outside and inside the container. The container is divided into two compartments by a rigid
partition made of a thermally conducting material that allows slow transfer of heat. the lower
compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the
upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. the heat capacities
3
5
per mole of an ideal monatomic gas are C V = R , C P = R , and those for an ideal diatomic
2
2
7
gas are C V = 5 R , C P = R .
2
2
Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final
temperature of the gases will be –
1.
(A) 550 K
2.
(B) 5525 K
(C) 513 K
(D) 490 K
Now consider the partition to be free to move without friction so that the pressure of gases in both compartments
is the same. Then total work done by the gases till the time they achieve equilibrium will be –
(A) 250 R
(B) 200 R
(C) 100 R
(D) –100 R
Subjective Questions
1.
One mole of an ideal monoatomic gas is taken round the cyclic process ABCA as P
3P 0
shown in figure. Calculate :
(i) The work done by the gas.
(ii) The heat rejected by the gas in the path CA and the
B
A
P0
heat absorbed by the gas in the path AB.
2V 0 V
V0
(iii) The net heat absorbed by the gas in the path BC.
[IIT-JEE 1998]
(iv) The maximum temperature attained by the gas during the cycle.
2.
C
Two moles of an ideal monoatomic gas initially at pressure P 1 and volume V1 undergo an adiabatic compression
until its volume is V2. Then the gas is given heat Q at constant volume V 2.
[IIT-JEE 1999]
(i) Sketch the complete process on a P–V diagram.
(ii) Find the total work done by the gas, the total change in internal energy and the final temperature of the
gas. (Give your answer in terms of P1, V1, V2, Q and R)
3.
Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in
the P–V diagram. During the process AB, pressure and temperature of the gas
very such that PT = constant. If T1 = 300K, calculate
[IIT-JEE 2000]
V
2P 1
A
P1
(i) The work done on the gas in the process AB and
(ii) The heat absorbed or released by the gas in each of the processes.
Give answers in terms of the gas constant P.
4.
C
B
T1
T
2T1
A monoatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure.
The volume ratio are
VB
V
= 2 and D = 4. If the temperature TA at A is 27°C. Calculate : [IIT-JEE 2001]
VA
VA
V
(i) The temperature of the gas at point B.
VD
D
C
(ii) Heat absorbed or released by the gas in each process.
(iii) The total work done by the gas during the complete cycle.
Express your answer in terms of the gas constant R.
70
B
VB
VA
O
A
TA
TB
T
KTG & Thermodynamics
5.
A cubical box of side 1m contains helium gas (atomic weight 4) at a pressure of 100 N/m 2. During an observation
time of 1s, an atom travelling with the root mean square speed parallel to one of the edges of the cube, was
found to make 5000 hits with a particular wall, without any collision with other atoms.
[IIT-JEE 2002]
Take : R = 25/3 J/mol–K and k = 1.38 × 10–23 J/K.
(i) Evaluate the temperature of the gas.
(ii) Evaluate the average kinetic energy per atom.
(iii) Evaluate the total mass of helium gas in the box.
6.
An insulated box containing a monoatomic gas of molar mass M moving with a speed v 0 is suddenly stopped.
Find the increment in gas temperature as a result of stopping the box.
[IIT-JEE 2003]
7.
The piston cylinder arrangement shown contains a diatomic gas at temperature 300 K. The cross–sectional
area of the cylinder is 1m2. Initially the height of the piston above the base of the cylinder is 1m. The temperature
is now raised to 400 K at constant pressure. Find the new height of the piston above the base of the cylinder.
If the piston is now brought back to its original height without any heat loss, find the new equilibrium temperature
[IIT-JEE 2004]
of the gas. You can leave the answer in fraction.
1m
8.
A metal of mass 1 kg at constant atmospheric pressure and at initial temperature 20°C is given a heat of
20000 J. Find the following :
[IIT-JEE 2005]
(i) change in temperature
(ii) work done and
(iii) change in internal energy.
(Given : Specific heat 400 J/kg/°C, coefficient of cubical expansion, g = 8 × 10–5 /°C, density r = 9000 kg/m3,
atmospheric pressure = 105 N/m2.
Integer Type
1.
A thermodynamic system is taken from an initial state i with internal energy Ui = 100 J to the final state f along
two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along the
paths af, ib and bf are Waf = 200 J, Wib = 50 J and Wbf = 100 J respectively. The heat supplied to the system
along the path iaf, ib and bf are Qiaf, Qib and Qbf respectively. If the internal energy of the system in the state b
is Ub = 200 J and Qiaf = 500 J, the ratio Qbf/Qib is –
[JEE(Adv.) 2014]
2.
One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight
times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 J mol –
1 –1
K , the decrease in its internal energy, in Joule, is............ .
[JEE(Adv.) 2018]
71
JEE-Physics
ANSWER KEY
EXERCISE-1
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
D
A
C
D
B
D
A
A
C
A
C
B
B
B
D
Que.
16
17
18
19
20
21
22
23
24
Ans.
B
C
B
A
A
D
D
D
C
EXERCISE-2
l
Que.
1
2
3
4
5
6
7
8
9
10
Ans.
B
B,C
A,B,C,D
B
A,B,C,D
A,B,D
A,B
A,C
B,D
C,D
Match the Column:
11. (A) q, (B) r (C) p (D) s
12. (A) s (B) r (C) q (D) p
13. (A) q (B) r (C) s (D) p
l
Comprehension Based Quesions
Comp-1 :
14. (A)
15. (C)
Comp-2:
16. (D)
17. (A)
18. (C)
Comp-3:
20. (B)
21. (B)
22. (B)
19. (D)
EXERCISE-3
3
5
R,
(iii) 450 R
2
3
2. (i) 1152 J (ii) 1152 J (iii) zero
1. (i) mono atomic, 3 (ii)
3. (i) P1<P2 , T1<T2 (ii) T1=T2<T3 (iii) V 1<V 2 (iv) P1>P2
4. AC, 170 J, 10 J
5. (i) 1.8 × 10 5 J (ii) 4.8 × 10 5 J (iii) 6.6 × 10 5 J (iv) 17.1 J/mole–K
6. (i) 189 K (ii) –2767 J (iii) 2767 J
7. –972 J
8. (i) 765 J (ii) 10.82 %
9. 5
10. 1.5
EXERCISE-4
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
D
A
D
A
A
C
B
B
A
D
B
B
D
D
B
Que.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
Que.
Ans.
A
31
C
B
32
B
C
33
B
D
34
D
C
35
D
B
36
A
A
37
D
B
38
C
B
39
A
A
40
C
B
41
D
C
42
C
C
43
C
D
44
D
C
45
A
72
KTG & Thermodynamics
EXERCISE-5
l
l
MCQ's (Single Correct answers)
1. (A)
2. (C)
3. (D)
4. (C)
5. (D)
6. (D)
7. (A)
8. (C)
9. (B)
10. (A)
11. (B)
12. (A)
13. (A)
14. (C)
15. (C)
16. (D)
17. (A)
18. (B)
19. (C)
20. (C)
21. (C)
22. (D)
MCQ's (one or more than one correct)
1. (C,D)
l
2. (B,C)
3. (B,D)
Match the column
4. (B,D)
5. (A,B,D) 6. (B,C,D) 7. (A,B)
8. (A,C,D)
1. (A) s, (B) p,r (C) r (D) q,s
2. (A) q, (B) p,r (C) p,s (D) q,s
3. (B)
l
Assertion–Reason
l
Comprehension Based Questions
l
4. (A)
5. (D)
1. (B)
Comprehension-1
1. (A
2. (D)
3. (C)
Comprehension-2
1. (D)
2. (B)
3. (B)
Comprehension-3
1. (D)
2. (D)
Subjective Questions
1. (i) P0V 0 (ii)
C
P
B
P2
2. (i)
5
1
25 P0 V0
P V , 3P0V 0 (iii) P0V 0 (iv)
R
2 0 0
2
8
A
P1
V1
V2
V
2/3
é æ V1 ö 2 / 3 ù
éæ V1 ö2 / 3
ù
3
3
Q
P1 V1 æ V1 ö
1
ê
ú
(ii) W= P1V1
, DU = P1V1 êç ÷ - 1ú + Q , T=
+
ç
÷
ç ÷
2
2
3R
2R è V2 ø
ëê è V2 ø ûú
úû
ëêè V2 ø
3. (i) 1200 R (ii) QAB = –2100 R, QBC = 1500 R, QCA = 1200R ln2
4. (i) 600 K (ii) 1500R, 831.6R, –900R, –831.6R (iii) 600R
5. (i) 160K (ii) 3.3 × 10 –21 J (iii) 0.3 g
æ 4ö
7. 400 çè ÷ø
3
l
6.
Mv 20
3R
0.4
8. (i) 50 °C (ii) 0.05 J (iii) 19999.95 J
K
Integer Questions
1. 2
2. 900
*****
73
THERMAL PHYSICS
Recap of Early Classes
W all have common-sense notions of heat and temperature. Temperature is a measure of ‘hotness’ of
We
a body. In this chapter, you will learn what heat is and how it is measured, and study the various processes by
which heat flows from one body to another.
1.0
TEMPERATURE
1.1
2.0
3.0
4.0
Temperature Scales
THERMAL EXPANSION
2.1
Linear expansion
2.2
Superficial (areal) expansion
2.3
Volume expansion
2.4
Application of thermal Expansion in Solids
2.5
Thermal Expansion in Liquids
2.6
Anomalous expansion of water
MODE OF HEAT TRANSFER
3.1
Thermal conduction
3.2
Convections
3.3
Thermal Radiation
HEAT
4.1
Mechanical Equivalent of Heat
4.2
Water Equivalent of a Body
4.3
Phase of a Substance
4.4
Phase Diagram for Water
4.5
Law of Mixtures
EXERCISE-1
EXERCISE-2
EXERCISE-3
EXERCISE-4
EXERCISE-5
Tinku
JEE-Physics
THERMAL PHYSICS
1.0 TEMPERATURE
Temperature is a macroscopic physical quantity related to our sense of hot and cold.
The natural flow of heat is from higher temperature to lower temperature, i.e. temperature determines the
thermal state of a body whether it can give or receive heat.
1.1
Temperature Scales
The Kelvin temperature scale is also known as thermodynamic scale. The SI unit of temperature is the kelvin
and is defined as (1/273.16) of the temperature of the triple point of water. The triple point of water is that
point on a P–T diagram where the three phase of water, the solid, the liquid and the gas, can coexist in
equilibrium.
In addition to Kelvin temperature scale, there are other temperature scales also like Celsius, Fahrenheit,
Reaumur, Rankine, etc. Temperature on one scale can be converted into other scale by using the following
identity
Hence
Reading on any scale - lower fixed point (LFP)
= constant for all scales
Upper fixed point (UFP) - lower fixed point (LFP)
C - 0°
F - 32°
K - 273.15
=
=
100° - 0° 212° - 32° 373.15 - 273.15
Different temperature scales :
Name of the
scale
Celsius
Symbol for
each degree
°C
Lower fixed
point (LFP)
0°C
Upper fixed
point (UFP)
100°C
Number of divisions
on the scale
100
Fahrenheit
Kelvin
°F
K
32°F
273.15 K
212°F
373.15 K
180
100
Illustration 1.
Express a temperature of 60°F in degree celsius and in kelvin.
Solution
By using
Þ
C - 0°
F - 32°
K - 273.15
=
=
100° - 0° 212° - 32° 373.15 - 273.15
C - 0°
60° - 32°
K - 273.15
=
=
100° - 0° 212° - 32° 373.15 - 273.15
Þ C= 15.15° C and K = 288.7 K
Illustration 2.
Solution
The temperature of an iron piece is heated from 30° to 90°C. What is the change in its temperature
on the fahrenheit scale and on the kelvin scale?
DC=90°–30° = 60°C
Temperature difference on Fahrenheit Scale
9
9
DC = (60°C ) = 108°F
5
5
Temperature difference on Kelvin Scale
DF =
DK = DC = 60K
2.0 THERMAL EXPANSION
When matter is heated without any change in it's state, it usually expands. According to atomic theory of matter,
asymmetry in potential energy curve is responsible for thermal expansion. As with rise in temperature the amplitude
of vibration increases and hence energy of atoms increases, hence the average distance between the atom
increases. So the matter as a whole expands.
74
Thermal Physics
Thermal expansion is minimum in case of solids but maximum in case of gases because intermolecular force is
maximum in solids but minimum in gases.
Solids can expand in one dimension (Linear expansion), two dimension (Superficial expansion) and three dimension (Volume expansion) while liquids and gases usually suffers change in volume only.
2.1
Linear expansion
l = l0 (1 + aDq) Þ Dl = l0aDq
0
T
Before heating
2.2
+D =
After heating
Superficial (areal) expansion
T+DT
T
A = A0 (1 + bDq)
2.3
0
T+DT
Also
A0 = l02 and A = l2
A0
So
l2 = l02(1 + bDq) = [l0(1 + aDq)]2 Þ b =2a
l0
l0
l
A
l
Volume expansion
V = V0 (1 + gDq) Also V = l and V0=l0 so g =3a
l0
Þ 6a = 3b = 2g or a : b : g = 1 : 2 : 3
T+DT
T
3
3
V0
l0
V
l
l0
l
l
l
l
Contraction on heating
Some rubber like substances contract on heating because transverse vibration of atoms of substance dominate
over longitudinal vibration which is responsible for expansion.
2.4
Application of thermal Expansion in Solids
(a)
Bi–metallic strip : Two strips of equal length but of different materials (different coefficient of linear
expansion) when join together, it is called "Bi–metallic strip" and can be used in thermostat to break or
make electrical contact. This strip has the characteristic property of bending on heating due to unequal
linear expansion of the two metals. The strip will bend with metal of greater a on outer side.
Steel
Brass
Steel
Room temperature
(b)
Brass
Higher temperature
ON
At room temperature
OFF
Bimetallic
strip
At high temperature
Effect of temperature on the time period of a simple pendulum : A pendulum clock keeps proper
time at temperature q. If temperature is increased to q' ( > q) then due to linear expansion, length of
pendulum increases and hence its time period will increase.
Fractional change in time period
DT 1 Dl
DT 1
=
)
= aDq (Q T µ l \
T
2 l
T
2
75
JEE-Physics
• Due to increment in its time period, a pendulum clock becomes slow in summer and will lose time.
1
a DqT
2
• The clock will lose time i.e. will become slow if q' > q (in summer) and will gain time i.e will become
fast if q' < q (in winter).
• Since coefficient of linear expansion (a) is very small for invar, hence pendulums are made of invar to
show the correct time in all seasons.
Loss of time in a time period DT=
(c)
When a rod whose ends are rigidly fixed such as to prevent expansion or contraction, undergoes a change
in temperature due to thermal expansion or contraction, a compressive or tensile stress is developed in it.
Due to this thermal stress the rod will exert a large force on the supports. If the change in temperature of
a rod of length L is Dq then :–
Heating
Cooling
DL 1
DL
´
Q a=
= aDq
Dq
L
L
So force on the supports F=YAaDq
So thermal stress = YaDq Q Y =
Thermal strain =
(d)
Error in scale reading due to expansion or contraction: If a scale gives correct reading at temperature
q. At temperature q'(>q) due to linear expansion of scale, the scale will expand and scale reading will be
lesser than true value so that,
0
a
0
SR
at q' > q
TV > SR
at q
TV = SR
(e)
stress
strain
a
0
a SR
at q' < q
TV < SR
Expansion of cavity: Thermal expansion of an isotropic object may be imagined as a photographic
enlargement.
a
A
r
B
r
a
D
Expansion of A
= Expansion of B
Expansion of C
= Expansion of D
C
b
(f)
76
b
Some other application
• When rails are laid down on the ground, space is left between the ends of two rails
• The transmission cable are not tightly fixed to the poles
• Test tubes, beakers and cubicles are made of pyrex–glass or silica because they have very low value of
coefficient of linear expansion
• The iron rim to be put on a cart wheel is always of slightly smaller diameter than that of wheel
• A glass stopper jammed in the neck of a glass bottle can be taken out by warming the neck of the
bottle.
Thermal Physics
Illustration 3.
A steel ruler exactly 20 cm long is graduated to give correct measurements at 200C.
( a steel = 1.2 ´ 10-5 °C -1 )
(a) Will it give readings that are too long or too short at lower temperatures?
(b) What will be the actual length of the ruler be when it is used in the desert at a temperature of
40°C?
(a) If the temperature decreases, the length of the ruler also decreases through thermal contraction.
Below 200C, each centimeter division is actually somewhat shorter than 1.0 cm, so the steel
ruler gives readings that are too long.
Solution
(b) At 400C, the increases in length of the ruler is
Dl=laDT = (20) (1.2 × 10–5) (400 – 200) = 0.48 × 10–2 cm
\ The actual length of the ruler is, l'=l+Dl=20.0048 cm
Illustration 4.
A second's pendulum clock has a steel wire. The clock is calibrated at 20 0C. How much time does
the clock lose or gain in one week when the temperature is increased to 30 0 C?
( a steel = 1.2 ´ 10-5 °C -1 )
Solution
The time period of second's pendulum is 2 second. As the temperature increases length time period
increases. Clock becomes slow and it loses the time. The change in time period is
DT =
2.5
1
æ 1ö
TaDq = ç ÷ (2) (1.2 ´ 10-5 ) ( 300 - 200 ) = 1.2 × 10–4 s
è 2ø
2
\ New Time period is
T ¢ = T + DT = ( 2 + 1.2 ´ 10-4 ) = 2.00012 s
\ Time lost in one week
(1.2 ´ 10-4 )
æ DT ö
Dt = ç
t=
÷
è T¢ ø
(2.00012) ( 7 ´ 24 ´ 3600) = 36.28 s
Thermal Expansion in Liquids
Liquids do not have linear and superficial expansion but these only have volume expansion.
Since liquids are always to be heated along with a vessel which contains them so
initially on heating the system (liquid + vessel), the level of liquid in vessel falls (as
vessel expands more since it absorbs heat and liquid expands less) but later on, it
starts rising due to faster expansion of the liquid.
PQ ® represents expansion of vessel
QR ® represents the real expansion of liquid.
The actual increase in the volume of the liquid
= The apparent increase in the volume of liquid + the increase in the volume of the vessel.
Liquids have two coefficients of volume expansion.
(i)
Co–efficient of apparent expansion (ga)
It is due to apparent (that appears to be, but in not) increase in the volume of liquid if expansion of vessel
containing the liquid is not taken into account.
ga =
(ii)
R
P
Q
Apparent expansion in volume
(DV)
=
Initial volume ´ Dq
V ´ Dq
Co–efficient of real expansion (gr)
It is due to the actual increase in volume of liquid due to heating.
gr =
Real increase in volume
( DV)
=
Initial volume ´ Dq
V ´ Dq
Also coefficient of expansion of flask g Vessel =
(DV) Vessel
V ´ Dq
77
JEE-Physics
gReal = gApparent + gVessel
Change (apparent change) in volume in liquid relative to vessel is
DVapp= V(gReal – gVessel) Dq = V(gr – 3a)Dq
a = Coefficient of linear expansion of the vessel.
Different level of liquid in vessel
g
DV
Level
g Real. > g Vessel ( = 3a ) Þ g app > 0
DVapp is positive
Level of liquid in Vessel will rise on heating
g Re al. < g Vessel ( = 3a ) Þ g app < 0 DVapp is negative
Level of liquid in vessel will fall on heating
g Real = g Vessel ( = 3a ) Þ g app = 0
Level of liquid in vessel will remain same
Illustration 5.
Solution
DVapp = 0
In figure shown, left arm of a U–tube is immersed in a hot
water bath at temperature t°C, and right arm is immersed in a
bath of melting ice; the height of manometric liquid in respective columns is ht and h0. Determine the coefficient of expan
sion of the liquid.
Water at
0
temperature t C
ht
The liquid is in hydrostatic equilibrium Þ rt gh t = r0 gh 0
Where, rt is density of liquid in hot bath, r0 is density of
Melting ice
h0
liquid in cold bath.
Volumes of a given mass M of liquid at temperatures t
and 00C
are related by Vt = V0(1+gt) Since rt Vt = r0 V0
r0 V0
r0
Þ rt = V = 1 + gt
(
)
t
Since ht =
2.6
( h t - h0 )
r0 h0
= h0 (1 + gt ) which on solving for g , yields g =
h0 t
rt
Anomalous expansion of water
maximum
Bahaviour
Anomalous
Vol/mass
Anomalous
Bahaviour
Vol/mass
min
0°C 4°C
0°C 4°C
(A)
Temperature
(B)
Temperature
Generally matter expands on heating and contracts on cooling. In case of water, it expands on heating if its
temperature is greater than 4°C. In the range 0°C to 4°C, water contracts on heating and expands on cooling, i.e.
g is negative. This behaviour of water in the range from 0°C to 4°C is called anomalous expansion.
This anomalous behaviour of water causes ice to form first at the surface of a lake in cold weather. As winter
approaches, the water temperature increases initially at the surface. The water there sinks because of its
increased density. Consequently, the surface reaches 0°C first and the lake becomes covered with ice. Aquatic
life is able to survive the cold winter as the lake bottom remains unfrozen at a temperature of about 4°C. At 4°C,
density of water is maximum while its specific volume is minimum.
78
Thermal Physics
Illustration 6.
Solution
The difference between lengths of a certain brass rod and of a steel rod is claimed to be constant at
all temperatures. Is this possible ?
If LB and LS are the lengths of brass and steel rods respectively at a given temperature, then the
lengths of the rods when temperature is changed by q °C.
L¢B= LB(1 + aB Dq) and L¢S= LS(1 + aB Dq)
So that L¢B= L¢S(LB – LS) + (LB aB – LSaS) Dq
So (L¢B– L¢S) will be equal to (LB – LS) at all temperatures if , LBaB – LSaS = 0 [as Dq ¹ 0]
or
L B aS
=
L S aB
i.e., the difference in the lengths of the two rods will be independent of temperature if the lengths are
in the inverse ratio of their coefficients of linear expansion.
Illustration 7.
Solution
There are two spheres of same radius and material at same temperature but one being solid while
the other hollow. Which sphere will expand more if
(a) they are heated to the same temperature, (b) same heat is given to them ?
(a) As thermal expansion of isotropic solids is similar to true photographic enlargement,
V
V
expansion of a cavity is same as if it had been a solid body of the same material
i.e. DV = Vg Dq
As here V, g and Dq are same for both solid and hollow spheres treated (cavity) ;
so the expansion of both will be equal.
(b) If same heat is given to the two spheres due to lesser mass, rise in temperature of hollow sphere
will be more [as Dq =
Q
] and hence its expansion will be more [as DV = VgDq].
mc
1.
Expansion during heating :
(A) occurs only in a solid
(C) decreases the density of the material
2.
Two identical beakers with neglegible thermal expansion are filled with water to the same level at 4°C. If one say
A is heated while the other B is cooled, then :
(A) water level in A must rise
(B) water level in B must rise
(C) water level in A must fall
(D) water level in B must fall
3.
If I is the moment of inertia of a solid body having a-coefficient of linear expansion then the change in I
correcponding to a small change in temperature DT is :
(A) aI DT
(B)
1
a IDT
2
(B) increases the density of the material
(D) occurs at the same rate for all liquids and solids
(C) 2a I DT
(D) 3aI DT
4.
A liquid with coefficient of volume expansion g is filled in a container of a material having the coefficient of
linear expansion a. If the liquid overflows on heating, then
(A) g > 3a
(B) g < 3a
(C) g = 3a
(D) None of these
5.
There is a rectangular metal plate in which two cavities in the shape of
rectangle and circle are made as shown with dimensions. P and Q are
centres of these cavities. On heating the plate, which of the following
quantities increase ?
(A) p r2
(B) ab
(C) R
(D) b
P
79
JEE-Physics
6.
Temperature of plate is increased by Dq then find new
(a) inner radius
(b) outer radius
(c) the difference in outer and inner radius and show that it is positive
(d) area of plate material (assume coefficient of expansion is a)
7.
We have a hollow sphere and a solid sphere of equal radii and of the same material. They are heated to raise
their temperature by equal amounts. How will the change in their volumes, due to volume expansions, be
related? Consider two cases (i) hollwo sphere is filled with air, (ii) there is vacuum inside the hollow sphere.
8.
Consider a cylindrical container of cross section area ‘A’, length ‘h’ having
coefficient of linear expansion ac. The container is filled by liquid of real
expansion coefficient lL up to height h1. When temperature of the system is
increased by Dq then
(a) Find out new height, area and volume of cylindrical container and new volume of liquid.
(b) Find the height of liquid level when expansion of container is neglected.
(c) Find the relation between lL and ac for which volume of container above the liquid level.
(i) increases
(ii) decreases
(iii) remains constant
(d) If ll > 3ac and h = h1 then calculate, the volume of liquid overflow
9.
The readings of a thermometer at 0° C and 100° C are 50 cm and 75 cm of mercury column respectively. Find
the temperature at which its reading is 80 cm of mercury column?
3.0 MODE OF HEAT TRANSFER
Heat is a form of energy which transfers from a body at higher temperature to a body at lower temperature.
The transfer of heat from one body to another may take place by any one of the following modes :.
•
Conduction
The process in which the material takes an active part by molecular action and energy is passed from one
particle to another is called conduction. It is predominant in solids.
•
Convection
The transfer of energy by actual motion of particle of medium from one place to another is called convection. It
is predominant is fluids (liquids and gases).
•
Radiation
Quickest way of transmission of heat is known as radiation. In this mode of energy transmission, heat is
transferred from one place to another without effecting the inter–venning medium.
Conduction
Convection
Heat Transfer due to
Temperature difference
Heat transfer due to density
difference
Heat transfer with out any
medium
Due to free electron or vibration
motion of molecules
Actual motion of particles
Electromagnetic radiation
Heat transfer in solid body (in
mercury also)
Heat transfer in fluids (Liquid +
gas)
All
Slow process
Slow process
Fast process (3 × 108 m/sec)
Irregular path
Irregular path
Straight line (like light)
3.1
Radiation
Thermal conduction
The process by which heat is transferred from hot part to cold part of a body through the transfer of energy from
one particle to another particle of the body without the actual movement of the particles from their equilibrium
positions is called conduction. The process of conduction only in solid body (except Hg) Heat transfer by
conduction from one part of body to another continues till their temperatures become equal.
80
Thermal Physics
The process of transmission of heat energy in which heat is transferred from one particle of the medium to the
other, but each particle of the medium stays at its own position is called conduction, for Illustration if you hold
an iron rod with one of its end on a fire for some time, the handle will get hot. The heat is transferred
from the fire to the handle by conduction along the length of iron rod.
L
The vibrational amplitude of atoms and electrons of the iron rod at
TC
the hot end takes on relatively higher values due to the higher
TH
temperature of their environment. These increased vibrational amplitude
Q2
are transferred along the rod, from atom to atom during collision between
adjacent atoms. In this way a region of rising temperature extends
Q1
itself along the rod to your hand.
O
A
B
Consider a slab of face area A, Lateral thickness L, whose faces have
x
dx
temperatures TH and TC(TH > TC).
Now consider two cross sections in the slab at positions A and B separated by a lateral distance of dx. Let
temperature of face A be T and that of face B be T + DT. Then experiments show that Q, the amount of heat
crossing the area A of the slab at position x in time t is given by
Here K is a constant depending on the material of the slab and is named thermal conductivity of the material,
æ dT ö
and the quantity ç ÷ is called temperature gradiant. The (–) sign in equation (2.1) shows heat flows from high
è dx ø
to low temperature (DT is a –ve quantity)
Steady state
If the temperature of a cross-section at any position x in the above slab remains constant with time (remember,
it does vary with position x), the slab is said to be in steady state.
Remember steady-state is distinct from thermal equilibrium for which temperature at any position (x) in the slab
must be same.
For a conductor in steady state there is no absorption or emission of heat at any cross-section. (as temperature
at each point remains constant with time). The left and right face are maintained at constant temperatures TH
and TC respectively, and all other faces must be covered with adiabatic walls so that no heat escapes through
them and same amount of heat flows through each cross-section in a given Interval of time. Hence
Q1 = Q = Q2. Consequently the temperature gradient is constant throughout the slab.
dT DT Tf - Ti TC - TH
=
=
=
dx
L
L
L
Hence,
and
æ TH - TC ö
Q
Q
DT
= -KA
= KA çè
Þ
L ÷ø
t
t
L
Here Q is the amount of heat flowing through a cross-section of slab at any position in a time interval of t.
Illustration 8.
Solution
One face of an aluminium cube of edge 2 metre is maintained at 100 °C and the other end is
maintained at 0 °C. All other surfaces are covered by adiabatic walls. Find the amount of heat
flowing through the cube in 5 seconds. (thermal conductivity of aluminium is 209 W/m–°C)
Heat will flow from the end at 100ºC to the end at 0°C.
Area of cross-section perpendicular to direction of heat flow, A = 4m 2 then
(T - TC )
Q
= KA H
t
L
Q=
(209W / mº C)(4m 2 )(100º C - 0º C)(5 sec)
= 209 kJ
2m
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JEE-Physics
Thermal Conductivity (K)
• It's depends on nature of material.
Order of thermal conductivity Ag > Cu > Au > Al K
•
•
•
•
For Ag maximum is (410 W/mK)
For Freon minimum is 12 (0.008 W/mK)
SI UNIT : J s–1 m–1 K–1 Dimensions : M1 L1 T–3 q–1
For an ideal or perfect conductor of heat the value of K = ¥
For an ideal or perfect bad conductor or insulator the value of K = 0
For cooking the food, low specific heat and high conductivity utensils are most suitable.
Application of Thermal Conduction
• In winter, the iron chairs appear to be colder than the wooden chairs.
• Cooking utensils are made of aluminium and brass whereas their handles are made of wood.
• Ice is covered in gunny bags to prevent melting of ice.
• We feel warm in woollen clothes and fur coat.
• Two thin blankets are warmer than a single blanket of double the thickness.
• Birds often swell their feathers in winter.
• A new quilt is warmer than old one.
Thermal Resistance to Conduction
If you are interested in insulating your house from cold weather or for that matter keeping the meal hot in your
tiffin-box, you are more interested in poor heat conductors, rather than good conductors. For this reason, the
concept of thermal resistance R has been introduced.
L
KA
For a slab of cross-section A, Lateral thickness L and thermal conductivity K,
R=
In terms of R, the amount of heat flowing though a slab in steady-state (in time t)
Q (TH - TL )
=
t
R
T - TL
Q
iT = H
as thermal current iT then,
R
t
This is mathematically equivalent to OHM’s law, with temperature donning the role of electric potential. Hence
results derived from OHM’s law are also valid for thermal conduction.
More over, for a slab in steady state we have seen earlier that the thermal current iL remains same at each crosssection. This is analogous to kirchoff’s current law in electricity, which can now be very conveniently applied to
thermal conduction.
If we name
Illustration 9.
Solution
Three identical rods of length 1m each, having cross-section area
of 1cm2 each and made of Aluminium, copper and steel respectively are maintained at temperatures of 12ºC, 4ºC and 50ºC
respectively at their separate ends. Find the temperature of their
common junction.
[KCu=400 W/m-K , KAl = 200 W/m-K , Ksteel = 50 W/m-K ]
L
1
104
=
=
RAl =
KA 10-4 ´ 200 200
104
104
and Rcopper =
50
400
Let temperature of common junction = T
then from Kirchoff;s current laws, iAl + isteel + iCu = 0
51ºC
Similarly Rsteel =
Þ
T - 12 T - 51 T - 4
+
+
=0
R Al
R steel
R Cu
Þ (T – 12) 200 + (T – 50) 50 + (T – 4) 400
Þ 4(T – 12) + (T – 50) + 8 (T – 4) = 0
Þ 13T = 48 + 50 + 32 = 130 Þ T = 10°C
82
iS
12ºC
Rs
iAl
RAl
T
RCu
iCu
4ºC
Thermal Physics
Slabs in parallel and series
Slabs in series (in steady state) : Consider a composite slab consisting of two materials having different
thickness L1 and L2 different cross-sectional areas A1 and A2 and different thermal conductivities K1 and K2.
The temperature at the outer surface of the states are maintained at TH and TC, and all lateral surfaces are
covered by an adiabatic coating.
L2
L1
Q
K2
Heat reservoir
at TC
K1
Heat reservoir
at temperature TH
adiabatic coating
Let temperature at the junction be T, since steady state has been achieved thermal current through each slab will
be equal. Then thermal current through the first slab.
I=
Q TH - T
=
Þ TH - T = IR1
t
R1
Q T - TC
Þ T - TC = IR 2
and that through the second slab, i = I = t = R
2
TH - TC
TH – TL = (R1 + R2)I Þ I = R + R
1
2
Adding equation
Thus these two slabs are equivalent to a single slab of thermal resistance R 1 + R2.
If more than two slabs are joined in series and are allowed to attain steady state, then equivalent thermal
resistance is given by
R = R1 + R2 + R3 + .......
Illustration 10. The figure shows the cross-section of the outer wall of a house built in a hill-resort to keep the house
insulated from the freezing temperature of outside. The wall consists of teak wood of thickness L1
and brick of thickness (L2 = 5L1), sandwitching two layers of an unknown material with identical
thermal conductivities and thickness. The thermal conductivity of teak wood is K1 and that of brick
is (K2 = 5K). Heat conduction through the wall has reached a steady state with the temperature of
three surfaces being known.
(T1 = 25°C, T2 = 20°C and T5 = –20°C). Find the interface temperature T4 and T3.
T1
T2
L1
T3
L
T4
L
T5
L2
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JEE-Physics
Solution
L1
Let interface area be A. then thermal resistance of wood,R1 = K A and that of brick wall
1
L2
5L1
R2= K A = 5K A = R1
2
1
Let thermal resistance of the each sand witch layer = R. Then the above wall can be visualised as a
circuit
iT
R1
25ºC
R
R
20ºC
T4
T3
iT
R1
–20ºC
Thermal current through each wall is same.
Hence
25 - 20 20 - T3 T3 - T4 T4 + 20
=
=
=
R1
R
R
R1
Þ 25 – 20 = T4 + 20 Þ T4 = –15°C
also, 20 – T3 = T3 – T4 Þ
T3 =
20 + T4
= 2.5°C
2
Equivalent conductivity for Heat flow through slabs in series
Req = R1 + R2
•
Q
t
L1
L2
L1 + L2
= K A +K A
K eq A
1
2
T1
Equivalent thermal conductivity of the system is
Keq =
K1
L1 + L 2
SL i
=
L1 L 2
L
+
S i
K1 K 2
Ki
L1
Q
t
K2
T0
T2
L2
R1
equ ivalent to
R2
Slabs in parallel (in steady state):
Consider two slabs held between the same heat reservoirs, their thermal conductivities K1 and K2 and crosssectional areas A1 and A2
L
Heat reservoir
at temperature TH
SLAB 1
K1 A1
Q1
SLAB 2
K2 A2
Q2
adiabatic coating
Heat reservoir
at temperature TC
then
L
L
R1 = K A , R2 = K A
1 1
2 2
thermal current through slab 1 : I1 =
TH - TC
R1
Net heat current from the hot to cold reservoir
84
and that through slab 2 : I2 =
æ 1
1ö
I = I1 + I2 = (TH - TC ) ç
+
è R1 R 2 ÷ø
TH - TC
R2
Thermal Physics
Comparing with I =
TH - TC
1
1
1
+
, we get,
=
R eq
R eq
R1 R2
If more than two rods are joined in parallel, the equivalent thermal resistance is given by
1
1
1
1
=
+
+
+ .....
R eq R1 R 2 R 3
•
Equivalent thermal condctivity for Heat flow through slabs in parallel
1
1
1
K eq
K 1A 1 K 2 A 2
L
=
+
+
R eq
R1 R 2 , R = KA ; L (A1 + A2) = L
L
A1
Q
t
A2
Equivalent thermal conductivity
K eq =
K1
K2
T1
SK i A i
K1 A1 + K 2 A 2
=
SA i
A1 + A 2
Q
t
T2
L
R1
equivaleent to
R2
Illustration 11. Three rods of material X and three rods of material Y are connected as shown in figure. All the rods
are identical in length and cross–sectional area. If the end A is maintained at 60 0C and the junction
E at at 100C, calculate the temp. of the junctions B,C,D. The thermal conductivity of X is 0.92 CGS
units and that of Y is 0.46 CGS units.
Solution
RX µ
1
KY
0.46
1
1
RX
, RY µ
=
=
=
Þ
KX
0.92
2
KY
KX
RY
Let RX = R \ RY=2R
C
The total resistance SR = R Y + effective resistance in the bridge
X
X
0
SR = 2R +
2R ´ 4R
4
10
R & Q Dq = l ´ R
R=
= 2R +
2R + 4R
3
3
Further IBCE (2R) = IBDE(4R) and IBCE + IBDE = I Þ IBCE =
A
60 C
0
Y
B
Y
Y
2
1
I and IBDE = I
3
3
For A and B
q A - qB = 600 - qB Þ 60 - qB = 2R ´ I
For B and C
qB - qC =
For A and E
q A - qE = 60 - 10 = 50 Þ
10 C
E
D
...(i)
2
( I ´ R ) ....(ii) qC - qE = 2 ´ R ´ I
3
3
10
(
)
3 R ´ I = 50
....(iii) \ R ´ I = 15
\
æ 2ö
q A - qB - 2 ´ 15 = 30 , qB = 60 - 30 = 300C, qB - qC = çè ÷ø ´ 15 = 10
3
\
qC = 30 - 10 = 200 C Obviously, qC = qD = 200 C
Growth of Ice on Lakes
In winter atmospheric temperature falls below 0°C and water in the lake start freezing.
Let at time t thickness of ice on the surface of the lake = x and air temperature = –q° C
The temperature of water in contact with the lower surface of ice = 0°C
Let area of the lake = A
Heat escaping through ice in time dt is
dQ = KA
[0 - (-q)]
dt
x
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JEE-Physics
Due to escape of this heat increasing extra thickness of ice = dx
Mass of this extra thickness of ice is m = rV = r A.dx
dQ = mL = (r A.dx) L
air at 0° C
dx
q
rL
dt = (r A.dx) L Þ dt =
x dx
x
Kq
So time taken by ice to grow a thickness x is
\
ice
x
water
KA
z
rL x
1 rL 2
x dx =
x
Kq 0
2 Kq
So time taken by ice to grow from thickness x 1 to thickness x2 is
t=
1 rL
(x 2 – x12) and t µ (x22 – x12)
2 KT 2
Time taken to double and triple the thickness ratio t 1 : t2 : t3 :: 12 : 22 : 32
t = t2 – t1 =
So t1 : t2 : t3 :: 1 : 4 : 9
Illustration 12. One end of a brass rod 2m long and having 1 cm radius is maintained at 250°C. When a steady
state is reached, the rate of heat flow across any cross–section is 0.5 cal s–1. What is the temperature
of the other end K = 0.26 cal s–1 cm–1 °C–1.
Solution
Q
= 0.5 cal s–1; r = 1 cm \ Area A = pr2 = 3.142 × 1 cm2 = 3.142 cm2
t
L = Length of rod = 2m = 200 cm, T1 = 250°C, T2 = ?
We know
0.5 ´ 200
Q
KA (T1 - T2 )
Q
Dx
or (T1 – T2) =
=122.4°C
=
×
=
0
26
.
C -1 ´ 3142
.
t
L
t
kA
\ T2 = 250°C – 122.4°C = 127.6°C
Illustration 13. Steam at 373 K is passed through a tube of radius 10 cm and length 2 m. The thickness of the tube
is 5 mm and thermal conductivity of the material is 390 W m–1 K–1, calculate the heat lost per
second. The outside temp. is 0°C.
Solution
KA (T1 - T2 )t
L
Here, heat is lost through the cylindrical surface of the tube.
A = 2pr (radius of the tube) (length of the tube) = 2p × 0.1 × 2 = 0.4 pm2
K = 390 W m–1 K–1
T1 = 373 K, T2 = 0°C = 273 K, L = 5 mm = 0.005 m
and t = 1 s
Using the relation Q =
\Q=
390 ´ 0.4 p ´ (373 - 273) ´ 1
390 ´ 0.4p ´ 100
=
= 98 × 105 J.
0.005
0.005
Illustration 14. The thermal conductivity of brick is 1.7 W m–1 K–1, and that of cement is 2.9 W m–1 K–1. What
thickness of cement will have same insulation as the brick of thickness 20 cm.
Solution
Since Q =
KA (T1 - T2 )t
. For same insulation by the brick and cement Q, A (T 1 – T2) and t do not
L
change. Hence,
K
remain constant. If K1 and K2 be the thermal conductivities of brick and cement
L
respectively and L1 and L2 be the required thickness then
\ L2 =
86
2.9
× 20 = 34.12 cm
17
.
17
.
2.9
K1 K 2
or
=
=
L2
20 L 2
L1
Thermal Physics
Illustration 15. Two vessels of different material are identical in size and wall–thickness. They are filled with equal
quantities of ice at 0°C. If the ice melts completely, in 10 and 25 minutes respectively then compare
the coefficients of thermal conductivity of the materials of the vessels.
Solution
Let K1 and K2 be the coefficients of thermal conductivity of the materials, and t1 and t2 be the time
in which ice melts in the two vessels. Since both the vessels are identical, so A and x in both the cases
is same.
Now, Q =
b
g
K 2 A q1 - q 2 t 2
K1A (q1 - q 2 )t1
=
L
L
K1
t2
25 min
5
ÞK =t =
=
10
min
2
2
1
Illustration 16. Two plates of equal areas are placed in contact with each other. Their thickness are 2.0 cm and
5.0 cm respectively. The temperature of the external surface of the first plate is –20°C and that of the
external surface of the second plate is 20°C. What will be the temperature of the contact surface if
the plate (i) are of the same material, (ii) have thermal conductivities in the ratio 2 : 5.
Solution
Rate of flow of heat in the plates is
plate 1
2cm
K1A (q1 - q)
K 2 A (q - q 2 )
Q
=
=
L1
L2
t
5cm
plate 2
0
0
-20 C
(i)
...(i)
20 C
Here q1 = –20°C,
q2 = 20°C,
L1 = 2 cm = 0.02 m, L2 = 5 cm = 0.05 m and K1 = K2 = K
\ equation (i) becomes
b
KA -20 - q
0.02
g = KAbq - 20g
0.05
\ 5(–20–q) = 2(q – 20) Þ –100 – 5q = 2q – 40 Þ 7q = –60 Þ q = –8.6°C
(ii)
K1
2
2
or
K
=
K
=
1
K2
5
5 2
\ from equation (i)
b
2 5 K 2 A -20 - q
0.02
g = K Abq - 20g –20 – q
2
0.05
= q – 20 or –2q = 0 \q = 0°C
Illustration 17. An ice box used for keeping eatables cold has a total wall area of 1 metre 2 and a wall thickness of
5.0 cm. The thermal conductivity of the ice box is K = 0.01 joule/metre–°C. It is filled with ice at 0°C
along with eatables on a day when the temperature is 30°C. The latent heat of fusion of ice is
334 × 103 joule/kg. Calculate the amount of ice melted in one day.
Solution
dQ KA
0.01 ´ 1
dQ
dq =
=
´ 30 = 6 joule@s, So
´ 86400 = 6 ´ 86400
dt
L
0.05
dt
Q = mL (L – latent heat),
m=
Q 6 ´ 86400
=
= 1552
.
kg
L 334 ´ 103
Illustration 18. A hollow spherical ball of inner radius a and outer radius 2a is
made of a uniform material of constant thermal conductivity K.
The temperature within the ball is maintained at 2T 0 and outside
the ball it is T0. Find, (a) the rate at which heat flows out of the
ball in the steady state, (b) the temperature at r = 3a/2, where r
is radial distance from the centre of shell. Assume steady
state condition.
T0
a
2T0
K
2a
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JEE-Physics
Solution
In the steady state, the net outward thermal current is constant, and does not depend on the radial
position.
dT
C 1
æ dQ ö
2 dT
=- 1 2
Þ
÷ = -K. ( 4pr )
dt ø
dr
dr
4pK r
Thermal current, C1 = çè
At r=a, T = 2T0 and r =2a, T=T0 Þ T =
3.2
T=
C1
+ C2
4pkr
2a
dQ
T0 (a)
= 8paKT0 (b) T ( r = 3a / 2) = 4T0 / 3
r
dt
Convections
Convection requires a medium and is the process in which heat is transferred
from one place to other by actual movement of heated substance (usually
fluid).The type of convection which results from difference in densities is called
natural convection (for Illustration, a fluid in a container heated through its
bottom). However, if a heated fluid is forced to move by a blower, fan or pump,
the process is called forced convection. The rate of heat convection from an
object is proportional to the temperature difference (Dq) between the object and
é dQ ù
convective fluid and the area of contact A, i.e., ê dt ú
=
ë
û convection
A
Flame
hA Dq where, h represents a constant of proportionality called convection
coefficient and depends on the properties of fluid such as density, viscosity,
specific heat and thermal conductivity, etc.
Phenomena Based on convection
(i)
Land and sea breezes
The heat from the Sun is absorbed more rapidly by land than by sea–water. Moreover, the specific heat
of land is low as compared to that of sea–water. Consequently, the rise in temperature of land is higher
as compared to that of sea–water. To sum–up, land is hotter than the sea during day time. As a result
of this, the colder air over the sea blows towards the land. This is called sea–breeze.
At night, air blows from land towards sea. This is called land breeze.
88
(ii)
Formation of trade winds
The surface of Earth near the equator gets heated strongly. So, the air in contact with the surface of
Earth at the expands and rises upwards. As a result of this, a low pressure is created at the equator.
At the poles, the air in the upper atmosphere gets cooled and comes down. So, a high pressure is created
at the poles. Due to difference of pressures at the poles and equator, the air at the poles moves towards
the equator, rises up, moves the poles and so on. In this way, a wind is formed in the atmosphere.
The rotation of the Earth also affects the motion of the wind. Due to anti–clockwise rotation of Earth
the warm wind blowing from equator to north drifts towards east. The steady wind blowing from
north–Earth to equator, near the surface of Earth, is called trade wind.
(iii)
Monsoons
In summer, the peninsular mass of central Asia becomes more strongly heated than the water of the
Indian Ocean. This is due to the fact that the specific heat of water is much higher than that of the soil
and rocks. Hot air from the heated land mass rises up and moves towards the Indian ocean. Air filled
with moisture flows over the Indian ocean on the south towards heated land mass. When obstructed by
mountains, the moist air rushes upwards to great height. In the process, it gets cooled. Consequently, the
moisture condenses and falls as rain.
(iv)
Ventilation
Ventilator of exhaust fan in a room help of remove impure and warm air from a room. The fresh air
from outside blows into the room. This is all due to the convection current set up in the room.
(v)
To regulate temperature in the human body
Heat transfer in the human body involves a combination of mechanisms. These together maintain a
remarkably uniform temperature in the human body inspite of large changes in environmental conditions.
Thermal Physics
The chief internal mechanism is forced convection. The heart serves as the pump and the blood as the
circulating fluid.
Some important points
l
Natural convection takes place from bottom to top while forced convection in any direction.
l
In case of natural convection, convection currents move warm air upwards and cool air downwards.
This is why heating is done from base, while cooling from the top.
l
Natural convection is not possible in a gravity free region such as a freely falling lift or an orbiting
satellite.
l
Natural convection plays an important role in ventilation, in changing climate and weather and in
forming land and sea breezes and trade winds.
l
The forced convection of blood in our body by a pump (heart) helps in keeping the temperature of body
constant.
Illustration 19. Water in a closed tube is heated with one arm vertically placed above the lamp. In
what direction water will begin the circulate along the tube ?
Solution
On heating the liquid at A will become lighter and will rise up. This will push the
liquid in the tube upwards and so the liquid in the tube will move clockwise
B
A
i.e. form B to A.
l
l
3.3
For heat propagation via convection, temperature gradient exists in vertical direction and not in horizontal
direction.
Most of heat transfer that is taking place on Earth is by convection, the contribution due to conduction and
radiation is very small.
Thermal Radiation
The process of the transfer of heat from one place to another place without heating the intervening medium is
called radiation. When a body is heated and placed in vacuum, it loses heat even when there is no medium
surrounding it. The heat can not go out from the body by the process of conduction or convection since both
of these process require the presence of a material medium between source and surrounding objects. The
process by which heat is lost in this case is called radiation. This does not require the presence of any material
medium.
It is by radiation that the heat from the Sun reaches the Earth. Radiation has the following properties:
(a)
Radiant energy travels in straight lines and when some object is placed in the path, it's shadow is formed
at the detector.
(b)
It is reflected and refracted or can be made to interfere. The reflection or refraction are exactly as in case
of light.
(c)
It can travel through vacuum.
(d)
Intensity of radiation follows the law of inverse square.
(e)
Thermal radiation can be polarised in the same way as light by transmission through a nicol prism.
All these and many other properties establish that heat radiation has nearly all the properties possessed by light
and these are also electromagnetic waves with the only difference of wavelength or frequency. The wavelength
of heat radiation is larger than that of visible light.
Types of thermal Radiation :– Two types of thermal radiation.
Plane Radiation
Diffuse Radiation
Radiations which are incident on a
surface at certain angle
Incident on the surface at all angles
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JEE-Physics
When radiation passes through any medium then radiations slightly absorbed by medium according to its
absorptive power so temperature of medium slightly increases.
In order to obtain a spectrum of radiation, a special prism used like KCl prism, Rock salt prism Flourspar prism.
Normal glass prism or Quartz prism can not be used (because it absorbed some radiation).
Radiation intensity measured with a specific device named as Bolometer.
Heat radiation are always obtained in infra–red region of electromagnetic wave spectrum so they are called Infra
red rays.
Thermal radiations is incident on a surface, it exerts pressure on the surface, which is known as Radiation
Pressure.
Basic Fundamental Definitions
Energy Density (u) : The radiation energy of whole wavelength (0 to ¥) present in unit volume at any point in
space is defined as energy density. S I UNIT : J/m3
¥
Spectral energy density (ul ) : Energy density per unit spectral region. u = ò ul dl
SI UNIT : J/m3 Å
0
Absorptive power or absorptive coefficient 'a' : The ratio of amount of radiation absorbed by a surface
(Qa) to the amount of radiation incident (Q) upon it is defined as the coefficient of absorption a =
Qa
.It is
Q
unitless
Spectral absorptive power (al ) : a l =
z
Qa l
: Also called monochromatic absorptive coefficient
Ql
¥
At a given wavelength a = a l dl . For ideal black body al and a = 1, a and al are unitless
0
Emissive power (e) : The amount of heat radiation emitted by unit area of the surface in one second at a
particular temperature. SI UNIT : J/m2s
Spectral Emmisive power (el) : The amount of heat radiation emitted by unit area of the body in one second
¥
in unit spectral region at a given wavelength. Emissive power or total emissive power e = ò el dl
0
SI UNIT : W/m2 Å
Emissivity (e)
Absolute emissivity or emissivity : Radiation energy given out by a unit surface area of a body in unit time
corresponding to unit temperature difference w.r.t. the surroundings is called Emissivity.
S I UNIT : W/m2 °K
emitted radiation by gray body
Q GB
e
= GB =
emitted radiation by ideal black body
Q IBB
E IBB
GB = gray or general body, IBB = Ideal black body
(i) No unit
(ii) For ideal black body er = 1
(iii) range 0 < er < 1
Relative emissivity (er) : er =
Spectral, emissive, absorptive and transmittive power of a given body surface
Due to incident radiations on the surface of a body following phenomena occur by which the radiation is divided
into three parts. (a) Reflection (b) Absorption (c) Transmission
From energy conservation
Q = Qr + Qa + Qt Þ
Qr Qa Qt
+
+
=1Þ r + a + t = 1
Q
Q
Q
Reflective Coefficient r =
90
Qr
Qa
, Absorptive Coefficient a =
,
Q
Q
amount of incident
radiation Q
amount of reflected
radiation Q r
amount of
absorbed
radiation Qa
amount of transmitted
radiation Q t
Thermal Physics
Transmittive Coefficient t =
r = 1 and
a = 1 and
t = 1 and
Qt
Q
a=0,
t = 0 Þ Perfect reflector
r = 0, t = 0 Þ Ideal absorber (ideal black body)
a = 0,r = 0 Þ Perfect transmitter (daithermanons)
é Qr
ù
éQ
ù
´ 100ú % , Absorption power (a) = ê a ´ 100ú %
Reflection power (r) = ê
ëQ
û
ëQ
û
é Qt
ù
´ 100ú %
Transmission power (t) = ê
Q
ë
û
Illustration 20. Total radiations incident on body = 400 J, 20% radiation reflected and 120 J absorbs. Then find out
% of transmittive power
Solution
Q = Qt + Qr + Qa Þ 400 = 80 + 120 + Qt Þ Qt = 200.
So % of transmittive power is 50%
Ideal Black Body
•
•
•
•
•
•
•
•
For a body surface which absorbs all incident thermal radiations at low temperature irrespective of their wave
length and emitted out all these absorbed radiations at high temperature assumed to be an ideal black body
surface.
The identical parameters of an ideal black body is given by
a = al = 1 and r = 0 = t, er = 1
The nature of emitted radiations from surface of ideal black body only depends
on its temperature
The radiations emitted from surface of ideal black body called as either full or
Ferry's ideal black body
white radiations.
At any temperature the spectral energy distribution curve for surface of an ideal
black body is always continuous and according to this concept if the spectrum of
a heat source obtained to be continuous then it must be placed in group of ideal
black body like kerosene lamp; oil lamp Heating filament etc.
There are two experimentally ideal black body
(a) Ferry's ideal black body(b) Wien's ideal black body.
At low temperature surface of ideal black body is a perfect absorber and at a high temperature it proves to be a
good emitter.
An ideal black body need not be black colour (eg. Sun)
Prevost's theory of heat energy exchange
According to Prevost at every possible temperature (Not absolute temperature) there is a continuous heat energy
exchange between a body and its surrounding and this exchange carry on for infinite time.
The relation between temperature difference of body with its surrounding decides whether the body experience
cooling effect or heating effect.
When a cold body is placed in the hot surrounding : The body radiates less energy and absorbs more
energy from the surrounding, therefore the temperature of body increases.
When a hot body placed in cooler surrounding : The body radiates more energy and absorb less energy
from the surroundings. Therefore temperature of body decreases.
When the temperature of a body is equal to the temperature of the surrounding
The energy radiated per unit time by the body is equal to the energy absorbed per unit time by the body, therefore
its temperature remains constant.
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JEE-Physics
l
At absolute zero temperature (0 kelvin) all atoms of a given substance remains in ground state, so, at this
temperature emission of radiation from any substance is impossible, so Prevost's heat energy exchange theory
does not applied at this temperature, so it is called limited temperature of prevosts theory.
l
With the help of Prevost's theory rate of cooling of any body w.r.t. its surroundings can be worked out (applied
to Stefen Boltzman law, Newton's law of cooling.)
Kirchhoff's Law
At a given temperature for all bodies the ratio of their spectral emissive power (e l) to spectral absorptive power
(al) is constant and this constant is equal to spectral emissive power (El) of the ideal black body at same
temperature
é el ù
é el ù
ê ú = ê ú = constant el µ al
a
ë l û1 ë a l û2
el
= E l = constant
al
Good absorbers are good emitters and bad absorbers are bad emitters
l
For a constant temperature the spectral emmisive power of an ideal black body is a constant parameter
l
The practical confirmation of Kirchhoff's law carried out by Rishi apparatus and the main base of this apparatus
is a Lessilie container.
l
The main conclusion predicted from Kirchhof's law can be expressed as
Good absorber
ƒ Good emitter
ƒ
Bad absorber
(at Low temperature)
Bad emitter
(at high temperature)
Applications of Kirchhoff Law
Fraunhoffer's lines
Fraunhoffer lines are dark lines in the spectrum of the Sun. When white light
emitted from the central core of the Sun (Photosphere) passes. Through its
atmosphere (Chromosphere) radiations of those wavelengths will be absorbed
by the gases present, resulting in dark lines in the spectrum of Sun. At the time
of total solar eclipse direct light rays emitted from photosphere cannot reach
on the Earth and only rays from chromosphere are able to reach on the
Earth surface. At that time we observe bright fraunhoffer lines.
•
K
Ca
Na
vapour
state
chromosphere
SUN
photo
sphere
7
10 K
6000K
In deserts days are hot and nights cold
Sand is rough and black, so it is a good absorber and hence in deserts, days (When radiation from Sun is
incident on sand) will be very hot. Now in accordance with Kirchhoff's Law, good absorber is a good emitter.
So nights (when send emits radiation) will be cold.
•
Stefan's Law
The amount of radiation emitted per second per unit area by a black body is directly proportional to the fourth
power of its absolute temperature.
Amount of radiation emitted E µ T4 where T = temperature of ideal black body (in K)
E = s T4
This law is true for only ideal black body
SI Unit : E = watt/m2
s = Stefen's constant = 5.67 x10–8 watt /m2 K4
Dimensions of s : M1 L0 T–3 q–4
Total radiation energy emitted out by surface of area A in time t :
Ideal black body
92
QIBB = s A T4 t
and for any other body QGB = ersA T4 t
Thermal Physics
Rate of emission of radiation
When Temperature of surrounding T0 (Let T0 < T)
Rate of emission of radiation from ideal black body surface E1 = s T4
Rate of emission of radiation from surrounding E2 = sT04
Net rate of loss of radiation from ideal black body surface is E = E1 – E2 = sT4– s T04 = s ( T4 – T04 )
Net loss of radiation energy from entire surface area in time t is Q BB = sA ( T4 – T04 ) t
For any other body QGB = er As ( T4 – T04 ) t
If in time dt the net heat energy loss for ideal black body is dQ and because of this its temperature falls by dq
dQ
= s A(T 4 - T04 )
dt
It is also equal to emitted power or radiation emitted per second
Rate of loss of heat
RH =
Rate of fall in temperature (Rate of cooling) RF =
dq
sA 4
(T - T04 )
=
dt ms J
dq ù
é dQ
êQ dt = m s J dt ú
ë
û
Note :
(i)
If all of T, T0, m, s, V, r, are same for different shape body then RF and RH will be maximum in the flat
surface.
(ii)
If a solid and hollow sphere are taken with all the parameters same then hollow will cool down at fast
rate.
(iii)
Rate of temperature fall , R F µ
1 dq
µ
so dt µ s. If condition in specific heat is Þ s1 > s2 > s3
s
dt
If all cooled same temperature i.e. temperature fall is also identical for all then required time
t µ s \ t1 > t2 > t3
When a body cools by radiation the cooling depends on :
(i)
Nature of radiating surface : greater the emissivity (er), faster will be the cooling.
(ii)
Area of radiating surface : greater the area of radiating surface, faster will be the cooling.
(iii)
Mass of radiating body : greater the mass of radiating body slower will be the cooling.
(iv)
Specific heat of radiating body : greater the specific heat of radiating body slower will be the cooling.
(v)
Temperature of radiating body : greater the temperature of radiating body faster will be the cooling.
Illustration 21. The operating temperature of a tungesten filament in an incandescent lamp is 2000 K and its
emissivity is 0.3. Find the surface area of the filament of a 25 watt lamp. Stefan's constant
s = 5.67 × 10–8 Wm–2 K–4
Solution
Q Rate of emission = wattage of the lamp
4
\ W = AesT Þ A =
25
W
2
4 = 0.3 ´ 5.67 ´ 10-8 ´ (200) 4 = 0.918 m
esT
Newton's Law of Cooling
Rate of loss of heat
FG dQ IJ is directly proportional to excess of temperature of the body over that of surrounding.
H dt K
[(when (q – q0) > 35°C]
dQ
µ (q – q0)
dt
Þ
dQ
dq
= ms
dt
dt
q = temperature of body [ in °C], qo = temperature of surrounding, q –q0 = excess of temperature ( q > q0 )
If the temperature of body decrease dq in time dt then rate of fall of temperature –
dq
µ ( q - q0 )
dt
Where negative sign indictates that the rate of cooling is decreasing with time.
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JEE-Physics
Excess of temperature
If the temperature of body decreases from q1 to q2 and temperature of surroundings is q0 then average excess of
é q1 + q2
ù
éq + q
ù
éq -q ù
- q0 ú Þ ê 1 2 ú = – K ê 1 2 - q0 ú
temperature = ê
t
2
2
ë
û
ë
û
ë
û
Illustration 22. If a liquid takes 30 seconds in cooling of 80°C to 70°C and 70 seconds in cooling 60°C to 50°C, then
find the room temperature.
Solution
q1 – q 2
t
=K
In first case,
FG q
H
1
+ q2
– q0
2
IJ
K
æ 80 + 70
ö
80 – 70
– q0 ÷
= K çè
ø
2
30
In second case ,
FG
H
60 + 50
60 – 50
– q0
=K
2
70
Equation (i) divide by equation (ii)
Þ 385 – 7q0 = 225 – 3q0 Þ q0 =
1
= K (75 – q0)
3
IJ
K
...(i)
1
= K (55 – q0) ... (ii)
7
(75 – q0 )
7
= (55 – q )
3
0
160
= 40°C
4
Limitations of Newton's Law
• Temperature difference should not exceed 35° C, (q – q0) >/ 35° C
• Loss of heat should only be by radiation.
• This law is an extended form of Stefan–Boltzman's law.
For Heating, Newton's law of heating
q1 - q2
q + q2 ù
é
= + H ê q0 - 1
where H heating constant.
t
2 úû
ë
Derivation of Newton's law from Steafen's Boltzman law
ì T - T0 = DT ü
dq
sA
4
(T 4 - T0 ) í
=
ý
dt m s J
î T = T0 + DT þ
sA
dq
é(T0 + DT)4 - T04 ùû
=
dt m s J ë
If x <<< 1 then (1 + x)n = 1 + nx
ù
ù
ù
dq
sA é 4
DT 4
sA 4 é
DT 4
sA 4 é
DT
) - T04 ú =
T0 ê(1 +
) - 1ú =
T0 ê1 + 4
- 1ú
=
ê T0 (1 +
dt m s J ë
T0
T0
T0
û msJ
ë
û msJ
ë
û
d q é sA 3 ù
dq
= 4
T0 DT Þ
= K DT
dt êë m s J úû
dt
Newton's law of cooling
dq
dt
constant K =
4s A T03
msJ
µ DT (for small temperature difference)
Application of Newton's Law of Cooling
To find out specific heat of a given liquid
If for the two given liquids their volume, radiating surface area, nature of surface, initial temperature are allowed
to cool down in a common environments then rate of loss of heat of these liquids are equal .
•
94
Thermal Physics
stirrer
q1 – q2
t1
water
ms
air
air
stirrer
q1 – q2
t2
liquid
m' s'
water out
air
water in
fall of temperature from q1 ® q2 in time t1 for water and t2 for liquid
é q - q2 ù
é q - q2 ù
é dQ ù
é dQ ù
=ê
= (m ' s ' + w) ê 1
(m s+w) ê 1
Q ê dt ú
\
ú
ú
ú
ë
û Water ë dt û Liquid
ë t2 û
ë t1 û
Þ
ms+w
m's'+ w
=
t1
t2
Cooling curve :
temperature (q)
where w = water equivalent of calorimeter.
–x
cooling curve e
q1
water
q2
t1
liquid
t2
time (t)
Illustration 23. When a calorimeter contains 40g of water at 50°C, then the temperature falls to 45°C in 10 minutes.
The same calorimeter contains 100g of water at 50°C, it takes 20 minutes for the temperature to
become 45°C. Find the water equivalent of the calorimeter.
Solution
m1s1 + W
m s +W
= 2 2
where W is the water equivalent
t1
t2
40 ´ 1 + W
100 ´ 1 + W
=
Þ 80 + 2W = 100 + W Þ W = 20 g
10
20
Þ
Spectral Energy distribution curve of Black Body radiations
E l3
T3
Mathematically given by : Plank
T3 > T2 > T 1
E n3
l m3< l m2< l m 1
E l2
different
wavelength
and emitted
radition
T2
E l1
T1
l m3 l m2 l m1
different
frequency
and emitted
radiation
T3 > T2 > T 1
n3 > n 2 > n1
E n2
E n or I l
intens ity
Spectral radiation
J sec-m2
Practically given by : Lumers and Pringshem
E n1
T3
T2
T1
l
n1
n 2 n3
n
95
JEE-Physics
height of peak µ T5
El
area µ T4
Am µ T
1
T
(i)
lm µ
(ii)
E lm µ T 5
¥
(iii)
4
Area ò E l dl = E = sT
0
lm
dl
l
spectral energy distribution curve (E l–l)
A1 é T1 ù
=ê ú
A2 ë T2 û
4
l
Spectral energy distribution curves are continuous. At any temperature in between possible wavelength
(0 – ¥) radiation emitted but for different wavelength quantity of radiations are different.
l
As the wave length increases, the amount of radiation emitted first increase, becomes maximum and then
decreases.
l
At a particular temperature the area enclosed between the spectral energy curve shows the spectral emissive
¥
power of the body. Area = ò E l dl = E = sT 4
0
Wein's Displacement Law
1ù
é
The wavelength corresponding to maximum emission of radiation decrease with increasing temperature êl m µ ú .
Tû
ë
This is known as Wein's displacement law. lmT = b where b Wein's constant
= 2.89 x 10–3 mK.
Dimensions of b : = M0 L1 T0 q1
Relation between frequency and temperature n m =
c
T
b
Illustration 24. The temperature of furnace is 2000°C, in its spectrum the maximum intensity is obtained at about
4000Å, If the maximum intensity is at 2000Å calculate the temperature of the furnace in °C.
Solution
By using lmT = b, 4000 (2000+273) = 2000(T) Þ T = 4546K
The temperature of furnace = 4546 – 273 = 4273 °C
Solar Constant 'S'
The Sun emits radiant energy continuously in space of which an in significant part reaches the Earth. The solar
radiant energy received per unit area per unit time by a black surface held at right angles to the Sun's rays and
placed at the mean distance of the Earth (in the absence of atmosphere) is called solar constant.
The solar constant S is taken to be 1340 watts/m2 or 1.937 Cal/cm2–minute
•
Temperature of the Sun
Let R be the radius of the Sun and 'd' be the radius of Earth's orbit around the
Sun. Let E be the energy emitted by the Sun per second per unit area. The total
energy emitted by the Sun in one second = E.A = E × 4pR2. (This energy is
falling on a sphere of radius equal to the radius of the Earth's orbit around the
Sun i.e., on a sphere of surface area 4pd2)
So, The energy falling per unit area of Earth =
8
R = 7× 10 m ,
Solar constant
96
4 pR 2 ´ E
2
=
E R2
S=
E R2
d2
R
T
2
4 pd
d
–8
–2 –4
d = 1.5 × 10 m, s = 5.7 × 10 W m K
11
d
A
A'
Thermal Physics
E = sT4
By Stefan's Law
S=
s T 4 R2
d2
1
Þ
1
é S ´ d 2 ù 4 é 1340 ´ (1.5 ´ 1011 )2 ù 4
=ê
T= ê
= 5732 K
2ú
8 2ú
-8
ës ´ R û
ë 5.7 ´ 10 ´ (7 ´ 10 ) û
1.
One end of a metal rod is kept in a furnace. In steady state, the temperature of the rod
(A) increases
(B) decreases
(C) remains constant
(D) is nonuniform
2.
In the figure shown AB is a rod of length 30 cm and area of crosssection 1.0 cm2 and thermal conductivity 336 S. I. units. The ends A
& B are maintained at temperatures 20°C and 40°C respectively. A
point C of this rod is connected to a box D, containing ice at 0°C,
through a highly conducting wire of negligible heat capacity. The rate
at which ice melts in the box is: [Assume latent heat of fusion for ice
Lf = 80 cal/gm ]
(A) 84 mg/s
(B) 84 g/s
(C) 20 mg/s
(D) 40 mg/s
3.
A wall is made of two layers A & B of the same thickness but different materials. The thermal conductivity
of A is twice that of B. In steady state, the temperature difference across the wall is 36 0C. The temperature
difference across the layer A is :
(A) 6°C
(B) 12°C
(C) 18°C
(D) 24°C
4.
All the rods have same conductance ‘K’ and same area of cross section ‘A’.
If ends A and B are maintained at temperature 2T 0 and T0 respectively then
which of the following is/are correct :
(A) Rate of heat flow through ABC, AOC and ADC is same
(B) Rate of heat flow through BO and OD is not same
3KAT0
(C) Total Rate of heat flow from A to C is
2a
(D) Temperature at junctions B, O and D are same
5.
A rod of length l and cross section area A has a variable thermal conductivity given by k = a T, where a is a
positive constant and T is temperature in kelvin. Two ends of the rod are maintained at temperatures T1 and T2
(T1 > T2). Heat current flowing through the rod will be
(A)
6.
A a (T12 - T22 )
l
(B)
Aa(T12 + T22 )
l
(C)
Aa(T12 + T22 )
3 l
(D)
Aa(T12 - T22 )
2 l
The three rods shown in figure have identical geometrical dimensions. Heat flows from the hot end at a rate of
40 W in the arrangement (a). Find the rates of heat flow when the rods are joined as in arrangement (b) and in
(c). Thermal conductivities of aluminium and copper are 200 W/m–°C and 400 W/m–°C respectively.
97
JEE-Physics
4.0 HEAT
When a hot body is put in contact with a cold one, the former gets colder and the latter warmer. From this
observation it is natural to conclude that a certain quantity of heat has passed from the hot body to the cold
one. Heat is a form of energy.
Heat is felt by its effects. Some of the effects of heat are :
(a) Change in the degree of hotness
(b) Expansion in length, surface area and volume
(c) Change in state of a substance
(d) Change in the resistance of a conductor
(e) Thermo e.m.f. effect
SI UNIT : J (joule) Also measured in the unit calorie.
•
Calorie
It is defined as the amount of heat required to raise the temperature of 1 g water by 1°C.
•
International calorie
International calorie is the amount of heat required to raise the temperature of 1g water from
14.5 °C to 15.5 °C rise of temperature.
•
Kilo Calorie
Kilo calorie is defined as the amount of heat required to raise the temperature of 1 kg water from 14.5 °C to
15.5 °C. (1 kcal = 1000 calorie).
•
British thermal unit (B. T. U.)
It is the amount of heat required to raise the temperature of one pound water by 1°F. (1 B.T.U. = 252 calorie).
4.1
Mechanical Equivalent of Heat
According to Joule, work may be converted into heat and vice–versa. The ratio of work done to heat produced
W
= constant (J) Þ W = J H
H
W must be in joule, irrespective of nature of energy or work and H must be in calorie.
J is called mechanical equivalent of heat. It is not a physical quantity but simply a conversion factor.
It converts unit of work into that of heat and vice–versa.
J = 4.18 joule/cal or 4.18 × 10³ joule per kilo–cal. For rough calculations we take J = 4.2 joule/cal
is always constant.
•
Specific Heat (s or c )
It is the amount of energy required to raise the temperature of unit mass of that substance by 1°C (or 1K) is
called specific heat. It is represented by s or c.
If the temperature of a substance of mass m changes from T to T + dT when it exchanges an amount of heat
1 dQ
m dT
The specific heat depends on the pressure, volume and temperature of the substance.
For liquids and solids, specific heat measurements are most often made at a constant pressure as functions of
temperature, because constant pressure is quite easy to produce experimentally.
SI UNIT : joule/kg–K
CGS UNIT : cal/g –°C
Specific heat of water : cwater = 1 cal/g–°C = 1 cal/g–K = 1 kcal/kg–K = 4200 joule/kg–K
When a substance does not undergo a change of state (i.e., liquid
The temperature dependence of the
remains liquid or solid remains solid), then the amount of heat required
specific heat of water at 1 atm
to raise the temperature of mass m of thesubstance by an amount Dq
1.008
is Q = msDq.
The temperature dependence of the specific heat of water at 1
1.004
atmospheric pressure is shown in figure. Its variation is less than 1%
1.000
over the interval from 0 to 100°C. Such a small variation is typical
for most solids and liquids, so their specific heats can generally be
taken to be constant over fairly large temperature ranges.
0
20
40
60
80
Temperature (°C)
• There are many processes possible to give heat to a gas.
A specific heat can be associated to each such process which depends on the nature of process.
• Value of specific heats can vary from zero (0) to infinity.
Specific heat (cal/g °C )
dQ with its surroundings then its specific heat is c =
98
Thermal Physics
• Generally two types of specific heat are mentioned for a gas –
(a) Specific heat at constant volume (Cv) (b) Specific heat at constant pressure (CP)
• These specific heats can be molar or gram.
•
Molar Heat Capacity
The amount of energy needed to raise the temperature of one mole of a substance by 1°C (or 1K) is called
molar heat capacity. The molar heat capacity is the product of molecular weight and specific heat i.e.,
Molar heat capacity C = Molecular weight (M) × Specific heat( c) Þ C =
1 æ dQ ö
ç
÷
µ è dT ø
If the molecular mass of the substance is M and the mass of the substance is m then number of moles of the
substance µ =
m
M æ dQ ö
ÞC=
ç
÷
m è dT ø
M
SI UNIT : J/mol–K
•
Thermal Capacity
The quantity of heat required to raise the temperature of the whole of that substance through 1°C is called
thermal capacity. The thermal capacity of mass m of the whole of substance of specific heat s is = ms
Thermal capacity = mass × specific heat
Thermal capacity depends on property of material of the body and mass of the body.
SI UNIT : cal/°C or cal/K,
Dimensions : ML2 T–2K–1
4.2
Water Equivalent of a Body
As the specific heat of water is unity so the thermal capacity of a body (ms) represents its water equivalent also.
Mass of water having the same thermal capacity as the body is called the water equivalent of the body
The water equivalent of a body is the amount of water that absorbs or gives out the same amount of heat as
is done by the body when heated or cooled through 1°C.
Water equivalent= mass of body × specific heat of the material Þ (w = ms).
•
Latent Heat or Hidden Heat
When state of a body changes, change of state takes place at constant temperature [melting point or boiling
point] and heat released or absorbed is Q = mL where L is latent heat. Heat is absorbed if solid converts into
liquid (at melting point) or liquid converts into vapours (at boiling point) and heat is released if liquid converts
into solid or vapours converts into liquid.
Latent heat of fusion
It is the quantity of heat (in kilocalories) required to change its 1 kg mass from solid to liquid state at its melting
point. Latent heat of fusion for ice : 80 kcal/kg = 80 cal /g.
Latent heat of vaporization
The quantity of heat required to change its 1 kg mass from liquid to vapour state at its boiling point.
Latent heat of vaporisation for water : 536 kcal/kg = 536 cal/g
Change of State
•
Melting
Conversion of solid into liquid state at constant temperature is known as melting.
•
Boiling
Evaporation within the whole mass of the liquid is called boiling. Boiling takes place at a constant
temperature known as boiling point. A liquid boils when the saturated vapour pressure on its surface is
equal to atmospheric pressure. Boiling point reduces on decreasing pressure.
•
Evaporation
Conversion of liquid into vapours at all temperatures is called evaporation. It is a surface phenomenon.
Greater the temperature, faster is the evaporation. Smaller the boiling point of liquid, more rapid is the
evaporation. Smaller the humidity, more is the evaporation. Evaporation increases on decreasing pressure
that is why evaporation is faster in vacuum.
99
JEE-Physics
•
Heat of evaporation
Heat required to change unit mass of liquid into vapour at a given temperature is called heat of evaporation
at that temperature.
•
Sublimation
Direct conversion of solid in to vapour state is called sublimation.
•
Heat of sublimation
Heat required to change unit mass of solid directly into vapours at a given temperature is called heat of
sublimation at that temperature.
Camphor and ammonium chloride sublimates on heating in normal conditions.
A block of ice sublimates into vapours on the surface of moon because of very–very low pressure on its
surface
•
•
4.3
•
Condensation
The process of conversion from gaseous or vapour state to liquid state is known as condensation .
These materials again get converted to vapour or gaseous state on heating.
•
Hoar frost
Direct conversion of vapours into solid is called hoar frost. This process is just reverse of the process of
sublimation.
Ex. : Formation of snow by freezing of clouds.
•
Regelation
Regelation is the melting of ice caused by pressure and its resolidification when the pressure is removed.
Ice shrinks when it melts, and if pressure is applied, deliberately promoting shrinkage, it is found that
melting is thereby assisted. In other words, melting of cold ice is ordinarily effected by raising the temperature,
but if pressure is present to help with the shrinkage the temperature need not be raised so much.
Ice heals up after being cut through by the wire. Melting takes
place
under the wire because pressure lowers the melting temperature.
Refreezing (regelation) occurs above the wire when the water
escapes to normal pressure again.
Increase of pressure lowers the melting (or freezing) point of
water. Conversely, if a substance expands on melting, the melting
point is raised by pressure.
Phase of a Substance
The phase of a substance is defined as its form which is homogeneous, physically distinct and mechanically
separable from the other forms of that substance.
Phase diagram
•
A phase diagram is a graph in which pressure (P) is represented along the y–axis and temperature (T) is
represented along the x–axis.
•
Characteristics of Phase diagram
(i)
Different phases of a substances can be shown on a phase diagram.
(ii)
A region on the phase diagram represents a single phase of the substance, a curve represents equilibrium
between two phases and a common point represents equilibrium between three phases.
(iii)
A phase diagram helps to determine the condition under which the different phases are in equilibrium.
(iv)
A phase diagram is useful for finding a convenient way in which a desired change of phase can be
produced.
4.4
Phase Diagram for Water
The phase diagram for water consists of three curves AB, AC and AD meeting each other at the point A, these
curves divide the phase diagram into three regions.
100
Thermal Physics
B
C
Normal melting point
Normal boiling point
101
Liquid
P(kPa)
fusion curve
or ice line
vaporisation curve
or steam line
Solid
A
0.61
Triple point
Vapour
sublimation curve
or hoar frost line
D
0.00
0.01
tc (°C)
100.00
Region to the left of the curve AB and above the curve AD represents the solid phase of water ( ice). The region
to the right of the curve AB and above the curve AC represents the liquid phase of water. The region below the
curves AC and AD represents the gaseous phase of water (i.e. water vapour). A curve on the phase diagram
represents the boundary between two phases of the substance.
Along any curve the two phases can coexist in equilibrium
•
Along curve AB, ice and water can remain in equilibrium.This curve is called fusion curve or ice line. This curve
shows that the melting point of ice decreases with increase in pressure.
•
Along the curve AC, water and water vapour can remain in equilibrium. The curve is called vaporisation curve
or steam line. The curve shows that the boiling point of water increases with increase in pressure.
•
Along the curve AD, ice and water vapour can remain in equilibrium.
This curve is called sublimation curve or hoar frost line.
•
Triple point of water
The three curves in the phase diagram of water meet at a single point A, which is called the triple point of water.
The triple point of water represents the co–existance of all the three phases of water ice water and water vapour
in equilibrium. The pressure corresponding to triple point of water is 6.03 × 10 –3 atmosphere or 4.58 mm of
Hg and temperature corresponding to it is 273.16K.
•
Significance of triple point of water
Triple point of water represents a unique condition and it is used to define the absolute temperature. While
making Kelvin's absolute scale, upper fixed point is 273.16 K and lower fixed point is 0 K. One kelvin of
1
temperature is fraction
of the temperature of triple point of water..
273.16
Heating Curve
If to a given mass (m) of a solid, heat is supplied at constant rate and a graph is plotted between temperature
and time as shown in figure is called heating curve.
•
In the region OA
Rate of heat supply P is constant and temperature of solid is changing with time
So, Q = mcS DTÞ P Dt = mcS DT [Q Q = P Dt] Q
heat of solid cS µ
DT
= The slope of temperature–time curve so specific
Dt
1
specific heat (or thermal capacity) is inversely proportional to the slope of
slope of line OA
temperature–time curve.
101
JEE-Physics
boiling
point
(heat is suplied at constant rate )
C
temprature
T1
melting
point
A
melting
he
of a tin
so g
lid
T2
O
a
B
he
g
in
at
of
liq
boiling
of
ting
hea g
D
ga s
d
ui
dy dq
slope = dx = dQ
b
mL vapourisation
dQ = ms dq
mL fusion
t1
t2
a > b >g
t3
tana > tanb > tang
t4
time
( He a ting ca pa c ity)vapour > ( He a ting c a pa c ity ) liquid > ( He a ting c a pa c ity )solid
•
In the region AB
Temperature is constant, so it represents change of state, i.e., melting of solid with melting point T 1. At point
A melting starts and at point B all solid is converted into liquid. So between A and B substance is partly solid
and partly liquid. If LF is the latent heat of fusion then
Q = mLF Þ L F =
P(t2 - t1 )
[as Q = P(t2 – t1] Þ LF µ length of line AB
m
i.e., Latent heat of fusion is proportional to the length of line of zero slope.
[In this region specific heat µ
•
1
= ¥]
tan 0°
In the region BC
Temperature of liquid increases so specific heat (or thermal capacity) of liquid will be inversely proportional to
the slope of line BC, c L µ
•
1
slope of line BC
In the region CD
Temperature in constant, so it represents change of state, i.e., liquid is boiling with boiling point T2. At C all
substance is in liquid state while at D is vapour state and between C and D partly liquid and partly gas. The
length of line CD is proportional to latent heat of vaporisation, i.e., LV µ Length of line CD.
1
=¥]
tan 0°
The line DE represents gaseous state of substance with its temperature increasing linearly with time. The
reciprocal of slope of line will be proportional to specific heat or thermal capacity of substance in vapour state.
[In this region specific heat µ
4.5
Law of Mixtures
When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat
will be transferred from body at higher temperature to a body at lower temperature till both acquire same
temperature. The body at higher temperature released heat while body at lower temperature absorbs it, so that
Heat lost = Heat gained. Principle of calorimetry represents the law of conservation of heat energy.
Temperature of mixture (T) is always ³ lower temperature (TL) and £ higher temperature (TH), TL £ T £ TH
The temperature of mixture can never be lesser than lower temperature (as a body cannot be cooled below the
temperature of cooling body) and greater than higher temperature (as a body cannot be heated above the
temperature of heating body). Further more usually rise in temperature of one body is not equal to the fall in
temperature of the other body though heat gained by one body is equal to the heat lost by the other.
102
Thermal Physics
Illustration 25. 5g ice at 0°C is mixed with 5g of steam at 100°C . What is the final temperature?
Solution
Heat required by ice to raise its temperature to 100°C,
Q1 = m1L1 + m1c1Dq1 = 5 × 80 + 5 × 1 × 100 = 400 + 500 + 900 = 1800 cal
Heat given by steam when condensed Q2 = m2L2 = 5 × 536 = 2680 cal
As Q2 > Q1. This means that whole steam is not even condensed.
Hence temperature of mixture will remain at 100°C.
Illustration 26. A calorimeter of heat capacity 100 J/K is at room temperature of 30°C. 100 g of water at 40°C of
specific heat 4200 J/kg–K is poured into the calorimeter. What is the temperature of water in
calorimeter?
Solution
Let the temperature of water in calorimeter is t. Then heat lost by water = heat gained by
calorimeter
(0.1) × 4200 × (40 – t) = 100 (t – 30) Þ 42 × 40 – 42t = 10t – 300 Þ t = 38.07°C
Illustration 27. Find the quantity of heat required to convert 40 g of ice at –20°C into water at 20°C.
Given Lice = 0.336 × 106 J/kg. Specific heat of ice = 2100 J/kg–K, specific heat of water =
4200 J/kg–K
Solution
Heat required to raise the temperature of ice from –20°C to 0°C = 0.04 × 2100 × 20 = 1680 J
Heat required to convert the ice into water at 0°C = mL = 0.04 × 0.336 × 106 = 13440 J
Heat required to heat water from 0°C to 20°C
= 0.04 × 4200 × 20 = 3360 J
Total heat required
= 1680 + 13440 + 3360 = 18480 J
Illustration 28. Steam at 100°C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg
at 15°C till the temperature of the calorimeter and its contents rises to 80°C. What is the mass of
steam condensed? Latent heat of steam = 536 cal/g.
Solution
Heat required by (calorimeter + water)
Q = (m1c1 + m2c2) Dq = (0.02 + 1.1 × 1) (80 – 15) = 72.8 kcal
If m is mass of steam condensed, then heat given by steam
Q = mL + mc Dq = m × 536 + m × 1 × (100 – 80) = 556 m \ 556 m = 72.8
\ Mass of steam condensed m =
1.
72.8
556
= 0.130 kg
When m g of water at 10°C is mixed with m g of ice at 0°C, which of the following statements are false?
(A) The temperature of the system will be given by the equation m × 80 + m × 1 × (T – 0) = m × 1 (10 – T)
(B) Whole of ice will melt and temperature will be more than 0°C but lesser than 10°C
(C) Whole of ice will melt and temperature will be 0°C
(D) Whole of ice will not melt and temperature will be 0°C
Comprehension
A 0.60 kg sample of water and a sample of ice are placed in two compartments A and B that are separated by
a conducting wall, in a thermally insulated container. The rate of heat transfer from the water to the ice though
the conducting wall is constant P, until thermal equilibrium is reached. The temperature T of the liquid water and
the ice are given in graph as functions of time t. Temperature of the compartments remain homogeneous during
whole heat transfer process.
Given specific heat of ice = 2100 J/kg-K
Given specific heat of water = 4200 J/kg-K
Latent heat of fussion of ice = 3.3 × 105 J/kg
103
JEE-Physics
2.
The value of rate P is :
(A) 42.0 W
3.
The initial mass of the ice in the container is equal to :
(A) 0.36 kg
(B) 1.2 kg
(C) 2.4 kg
4.
The mass of the ice formed due to conversion from the water till thermal equilibrium is reached is equal to:
(A) 0.12 kg
(B) 0.15 kg
(C) 0.25 kg
(D) 0.40 kg
5.
A copper cube of mass 200 g slides down on a rough inclined plane of inclination 37° at a constant speed.
Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increases in
the temperature of the block as it slides down through 60 cm. Specific heat capacity of copper = 420 J/kg-K.
6.
Figure shown a paddle wheel coupled to a mass of 12 kg through fixed
frictionless pulleys. The paddle is immersed in a liquid of heat capacity
4200 J/K kept in an adiabatic container. Consider a time interval in which
the 12 kg block falls slowly through 70 cm. (a) How much heat is given to
the liquid ? (b) How much work is done on the liquid ? Calculate the rise
in the temperature of the liquid neglecting the heat capacity of the con
tainer and the paddle.
7.
A thermally insulated, closed copper vessel contains water at 15°C. when the vessel is shaken vigorously for
15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g and that of the water is 200 g. The
specific heat capacities of copper and water are 420 J/kg-K and 4200 J/kg-K respectively. Neglect any thermal
expansion. (a) How much heat is transferred to the liquid-vessel system ? (b) How much work has been done on
this system ? (c) How much is the increase in internal energy of the system ?
8.
The temperatures of equal masses of three different liquids A, B and C are 12ºC, 19ºC and 28ºC respectively.
The temperature when A and B are mixed is 16ºC, and when B and C are mixed, it is 23ºC. What will be the
temperature when A and C are mixed ?
9.
1kg of ice at 0ºC is mixed with 1 kg of steam at 100ºC. What will be the composition of the system when
thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 105 J/kg and latent heat of
vaporization of water = 2.26 × 10 6 J/kg.
104
(B) 36.0 W
(C) 21.0 W
(D) 18.0 W
(D) 3.6 kg
Thermal Physics
SOME WORKED OUT ILLUSTRATIONS
Illustration 1.
In a 20m deep lake, the bottom is at a constant temperature of 4°C. The air temperature is constant at
–10°C. The thermal conductivity of ice is 4 times that water. Neglecting the expansion of water on freezing,
the maximum thickness of ice will be
(A)
20
m
11
(B)
200
m
11
(C) 20 m
(D) 10 m
Ans. (B)
Solution
The rate of heat flow is the same through water and ice in the steady state so
-10°C
0°C
4°C
x
20-x
4K
ice
K
water
KA ( 4 - 0) 4KA [ 0 - ( -10) ]
200
=
Þx=
m
20 - x
x
11
Illustration 2.
The figure shows two thin rods, one made of aluminum [a = 23 × 10-6
(C°)-1] and the other of steel [a = 12 × 10-6 (C°)–1]. Each rod has the same
Midpoint
length and the same initial temperature. They are attached at one end to
Steel
Aluminum
two separate immovable walls. Temperature of both the rods is increased by
the same amount, until the gap between the rods vanishes.
Where do the rods meet when the gap vanishes?
(A) The rods meet exactly at the midpoint.
(B) The rods meet to the right of the midpoint .
(C) The rods meet to the left of the midpoint.
(D) Information insufficient
Ans. (B)
Solution
As a Al > a steel so expansion in aluminum rod is greater..
(B) 1.5 × 10–4 K–1
(C) 3 × 10–5 K–1
Ans. (B)
(D) 1.5 × 10–5 K–1
2.0
D (mm)
(A) 3 × 10–4 K–1
2.5
1.5
1.0
0.5
0
Illustration 3.
Figures shows the expansion of a 2m long metal rod with
temperature. The volume expansion coefficient of the metal
is :
5
10
15
D T(K)
20
25
Solution
Qa =
Dl
0.5 ´ 10-3
=
= 5 × 10–5 K–1 Þ g = 3a = 1.5 × 10–4 K–1
lDT
2 ( 5)
Illustration 4.
The temperature of a body rises by 44°C when a certain amount of heat is given to it. The same heat when supplied
to 22 g of ice at – 8°C, raises its temperature by 16°C. The water eqivalent of the body is
[Given : swater = 1 cal/g°C & Lf = 80 cal/g, sice = 0.5 cal/g°C]
(A) 25g
(B) 50 g
(C) 80 g
(D) 100 g
Ans. (B)
105
JEE-Physics
Solution
Supplied heat = (22) (0.5) (8) + (22) (80) + (22) (1) (16) = 88 + 1760 + 352 = 2200 cal
Heat capacityof the body =
2200 cal
= 50 cal/°C
44°C
Water equavalent of the body =
Heat capacity of the body
50 cal/°C
= 50g
=
spcific heat capacity of water
1 cal/g°C
Illustration 5.
A fine steel wire of length 4m is fixed rigidly in a heavy brass frame as shown in figure. It is just taut at
20°C. The tensile stress developed in steel wire if whole system is heated to 120°C is :–
(Given abrass = 1.8 × 10–5 °C–1, asteel=1.2 × 10–5 °C–1,Ysteel =2 × 1011 Nm–2,Ybrass=1.7 × 107 Nm–2)
(A) 1.02 × 104 Nm–2
(B) 1.2 × 108 Nm–2
(C) 1.2 × 106 Nm–2
(D) 6 × 108 Nm–2
Ans. (B)
Solution
Stress = Y (strain) = YS(ab–as)DT= (2 × 1011) (0.6 × 10–5) (100) = 1.2 × 108 Nm–2
Illustration 6.
540 g of ice at 0°C is mixed with 540 g of water at 80°C. The final temperature of the mixture is
(Given latent heat of fusion of ice = 80 cal/g and specific heat capacity of water = 1 cal/g0C)
(A) 0°C
(B) 40°C
(C) 80°C
(D) less than 0°C
Ans. (A)
Solution
Heat taken by ice to melt at 0°C is Q2 = mL = 540 × 80 = 43200 cal
Heat given by water to cool upto 0°C is Q2 = msDq = 540 × 1× (80–0) = 43200 cal
Hence heat given by water is just sufficient to melt the whole ice and final temperature of mixture is 0°C.
Illustration 7.
A refrigerator converts 100 g of water at 25°C into ice at – 10°C in one hour and 50 minutes. The quantity
of heat removed per minute is (specific heat of ice = 0.5 cal/g°C, latent heat of fusion = 80 cal/g)
(A) 50 cal
(B) 100 cal
(C) 200 cal
(D) 75 cal
Ans. (B)
Solution
Heat removed in cooling water from 25°C to 0°C = 100 × 1 × 25 = 2500 cal
Heat removed in converting water into ice at 0°C = 100 × 80 = 8000 cal
Heat removed in cooling ice from 0° to –15°C = 100 × 0.5 × 10 = 500 cal
Total heat removed in1 hr 50min = 2500 + 8000 + 500 = 11000 cal
Heat removed per minute =
11000
=100 cal/min
110
Illustration 8.
An irregular rod of same uniform material as shown in figure is conducting heat at a
steady rate. The temperature gradient at various sections versus area of cross
section graph will be
106
Heat
Thermal Physics
dT/dx
dT/dx
(A)
(B)
A
A
dT/dx
dT/dx
(C)
(D)
A
A
Ans. (B)
Solution
H = KA
dT
dT
is same in steady state condition, \ A
= constant \ rectangular hyperbolic graph
dx
dx
Illustration 9.
Which of the following graph(s) shows the correct variation in intensity of heat radiations by black body and
frequency at a fixed temperature–
El
El
UV Visible
Infra-red
UV
3500K
2500K
(A)
Visible Infra-red
1500K
2500K
(B)
1500K
3500K
l
l
El
El
Infra-red
Visible
Ultra-voilet
Infra-red
Visible
3500K
(C)
Ultra-voilet
1500K
(D)
2500K
2500K
1500K
Ans. (A,C)
Solution
3500K
n
n
1
Þ nmµT. As the temperature of body increases, frequency corresponding to
T
maximum energy in radiation (nm) increases.
According to Wien's law lmµ
Also area under the curve
òE
l
dl µò E n dn µ T 4
Illustration 10.
The temperature drop through a two layer furnace wall is 900°C. Each layer
is of equal area of cross–section. Which of the following action(s) will result
in lowering the temperature q of the interface?
(A) By increasing the thermal conductivity of outer layer.
(B) By increasing the thermal conductivity of inner layer.
(C) By increasing thickness of outer layer.
(D) By increasing thickness of inner layer.
Ans. (A,D)
107
JEE-Physics
Solution
Rate of heat flow
K i A (1000 - q) K 0 A (q - 100)
q - 100
=
Þ
=
900
li
l0
1
K l
1+ 0 i
Kil0
Now, we can see that q can be decreased by increasing thermal conductivity of outer layer (K o) and thickness of
inner layer (li).
Illustration 11.
5 kg of steam at 100°C is mixed with 10 kg of ice at 0°C. Choose correct alternative/s
(Given swater = 1 cal/g°C, LF = 80 cal/g, LV = 540 cal/g)
(A) Equilibrium temperature of mixture is 160°C
(B) Equilibrium temperature of mixture is 100°C
(C) At equilibrium, mixture contains 13
(D) At equilibrium, mixture contains 1
Ans. (B,C,D)
Solution
Required heat
10 kg ice (0°C)
2
kg of steam
3
Available heat
5 kg steam (100°C)
800 kcal
10 g water (0°C)
1
kg of water
3
2700 cal
5 g water (100°C)
1000 kcal
10 g water (100°C)
So available heat is more than required heat therefore final temperature will be 100°C.
Mass of heat condensed =
Total mass of steam = 5 -
800 + 1000 10
10 40
1
=
kg. Total mass of water = 10 +
=
= 13 kg
540
3
3
3
3
10 5
2
= = 1 kg
3
3
3
Illustration 12.
Water contained in a jar at room temperature (20°C) is intended to be cooled by method-I or method-II given
below :
Method -I : By placing ice cubes and allowing it to float.
Method-II : By wrapping ice cubes in a wire mesh and allowing it to sink.
Choose best method(s) to cool the water.
(A) Method-I from 20°C to 4°C
(B) Method-I from 4°C to 0°C
(C) Method-II from 20°C to 4°C
(D) Method-II from 4°C to 0°C
Ans. (A,D)
Solution
Initially (above 4°C), a decrease in temperature, increases the density of water and consequently it descends,
replacing the relatively warm water. Convention currents set up in this way demands the location of ice to be on
the water surface.
Below 4°C, a decrease in temperature decreases the water density and as a result it ascends up displacing the
relatively warm water. To setup convention currents in this way, the position of ice cubes should be at the
bottom.
108
Thermal Physics
Illustration 13.
dq
= – k(q–q0), the constant k is proportional to
dt
(A) A, surface area of the body
(B) S, specific heat of the body
In Newton's law of cooling,
1
, m being mass of the body
m
Ans. (A,C)
Solution
(C)
(D) e, emissivity of the body
dQ
= esA (q 4 - q40 ) » ( 3esAq30 ) Dq
dt
Illustration 14 to 16.
At 20°C a liquid is filled upto 10 cm height in a container of glass of length
20 cm and cross-sectional area 100 cm2 Scale is marked on the surface of
container. This scale gives correct reading at 20°C.
Given gL = 5 × 10–5 k–1, ag = 1 × 10–5 °C–1
14.
15.
16.
20cm
dQ
= esA (q 4 - q40 ) » ( 3esAq30 ) Dq
dt
10cm
A0 = 100cm
The volume of liquid at 40°C is :–
(A) 1002 cc
(B) 1001 cc
(C) 1003 cc
(D) 1000.5 cc
The actual height of liquid at 40°C is(A) 10.01 cm
(B) 10.006 cm
(C) 10.6 cm
(D) 10.1 cm
The reading of scale at 40°C is(A) 10.01 cm
(B) 10.004 cm
(C) 10.006cm
(D) 10.04 cm
2
Solution
14. Ans. (B)
V = V0 (1+gLDT) = (10) (100) [1 + 5 × 10–5 × 20] = 1000 (1 + 0.001) = 1001 cm3 = 1001 cc
15.
Ans. (B)
Cross sectional area of vessel at 40°C
A = A0(1 + 2 agDT) = 100 (1 + 2 × 10–5 × 20) = 100.04 cm2
Actual height of liquid =
0.04 ö
æ 1001ö æ
=ç
1+
è 100 ÷ø çè
100 ÷ø
16.
Actual volume of liquid
1001
=
= (1001) (100 + 0.04)–1
cross - sec tional area of vessel
100.04
-1
=
(1001) æ
0.04 ö
1
=
1(1001 – 0.4) = 10.006 cm
ç
÷
è
ø
100
100
100
Ans. (B)
Q TV = SR (1 + agDT) = (TV) (1 – agDT)
\ SR = (TV) (1 + ag DT)–1 = (10.006) (1 – 10–5 × 20) = 10.006 – 0.002 = 10.004 c
Illustration 17 to 19.
A body cools in a surrounding of constant temperature 30°C. Its heat capacity is 2J/°C. Initial temperature of the
body is 40°C. Assume Newton’s law of cooling is valid. The body cools to 36°C in 10 minutes.
17.
In further 10 minutes it will cool from 36°C to :
(A) 34.8°C
(B) 32.1°C
(C) 32.8°C
(D) 33.6°C
109
JEE-Physics
18.
The temperature of the body in °C denoted by q the variation of q versus time t is best denoted as
40°C
40°C
q
q
(A) 30°C
(B) 30°C
t
(0,0)
(0,0)
40°C
t
40°C
q
(C)
q
(0,0)
(D) 30°C
t
(0,0)
t
19.
When the body temperature has reached 36°C, it is heated again so that it reaches to 40°C in 10 minutes.
Assume that the rate of loss of heat at 38°C is the average rate of loss for the given time. The total heat required
from a heater by the body is :
(A) 7.2 J
(B) 0.728 J
(C) 16 J
(D) 32 J
Solution
17. Ans. (D)
36 - x
æ 36 + x
ö
40 - 36
= kç
- 30÷ Þ x = 33.6
= k(38 – 30) and
è
ø
10
2
10
18.
Ans. (D)
dq
kA
=(T – T0) Magnitude of slope will decrease with time.
dt
ms
19.
Ans. (C)
Q
4
1
40 - 36
=
= k (38 – 30) Þ k =
10 ´ 8 20
10
when the block is at 38°C and room temperature is at 30°C the rate of heat loss ms ×
dq
= ms k (38 – 30)
dt
1
× 8 × 10 = 8 J
20
Now heat agained by the object in the said 10 minutes. Q = ms Dq = 2 × 4 = 8 J
Total heat required = 8 + 8 = 16 J
Total heat loss in 10 minute Þ DQ = ms k (38 – 30) × 10 = 2 ×
Illustration 20.
Column I (Questions)
(A) The temperature of an iron piece is increased from 20° to 70°.
What is change in its temperature on the Fahrenheit scale (in °F)?
Column II (Answers)
(p) 20
(B) At what temperature (in °C) do the Celsius and Fahrenheit readings
have the same numerical value?
(q) 40
(C) 100 g ice at 0°C is converted into steam at 100 °C.
Find total heat required (in kcal)
(Lf = 80 cal/g, sw = 1 cal/g°C,Lv = 540 cal/g)
(r)
(D) A ball is dropped on a floor from a height of 5 m.
After the collision it rises upto a height of 3m. Assume that
50% of the mechanical energy lost goes as thermal energy
into the ball. Calculate the rise in temperature (in milli centigrade)
of the ball in the collision. (sball = 500 J/K, g = 10 m/s2)
Ans. (A) s; (B) r; (C) t; (D) p
110
– 40
(s) 72
(t)
90
Thermal Physics
Solution
(A)
C - 0 F - 32
9
9
=
Þ DF = DC = ´ 50 = 90°F
100
180
5
5
(B)
x - 0 x - 32
=
Þ x = -40
100
180
(C) Total heat required = mLf + msDq + mLv = 72000 cal = 72 kcal
(D) msDq =
50
1
´ mg ( h1 - h2 ) Þ Dq =
°C = 20 ´ 10 -3°C
100
50
Illustration 21.
In an industrial process 10 kg of water per hour is to be heated from 20°C to 80°C . To do this steam at 150°C
is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned
to the boiler as water at 90°C. How many kg of steam is required per hour.
(Specific heat of steam = specific heat of water = 1 cal/g°C, Latent heat of vaporisation = 540 cal/g)
Ans. 1
Solution
Suppose m kg steam is required per hour
Heat is released by steam in following three steps
® 100°C steam
(i) When 150°C steam ¾¾¾
Q = mc
Dq = m × 1 (150–100) = 50 m cal
Q1
1
steam
® 100°C water
(ii) When 100°C steam ¾¾¾
Q2
Q2 = mLv = m× 540= 540 m cal
® 90°C water
(iii) When 100 °C water ¾¾¾
Q2
Q3 = mcWDq = m × 1 × (100 –90) = 10 m cal
Hence total heat given by the steam
Q = Q1 + Q2 + Q3 = 600 m cal ....(i)
Heat taken by 10 kg water
Q' = mcWDq = 10 × 103 × 1 × (80–20) = 600 × 103 cal
Hence Q = Q' Þ 600 m = 600 × 103 Þ m = 103 gm = 1 kg
Illustration 22.
The specific heat of a metal at low temperatures varies according to S = (4/5)T3 where T is the absolute
temperature. Find the heat energy needed to raise unit mass of the metal from T = 1 K to T = 2 K.
Ans. 3
Solution
Q=
ò
mSdT =
mT 4
Q 15
Þ
=
=3
5
m
5
Illustration 23.
A rod has variable co-efficient of linear expansion a =
x
. If length of the rod is 1m. Determine increase in
5000
length of the rod in (cm) on increasing temperature of the rod by 100°C.
x
Ans. 1
Solution
x
æ x ö
Increase in length of dx = a l0DT = ç
( dx )(100) = 50 dx
è 5000 ÷ø
1
æ x2 ö
x
1
=
dx = ç
m = 1cm
Þ Total thermal expansion = ò0
÷
50
è 100 ø 0 100
1
111
JEE-Physics
Illustration 24.
A clock pendulum made of invar has a period of 2 s at 20°C. If the clock is used in a climate where average
temperature is 40°C, what correction (in seconds) may be necessary at the end of 10 days to the time given by
clock ? (ainvar = 7 × 10–7 °C–1, 1 day = 8.64 × 104 s)
Ans. 6
Solution
æ 10 ´ 8.64 ´ 104 ö
l
DT 1 Dl 1
T
-7
Þ
=
= aDq Þ DT = ( aDq ) = ç
÷ø (7 ´ 10 ) (20) = 6 s
è
g
T
2 l
2
2
2
T = 2p
Illustration 25.
Two different rods A and B are kept as shown in figure.
Temp.(°C)
100°
100°C
70°
A
35°
0
30
100
70°C
35°C
B
Distance
(cm)
The variation of temperature of different cross sections is plotted in a graph shown in figure.
Find the ratio of thermal conductivities of B to A.
Ans. 2
Solution
KB
K A (100 - 70) K B A ( 70 - 35)
K
æ DQ ö
æ DQ ö
=
\ A
=
Þ KA =
Þ B =2
Qç
è Dt ÷ø A çè Dt ÷ø B
30
70
2
KA
Illustration 26
A body cools from 50°C to 49.9°C in 5 sec. It cools from 40°C to 39.9°C in t sec. Assuming Newtons law of
cooling to be valid and temperature of surrounds at 30°C, value of t/5 will be?
Ans. 2
Solution
From
q1 - q2
æ q + q2
ö
= kç 1
- q0 ÷
è 2
ø
t
We have
0.1
0.1
æ 39.5 ö
æ 19.5 ö
t 39.5
t
= kç
= kç
Þ =
= 2 Þ t = 10 Þ = 2
and
÷
è 2 ø
è 2 ÷ø
5
t
5 19.5
5
Illustration 27
Two taps A and B supply water at temperatures 10º and 50° C respectively. Tap A alone fills the tank in 1 hour
and tap B alone fills the tank in 3 hour. If we open both the taps together in the empty tank, if the final
temperature of the water in the completely filled tank is found to be 5a (in °C). Find the value of a. Neglect loss
of heat to surrounding and heat capacity of the tank.
Ans. 4
Solution
m (T–10) s =
112
m
S ( 50 - T ) Þ T = 20°
3
Thermal Physics
Illustration 28
Two identical metal plates are welded end to end as shown in figure-(i). 20 cal of heat flows through it in 4 minutes.
if the plates are welded as shown in figure-(ii), find the time (in minutes) taken by the same amount of heat to
flow through the plates.
T1
T2
Fig. (i)
T1
T2
Fig. (ii)
Ans. 1
Solution
Rate of heat flow
l
DQ kA ( T1 - T2 )
=
Þ Dt µ
Dt
A
l
113
JEE-Physics
ANSWERS
BEGINNER'S BOX-1
1. (C)
2. (A,B)
3. (C)
4. (A)
6. (a) R1' = R1 (1 + aDq)
5. (A,B,C,D)
(b) R '2 = R 2 (1 + aDq)
(d) A ' = (pR 22 - pR12 )(1 + 2aDq) = A(1 + 2aDq)
(c) R '2 = R1' = (R 2 - R1 )(1 + aDq)
7. (i) hollow sphere > solid sphere
(ii) hollow sphere = solid sphere
8. (a) hf = h{1 + ac Dq}; Af = A{1 + 2ac Dq}; vf = Ah{1 + 3ac Dq};
Volume of liquuid Vw = Ah1 (1 + gL Dq)
(b) hf = h1 {1 + gL Dq}
(c) (i) 3h ac > h1 gl, (ii) 3h ac < h1 gl (iii) 3h ac = h1 gl
(d) DV = Ah(gL – 3ac)Dq
9. 120° C
BEGINNER'S BOX-2
1. (D)
2. (D)
3. (B)
4. (D)
5. (D)
3. (C)
4. (B)
5.
6. 75 W, 400 W
BEGINNER'S BOX-3
1. (A,B,C)
2. (A)
6. (a) zero (b) 84 J (c)
1
°C
50
9. 665 g steam and 1.335 kg water
114
3
= 8.6 ´ 10 -3 °C
350
7. (a) zero, (b) 1764 J (c) 1764 J
8. 20.3ºC
Thermal Physics
1.
On an X temperature scale, water freezes at –125.0° X and boils at 375.0° X. On a Y temperature scale,
water freezes at –70.0° Y and boils at –30.0° Y. The value of temperature on X–scale equal to the temperature
of 50.0°Y on Y–scale is :–
(A) 455.0°X
(B) –125.0°X
(C) 1375.0°X
(D) 1500.0°X
2.
A centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until
the Fahrenheit thermometer registers 140°F What is the temperature as registered by the centigrade thermometer:–
(A) 30°
(B) 40°
(C) 60°
(D) 80°
3.
The graph AB shown in figure is a plot of temperature of a body in degree
Celsius and degree Fahrenheit. Then :–
(A) Slope of line AB is 9/5
(B) Slope of line AB is 5/9
(C) Slope of line AB is 1/9
(D) slope of line AB is 3/9
4.
Centigrade
Fahrenheit
32°F 212°F
A
Two absolute scales X and Y assigned numerical values 200 and 450 to the triple point of water. What is the
relation between TX and TY ?
(A) 9TX = 4TY
5.
B
100°C
(B) 4TX = 9TY
(C) TX = 3TY
(D) None of these
A faulty thermometer reads freezing point and boiling point of water as –5°C and 95°C respectively.
What is the correct value of temperature as it reads 60°C on faulty thermometer?
(A) 60°C
6.
(D) 62°C
(B) 4.5°C
(C) 5.0°C
(D) 5.5°C.
A meter washer has a hole of diameter d1 and external diameter d2 , where d2=3d1. On heating, d2 increases by
0.3%. Then d1 will :–
(A) decrease by 0.1%
8.
(C) 64°C
A steel scale is to be prepared such that the millimeter intervals are to be accurate within 6 × 10–5 mm. The
maximum temperature variation during the ruling of the millimeter marks is (a=12×10–6C–1):–
(A) 4.0°C
7.
(B) 65°C
(B) decrease by 0.3%
(C) increase by 0.1%
(D) increase by 0.3%.
At 4°C, 0.98 of the volume of a body is immersed in water. The temperature at which the entire body gets
immersed in water is (neglect the expansion of the body) ( g w = 3.3 ´ 10-4 K -1 ) :–
(A) 40.8°C
9.
(C) 60.6°C
(D) 58.8°C
Two metal rods of the same length and area of cross–section are fixed ends to end between rigid supports.
The materials of the rods have Young moduli Y 1 and Y2, and coefficients of linear expansion a1 and a2. When
rods are cooled the junction between the rods does not shift if:–
(A) Y1a1 = Y2a2
10.
(B) 64.6°C
(B) Y1a2 = Y2a1
(C) Y1a21 = Y2a22
(D) Y21a1 = Y22a2
A steel rod of length 1 m is heated from 25°C to 75°C keeping its length constant. The longitudinal strain
developed in the rod is:– (Given : Coefficient of linear expansion of steel = 12 × 10 –6/°C)
(A) 6 × 10–6
(B) –6 × 10–5
(C) –6 × 10–4
(D) zero
115
JEE-Physics
12.
A brass disc fits simply in a hole of a steel plate. The disc from the hole can be loosened if the system
(abrass > asteel)
(A) First heated then cooled
(B) First cooled then heated
(C) Is heated
(D) Is cooled
The variation of lengths of two metal rods A and B with change in temperature
a
is shown in figure. The ratio of A is:–
aB
(A)
3
2
(B)
2
3
(C)
4
3
(D)
3
4
106
length(cm)
11.
B
104
A
100
0
T
temp. (°C)
13.
A steel tape is placed around the earth at the equator when the temperature is 10°C. What will be the clearance
between the tape and the ground (assumed to be uniform) if the temperature of the tape rises to 40°C ? Neglect
expansion of the earth. Radius of earth at equator is 6400 km & asteel = 1.2 × 10–5 K–1
(A) 2.3 m
(B) 2.1 m
(C) 2.3 km
(D) 230 m
14.
Bars of two different metals are bolted together, as shown in figure.The distance x does not change with temperature
if:–
15.
l A aA
(A) l = a
B
B
l A aB
(B) l = a
B
A
l2A a A
(C) l2 = a
B
B
l2A a B
(D) l2 = a
B
A
17.
18.
5
4
l0, l0
3
3
aB
aA
(B) l0, 2l0
(C)
3l 0 3l 0
,
2
2
(D)
2
7
l0, l0
3
3
The coefficient of linear expansion a of the material of a rod of length l0 varies with absolute temperature as
a=aT – bT2 where a & b are constants. The linear expansion of the rod when heated from T1 to T2 = 2T1
is:–
7b 3 ö
æ 3 2 7b 3 ö
æ
7b ö
æ
2
T1 ÷ L 0
(A) ç aT1 (B) ç 4a (D) None of these
(C) ç 2aT1 - 3 T1 ÷ L 0
÷ T1 L 0
2
3
3 ø
è
è
ø
è
ø
A clock with a metallic pendulum gains 6 seconds each day when the temperature is 20°C and loses 6 second
when the temperature is 40°C. Find the coefficient of linear expansion of the metal.
(A) 1.4 × 10–5 °C–1
(B) 1.4 × 10–6 °C–1
(C) 1.4 × 10–4 °C–1
(D) 0.4 × 10–6 °C–1
A steel scale measures the length of a copper rod as l0 when both are at 20°C, which is the calibration temperature for the scale. The scale reading when both are at 40°C, is:–
(A) (1 + 20a C ) l 0
19.
A
A metal rod A of length l0 expands by Dl when its temperature is raised by 100°C. Another rod B of different
metal of length 2l0 expands by Dl/2 for same rise in temperature. A third rod C of length 3l0 is made up
of pieces of rods A and B placed end to end expands by 2Dl on heating through 100 K. The length of
each portion of the composite rod is:–
(A)
16.
B
(B) (1 + 20a S ) l 0
æ 1 + 20a S ö
(C) ç 1 + 20a ÷ l 0
è
Cø
æ 1 + 20a C ö
(D) ç 1 + 20a ÷ l 0
è
Sø
The coefficient of apparent expansion of a liquid when determined using two different vessels A and
B are g1 and g2 respectively. If the coefficient of linear expansion of the vessel A is a 1, the coefficient
of linear expansion of the vessel B is:–
a1 g 1 g 2
(A) g + g
1
2
116
(B)
g1 - g2
2a1
(C)
g1 + g 2 + a
3
(D)
g 1 - g 2 + 3a1
3
Thermal Physics
20.
50°C
Three rods of the same dimensions have thermal conductivities 3k, 2k
and k. They are arranged as shown, with their ends at 100°C, 50°C and
0°C. The temperature of their junction is:–
(A) 75°C
200
°C
(B)
3
(C) 40°C
100
°C
(D)
3
2k
100°C
3k
k
0°C
21.
A cup of tea cools from 80°C to 60°C in one minute. The ambient temperature is 30°C. In cooling from 60°C to
50°C. It will take :–
(A) 50 s
(B) 90 s
(C) 60 s
(D) 48 s
22.
Ice starts forming in lake with water at 0°C when the atmospheric temperature is –10°C. If the time taken for
1 cm of ice be 7 hours, then the time taken for the thickness of ice to change from 1 cm to 2 cm is :–
(A) 7 hours
(B) 14 hours
(C) less than 7 hours
(D) more than 7 hours
23.
There is a small hole in a container. At what temperature should it be maintained in order that it emits one
calorie of energy per second per meter2 :–
(A) 10K
(B) 500K
(C) 200K
(D) 100K
24.
A blackened metallic foil is kept at a distance d from a spherical heater. The power absorbed by the foil is P. If
the temperature of heater and distance both are doubled, then the power absorbed by the foil will be:–
(A) 8P
(B) 4P
(C) 2P
(D) P
25.
Two different rods A and B are kept as shown in figure. The variation of temperature of different cross sections
with distance is plotted in a graph shown in figure. The ratio of thermal conductivities of A and B isTemp.(°C)
100°
70°
100°C
70°C
A
35°C
B
35°
0
(A) 2
26.
(B) 0.5
30
100
(C) 1
Distance
(cm)
(D) 2/3
Which of the following graph shows the correct variation in intensity of heat radiations by black body and frequency
at a fixed temperature:–
En
En
UV Visible
Infra-red
UV
3500K
2500K
(A)
1500K
Visible Infra-red
1500K
2500K
(B)
3500K
n
n
En
En
Infra-red
Visible
Ultra-voilet
Infra-red
3500K
(C)
2500K
Visible
Ultra-voilet
1500K
(D)
2500K
1500K
n
27.
3500K
n
A red star and a green star radiate energy at the same rate which star is bigger.
(A) Red
(B) Green
(C) Both have same size
(D) Can't be say anything
117
JEE-Physics
28.
Two identical masses of 5 kg each fall on a wheel from a height of 10m. The wheel disturbs a mass of 2 kg water,
the rise in temperature of water will be :–
(A) 2.6° C
(B) 1.2° C
(C) 0.32° C
(D) 0.12° C
29.
Hailstone at 0°C falls from a height of 1 km on an insulating surface converting whole of its kinetic energy into
heat. What part of it will melt:– [g = 10 m/s2, Lice = 330 × 103 J kg–1]
(A)
1
33
(B)
1
8
(C)
1
´ 10 -4
33
(D) All of it will melt
30.
Steam at 100°C is passed through 1.1 kg of water contained in a calorimeter of water equivalent
0.02 kg at 15°C till the temperature of the calorimeter and its contents rises to 80°C. The mass of the steam
condensed in kg is :–
(A) 0.130
(B) 0.065
(C) 0.260
(D) 0.135
31.
Water is used to cool the radiators of engines in cars because :–
(A) of its low boiling point
(B) of its high specific heat
(C) of its low density
(D) of its easy availability
118
Thermal Physics
1.
2.
A triangular plate has two cavities, one square and one rectangular as shown
in figure. The plate is heated.
(A) a increase, b decrease
(B) a and b both increase
(C) a and b increase, x and l decrease
(D) a, b, x and l all increase
4.
5.
a
b
b
x
Three rods of equal length are joined to form an equilateral triangle ABC.D
is the midpoint of AB. The coefficient of linear expansion is a1 for AB, and
a2 for AC and BC. If the distance DC remains constant for small changes
in temperature:–
(A) a1 = a2
(B) a1 = 2a2
A
D
a2
a1
B
a2
1
C
a
2 2
If water at 0°C, kept in a container with an open top, is placed in a large evacuated chamber:–
(A) All the water will vaporize.
(B) All the water will freeze.
(C) Part of the water will vaporize and the rest will freeze.
(D) Ice, water and water vapour will be formed and reach equilibrium at the triple point.
(C) a1 = 4a2
3.
a
(D) a1 =
In the previous question, if the specific latent heat of vaporization of water at 0°C is h times the specific latent
heat of freezing of water at 0°C, the fraction of water that will ultimately freeze is:–
1
h
h-1
h -1
(A)
(B)
(C)
(D)
h
h+1
h
h+1
When two samples at different temperatures are mixed, the temperature of the mixture can be :–
(A) lesser than lower or greater than higher temperature
(B) equal to lower or higher temperature
(C) greater than lower but lesser than higher temperature
(D) average of lower and higher temperatures.
6.
Two identical beakers are filled with water to the same level at 4°C. If one say A is heated while the other B is
cooled, then:–
(A) water level in A will rise
(B) water level in B will rise
(C) water level in A will fall
(D) water level in B will fall
7.
Two substances A and B of equal mass m are heated by uniform rate
of 6 cal s–1 under similar conditions. A graph between temperature
and time is shown in figure. Ratio of heat absorbed HA /HB by them for
complete fusion is:–
8.
8
(C)
5
5
(D)
8
A
B
0 1 2
3 4 5
6 7
Time(s)
Radiation from a black body at the thermodynamic temperature T1 is measured by a small detector at distance
d1 from it. When the temperature is increased to T 2 and the distance to d2, the power received by the detector
is unchanged. What is the ratio d2/d1?
T2
(A) T
1
9.
4
(B)
9
Temperature (°C)
9
(A)
4
100
80
60
40
20
æ T2 ö
(B) ç ÷
è T1 ø
2
æ T1 ö
(C) ç ÷
èT ø
2
2
2
æT ö
(D) ç 2 ÷
è T1 ø
4
A point source of heat of power P is placed at the center of a spherical shell of mean radius R. The material of
the shell has thermal conductivity k. If the temperature difference between the outer and the inner surface of the
shell is not to exceed T, then the thickness of the shell should not be less than :–
(A)
2pR 2kT
P
(B)
4pR2 kT
P
(C)
pR 2kT
P
(D)
pR2 kT
4P
119
JEE-Physics
10.
A black body emits radiation at the rate P when its temperature is T. At this temperature the wavelength
at which the radiation has maximum intensity is l0. If at another temperature T' the power radiated is 'P'
l0
then:–
2
(B) P' T' = 16 PT
(C) P' T' = 8 PT
and wavelength at maximum intensity is
(A) P' T' = 32 PT
(D) P' T' = 4 PT
Match the column
11.
Three liquids A, B and C having same specific heat and mass m, 2m and 3m have temperature 20°C, 40°C
and 60°C respectively. Temperature of the mixture when :
Column I
Column II
(A)
A and B are mixed
(p) 35°C
(B)
A and C are mixed
(q) 52°C
(C)
B and C are mixed
(r)
50°C
(D)
A, B and C all three are mixed
(s)
45°C
(t)
None
12.
Three rods of equal length of same material are joined to form an equilateral
triangle ABC as shown in figure. Area of cross–section of rod AB is S, of
rod BC is 2S and that of AC is S, then
Column I
(A)
(B)
(C)
Column II
Temperature of junction B
Heat current in AB
Heat current in BC
(p)
(q)
(r)
(s)
(t)
13.
B
A
100°C
C
0°C
Greater than 50°C
Less than 50°C
Is equal to heat current in BC
2
times heat current in AC
3
None
Is
A copper rod (initially at room temperature 20°C) of non–uniform cross section is placed between a steam
chamber at 100°C and ice–water chamber at 0°C. A and B are cross sections as shown in figure. Then match
the statements in column–I with results in column–II using comparing only between cross section A and B.
(The mathematical expressions in column–I have usual meaning in heat transfer).
100°C
Steam
Chamber
A
B
0°C
Ice Water
Chamber
Column I
(A)
(B)
æ dQ ö
Initially rate of heat flow çè
÷ will be
dt ø
æ dQ ö
At steady state rate of heat flow çè
÷ will be
dt ø
Column II
(p)
Maximum at section A
(q)
Maximum at section B
(C)
æ dT ö
At steady state temperature gradient çè ÷ø will be
dx
(r)
Minimum at section B
(D)
At steady state rate of change of temperature
(s)
Same for all section
æ dT ö
çè ÷ø at a certain point will be
dt
120
Thermal Physics
Comprehension Based Questions
Comprehension-1.
A certain amount of ice is supplied heat at a constant rate for 7 min. For the first one minute the temperature
rises uniformly with time. Then it remains constant for the next 4 min and again the temperature rises at uniform
rate for the last 2 min. Given Sice = 0.5 cal/g°C, Lf = 80 cal/g :
14.
15.
The initial temperature of ice is :–
(A) –10°C
(B) –20°C
(C) –30°C
(D) –40°C
Final temperature at the end of 7 min is :
(A) 10°C
(B) 20°C
(C) 30°C
(D) 40°C
Comprehension-2
A substance is in the solid form at 0°C. The amount of heat added to this substance and its temperature is
plotted in the graph. The specific heat capacity of the solid substance is 0.5 cal/g°C.
Temperature
(°C)
240
150
0
16.
17.
The mass of the substance is(A) 6g
(B) 12g
800
1000
Q(cal)
(C) 3g
(D) Can't be calculated
Latent heat capacity in melting process is(A)
18.
450
350
cal/g
3
(B)
175
cal/g
3
(C)
400
cal/g
3
(D) Can't say
(C)
10
cal/g°C
27
(D) Can't say
Specific heat capacity in the liquid state is(A)
5
cal/g°C
27
(B)
5
cal/gK
27
121
JEE-Physics
1.
Two rods each of length L2 and coefficient of linear expansion a2 each are connected freely to a third rod of
length L1 and coefficient of expansion a1 to form an isosceles triangle. The arrangement is supported on a knife–
edge at the midpoint of L1 which is horizontal. What relation must exist between L 1 and L2 so that the apex of
the isosceles triangle is to remain at a constant height from the knife edge as the temperature changes?
2.
A bimetallic strip of thickness d and length L is clamped at one end at temperature t1. Find the radius of
curvature of the strip if it consists of two different metals of expansivity a1 and a2 (a1>a2) when its temperature
rises to t2°C.
3.
Two metal cubes with 3 cm–edges of copper and aluminium are arranged as shown
in figure. Find
(i)
The total thermal current from one reservoir to the other.
(ii)
The ratio of the thermal current carried by the copper cube to that carried by
the aluminium cube.
Thermal conductivity of copper is 60 W/m–K and that of aluminium is 40 W/m–K.
4.
Calculate q1 and q2 in shown situation.
q1
200°C
q2
l
18°C
l
l
5.
A 'thermacole' icebox is a cheap and efficient method for storing small quantities of cooked food in summer in
particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate
the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co–efficient of thermal conductivity
of thermacole is 0.01 J s–1 m 1 °C–1. [Heat of fusion of water = 335 × 103 J kg–1]
6.
An electric heater is used in a room of total wall area 137 m 2 to maintain a temperature of +20°C inside it,
when the outside temperature is –10°C. The walls have three different layers materials. The innermost layer is
of wood of thickness 2.5 cm, the middle layer is of cement of thickness 1.0 cm and the outermost layer is of
brick of thickness 25.0 cm. Find the power of the electric heater. Assume that there is no heat loss through the
floor and the ceiling. The thermal conductivities of wood, cement and brick are 0.125, 1.5 and 1.0 W/m/°C
respectively.
7.
The figure shows the face and interface temperature of a composite slab containing of four layers of two
materials having identical thickness. Under steady state condition, find the value of temperature q.
0
0
20 C 10 C
k
2k
0
q
0
-5 C -10 C
k
2k
k=thermal conductivity
8.
A lagged stick of cross section area 1 cm2 and length 1m is initially at a temperature of 00C. It is then kept
between 2 reservoirs of temperature 1000C and 00C. Specific heat capacity is 10 J/kg0C and linear mass density
is 2kg/m. Find
100 0C
x
(i) Temperature gradient along the rod in steady state
(ii) Total heat absorbed by the rod to reach steady state
122
00 C
Thermal Physics
9.
Calculate the temperature of the black body from given graph.
10.
Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two
bodies are same, the two bodies emit total radiant power at the same rate. The wavelength lB corresponding to
maximum spectral radiance from B is shifted from the wavelength corresponding to maximum spectral radiancy
in the radiation from A by 1.0 mm. If the temperature of A is 5802K, Calculate :–
(i)
The temperature of (ii) Wavelength lB
123
JEE-Physics
1.
The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when
its temperature is raised by 100°C is (For steel Young's modulus is 2 × 1011 Nm–2 and coefficient of thermal
expansion is 1.1 × 10–5 K–1)
[JEE(Main) 2014]
(A) 2.2 × 108 Pa
2.
4.
(D) 6.0 cal/s
(C) 60°C ; a = 1.85 × 10–4 / °C
(D) 30°C ; a = 1.85 × 10–3 / °C
A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled
with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T
is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°C [JEE(Main) 2017]
(B) 825°C
(C) 800°C
(D) 885° C
An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides.
K is the bulk modulus of the material of the cube and a is its coefficient of linear expansion. Suppose we want
to bring the cube to its original size by heating. The temperature should be raised by :
[JEE(Main) 2017]
3a
PK
(B) 3PKa
(C)
P
3aK
(D)
P
aK
An unknown metal of mass 192 g heated to a temperature of 100°C was immersed into a brass calorimeter of
mass 128 g containing 240 g of water a temperature of 8.4°C Calculate the specific heat of the unknown metal
if water temperature stabilizes at 21.5°C (Specific heat of brass is 394 J kg –1 K–1)
[JEE(Main) 2019]
(B) 458 J kg–1 K–1
(C) 654 J kg–1 K–1
(D) 916 J kg–1 K–1
When 100 g of a liquid A at 100°C is added to 50 g of a liquid B at temperature 75°C, the temperature of the
mixture becomes 90°C. The temperature of the mixture, if 100 g of liquid A at 100°C is added to 50 g of liquid
B at 50°C, will be :
[JEE(Main) 2019]
(A) 80°C
8.
(C) 4.8 cal/s
(B) 25°C ; a = 1.85 × 10–5 / °C
(A) 1232 J kg–1 K–1
7.
(B) 2.4 cal/s
(A) 55°C ; a = 1.85 × 10–2 / °C
(A)
6.
(D) 2.2 × 106 Pa
A pendulume clock loses 12s a day if the temperature is 40°C and gains 4s a day if the temperature is 20°C. The
temperature at which the clock will show correct time, and the co-effecient of linear expansion (a) of the metal
of the pendulum shaft are respectively :[JEE(Main) 2016]
(A) 1250°C
5.
(C) 2.2 × 107 Pa
Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of
each rod = 4 cm2. End of copper rod is maintained at 100°C where as ends of brass and steel are kept at 0°C.
Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated
from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12
CGS units rexpectively. Rate of heat flow through copper rod is :
[JEE(Main) 2014]
(A) 1.2 cal/s
3.
(B) 2.2 × 109 Pa
(B) 60°C
(C) 70°C
(D) 85°C
A metal ball of mass 0.1 kg is heated upto 500°C and dropped into a vessel of heat capacity 800 JK—1 and
containing 0.5 kg water. The initial temperature of water and vessel is 30°C. What is the appro×imate percentage
increment in the temperature of the water ?
[JEE(Main) 2019]
[Specific Heat Capacities of water and metal are, respectively, 4200 Jkg–1K–1 and 400 JKg–1K–1]
(A) 30%
9.
(B) 20%
(C) 25%
(D) 15%
Ice at –20° C os added tp 50 g of water at 40°C. When the temperature of the mixture reaches 0°C, it is found that
20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2 J/g/°C)
Specific heat of Ice = 2.1 J/g/°C
JEE(Main) 2019]
Heat of fusion of water at 0°C = 334 J/g)[
(A) 50 g
124
(B) 40 g
(C) 60 g
(D) 100 g
Thermal Physics
10.
11.
A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K2, K3, K4
arranged as shown in the figure. The effective dielectric constant K will be :
[JEE(Main) 2019]
(A) K =
(K 1 + K 2 )(K 3 + K 4 )
2(K 1 + K 2 + K 3 + K 4 )
(B) K =
(K 1 + K 2 )(K 3 + K 4 )
(K1 + K 2 + K 3 + K 4 )
(C) K =
(K 1 + K 4 )(K 2 + K 3 )
2(K 1 + K 2 + K 3 + K 4 )
(D) K =
(K 1 + K 3 )(K 2 + K 4 )
(K1 + K 2 + K 3 + K 4 )
Temperature difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another
3L
, is connected across AB (See figure). In steady state,
2
temperature difference between P and Q will be close to :
[JEE(Main) 2019]
bent rod PQ, of same cross–section as AB and length
(A) 60°C
12.
(B) 75°C
(C) 35°C
(D) 45°C
A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal
conductivity of the material of the inner cylinder is K1 and that of the outer cylinder is K2. Assuming no loss of
heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is :
[JEE(Main) 2019]
(A) K1 + K 2
13.
(B)
(C)
2K 1 + 3K 2
5
(D)
K 1 + 3K 2
4
A heat source at T = 103 K is connected to another heat reservoir at T=102 K by a copper slab which is 1 m
thick. Given that the thermal conductivity of copper is 0.1 WKr1 m_1, the energy flux through it in the steady state
is :
[JEE(Main) 2019]
(A) 90 Wm–2
14.
K1 + K 2
2
(B) 200 Wnr2
(C) 65 Wm–2
(D) 120 Wm–2
Two identical breakers A and B contain equal volumes of two different liquids at 60°C each and left to cool
down. Liquid in A has density of 8 × 102 kg/m3 and specific heat of 2000 J kg –1 K–1 while liquid in B has density
of 103 kg m–3 and specific heat of 4000 J kg–1 K–1. Which of the following best describes their temperature
versus time graph schematically? (assume the emissivity of both the beakers to be the same)[JEE(Main) 2019]
60°C
T
60°C
T
B
(A)
A
(B)
A
B
t
t
60°C
T
60°C
T
A and B
A
(C)
(D)
t
B
t
125
JEE-Physics
15.
A thermally insulated vessel contains 150g of water at 0°C. Then the air from the vessel is pumped out adiabatically.
A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be
closest to :
[JEE(Main) 2019]
(Latent heat of vaporization of water = 2.10 × 106 J kg–1 and Latent heat of Fusion of water = 3.36 × 105 J kg–1)
(A) 130 g
16.
(B) 35 g
(C) 20 g
Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness 'd' and '3d', respectively, are
joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ' q2' and 'q1' respectively,
(q2 > q1). The temperature at the interface is :[JEE(Main) 2019]
q2
(A)
17.
q 2 + q1
2
(B)
d
3d
3K
3K
q1 9q 2
+
10 10
(C)
q1
q1 2q 2
+
3
3
(D)
q1 5q 2
+
6
6
A uniform rod of length l is being rotated in a horizontal plane with a constant angular speed about an axis
passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the
axis, then which of the following graphs depicts it most closely ?
[JEE(Main) 2019]
T(x)
T(x)
(A)
(B)
l
x
T(x)
l
x
l
x
T(x)
(C)
(D)
l
18.
(D) 150 g
x
At 40ºC, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of the
wire. When the wire is cooled down from 40ºC to 20ºC it regains its original length of 0.2 m. The value of M is
close to :
[JEE(Main) 2019]
(Coefficient of linear expansion and Young's modulus of brass are 10 –5/ºC and 1011 N/m2, respectively; g = 10 ms–2)
(A) 1.5 kg
19.
(C) 0.9 kg
(D) 0.5 kg
When M1 gram of ice at –10ºC (specific heat = 0.5 cal g –1ºC–1) is added to M2 gram of water at 50ºC, finally
no ice is left and the water is at 0ºC. The value of latent heat of ice, in cal g–1 is :
[JEE(Main) 2019]
(A)
20.
(B) 9 kg
5M1
- 50
M2
(B)
50M 2
M1
(C)
50M 2
-5
M1
(D)
5M 2
-5
M1
One kg of water, at 20ºC, is heated in an electric kettle whose heating element has a mean (temperature
averaged) resistance of 20 W. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time
taken for water to evaporate fully, is close to :
[JEE(Main) 2019]
[Specific heat of water = 4200 J/kg ºC), Latent heat of water = 2260 kJ/kg]
(A) 3 minutes
126
(B) 22 minutes
(C) 10 minutes
(D) 16 minutes
Thermal Physics
MCQ's with One Correct Answer
1.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this object with wavelength
between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm
is U3. The Wien constant, b = 2.88 × 106 nm–K. Then :–
[IIT-JEE 1998]
(A) U1 = 0
2.
(B) U3 = 0
(C) U1 > U2
(D) U2 > U1
The plots of intensity versus wavelength for three black bodies at temperature T 1, T2 and T3 respectively are as
shown. Their temperatures are such that:–
[IIT-JEE 2000]
I
T3
T1
T2
l
(A) T1 > T2 > T3
(D) T3 > T2 > T1
(D)
H eat sup p lied
H eat sup p lied
Tem p.
(C)
Tem p .
(B)
Tem p .
(A)
4.
(C) T2 > T3 > T1
A block of ice at –10°C is slowly heated and converted to steam at 100°C. Which of the following curves
represents the phenomena qualitatively:–
[IIT-JEE 2000]
Tem p .
3.
(B) T1 > T3 > T2
H eat sup p lied
H eat sup p lied
Three rods made of the same material and having the same cross–section have been joined as shown in the
figure. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C respectively. The
temperature of junction of the three rods will be :
[IIT-JEE 2001]
90°C
0°C
90°C
(A) 45°C
5.
(B) 60°C
(C) 30°C
(D) 20°C
An ideal black–body at room temperature is thrown into a furnance. It is observed that
[IIT-JEE 2003]
(A) initially it is the darkest body and at later times the brightest
(B) it is the darkest body at all times
(C) it cannot be distinguished at all times
(D) initially it is the darkest body and at later times it cannot be distinguished
127
JEE-Physics
6.
The graph, shown in the diagram, represents the variation of temperature (T) of the bodies, x and y having same
surface area, with time (t) due to the emission of radiation. Find the correct relation between the emissivity and
[IIT-JEE 2003]
absorptivity power of the two bodies :–
T
y
x
(A) ex > ey and ax < ay
(B) ex < ey and ax > ay
t
(C) ex > ey and ax > ay
(D) ex < ey and ax < ay
Two rods, one of aluminium and the other made of steel, having initial length l1 and l2 are connected together
to form a single rod of length l1 + l2. The coefficient of linear expansion for aluminium and steel area aa and
as respectively. If the length of each rod increases by the same amount when their temperature are raised by t°C,
7.
l1
then find the ratio l + l :–
1
2
as
(A) a
a
8.
[IIT-JEE 2003]
as
(C) (a + a )
a
s
aa
(B) a
s
aa
(D) (a + a )
a
s
2 kg of ice at –20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible heat capacity.
Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice
are 1 kcal/kg/°C while the latent heat of fusion of ice is 80 kcal/kg :–
[IIT-JEE 2003]
(A) 7 kg
9.
(B) 6 kg
(C) 4 kg
(D) 2 kg
Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which
of the following graphs represent the variation of temperature with time :–
[IIT-JEE 2004]
Temp.
Temp.
(A)
10.
Time
(B)
Time
Temp.
(C)
Time
Temp.
(D)
Time
Two identical conducting rods are first connected independently to two vessels, one containing water at 100°C
and the other containing ice at 0°C. In the second case, the rods are joined end to end and connected to the
same vessels. Let q1 and q2 g/s be the rate of melting of ice in the two cases respectively. The ratio
(A)
11.
1
2
2
1
(C)
4
1
(D)
1
4
[IIT-JEE 2004]
Three discs, A, B and C having radii 2m, 4m and 6m respectively are coated with carbon black on their outer
surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and
500 nm respectively. The power radiated by them are Q A, QB and QC respectively :–
[IIT-JEE 2004]
(A) QA is maximum
12.
(B)
q1
is :–
q2
(B) QB is maximum
(C) QC is maximum
(D) QA = QB = QC
Water of volume 2L in a container is heated with a coil of 1 kW at 27°C. The lid of the container is open and
energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27°C to 77°C ? (Give specific
heat of water is 4.2 kJ/kg) :–
[IIT-JEE 2005]
(A) 8 min 20 s
128
(B) 6 min 2 s
(C) 7 min
(D) 14 min
Thermal Physics
13.
14.
[IIT-JEE 2005]
In which of the following process, convection does not take place primarily :–
(A) Sea and land breeze
(B) Boiling of water
(C) Warming of glass of bulb due to filament
(D) Heating air around a furnace
Variation of radiant energy emitted by sun, filament of tungsten lamp and welding arc as a function of its
wavelength is shown in figure. Which of the following option is the correct match :–
[IIT-JEE 2005]
E1
T3
T2
T1
(A) Sun–T1, tungsten filament–T2, welding arc–T3
(C) Sun–T3, tungsten filament–T2, welding arc–T1
15.
16.
Calorie is defined as the amount of heat required to raise temperature of 1 g of water by 1°C and it is defined
under which of the following conditions :–
[IIT-JEE 2005]
(A) From 14.5°C to 15.5°C at 760 mm of Hg
(B) From 98.5°C to 99.5°C at 760 mm of Hg
(C) From 13.5°C to 14.5°C at 76 mm of Hg
(D) From 3.5°C to 4.5°C at 76 mm of Hg
A body with area A and temperature T and emissivity e = 0.6 is kept inside a spherical black body. What will be
the maximum energy radiated :–
[IIT-JEE 2005]
(A) 0.60 eAT4
17.
I
(B) Sun–T2, tungsten filament–T1, welding arc–T3
(D) Sun–T3, tungsten filament–T1, welding arc–T2
(B) 0.80 eAT4
(C) 1.00 eAT4
(D) 0.40 eAT4
Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration
II as shown in the figure. One of the blocks has thermal conductivity k and the other 2k. The temperature
difference between the ends along the x-axis is the same in both the configurations. It takes 9s to transport a
certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same
amount of heat in the configuration II is :[IIT-JEE 2013]
C o n fig u rat io n II
C o n fig u rat io n I
k
2k
k
2k
x
(A) 2.0 s
18
(C) 4.5 s
(D) 6.0 s
Parallel rays of light of intensity I = 912 Wm –2 are incident on a spherical black body kept in surroundings of
temperature 300 K. Take Stefan-Boltzmann constant s = 5.7×10–8 Wm–2 K–4 and assume that the energy
exchange with the surrounding is only through radiation. The final steady state temperature of the black body is
close to–
[JEE(Adv.) 2014]
(A) 330 K
19.
(B) 3.0 s
(B) 660 K
(C) 990 K
(D) 1550 K
A water cooler of storage capacity 120 litres can cool water at constant rate of P watts. In a closed circulation
system (as shown schematically in the figure), the water from the cooler is used to cool an external device that
generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed
30°C and the entire stored 120 litres of water is initially cooled to 10°C. The entire system is thermally insulated.
The minimum value of P (in watts) for which the device can be operated for 3 hours is : [JEE(Adv.) 2016]
129
JEE-Physics
Cooler
Device
Hot
Cold
(Specific heat of water is 4.2 kJ kg –1 K–1 and the density of water is 1000 kg m –3)
(A) 1600
20.
(C) 2533
(D) 3933
The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a
length of 1m at 10°C. Now the end P is maintained at 10°C, while the end S is heated and maintained at 400°C.
The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the
wire RS and the coefficient of linear thermal expansion of PQ is 1.2 × 10 –5 K–1, the change in length of the wire
PQ is
[JEE(Adv.) 2016]
(A) 0.78 mm
21.
(B) 2067
(B) 0.90 mm
(C) 1.56 mm
(D) 2.34 mm
A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The metal rod is
enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t)
as :
T(t) = T0 (1 + bt )
1/4
where b is a constant with appropriate dimension while T0 is a constant with dimension of temperature. The
heat capacity of the metal is :
[IIT-JEE 2019]
(A)
4P(T(t) - T0 )3
b4 T04
(B)
4P(T(t) - T0 )
b4 T02
(C)
4P(T(t) - T0 )4
b4 T05
(D)
4P(T(t) - T0 )2
b4 T03
MCQs with one or more than one correct answer
1.
2.
A bimetallic strip is formed out of two identical strips one of copper and the other of brass. The coefficients of
linear expansion of the two metals are aC and aB. On heating, the temperature of the strip goes up by DT and
the strip bends to form an arc of radius of curvature R. Then R is :–
[IIT-JEE 1999]
(A) Proportional to DT
(B) Inversely proportional to DT
(C) Proportional to |aB – aC|
(D) Inversely proportional to |aB – aC|
The figure below shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). The
temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, the
following statement(s) is (are) correct to a reasonable approximation :[IIT-JEE 2013]
C
100
200
300 400 500
T(K)
(A) The rate at which heat is absorbed in the range 0–100 K varies linearly with temperature T.
(B) Heat absorbed in increasing the temperature from 0–100 K is less than the heat required for increasing the
temperature from 400–500 K.
(C) There is no change in the rate of heat absorption in the range 400–500 K
(D) The rate of heat absorption increases in the range 200–300 K
130
Thermal Physics
3.
An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric
current. The hot filament emits black-body radiation. The filament is observed to break up at random locations
after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the
bulb is powered at constant voltage, which of the following statement(s) is(are) true?
[IIT-JEE 2016]
(A) The temperature distribution over the filament is uniform
(B) The resistance over small sections of the filament decreases with time
(C) The filament emits more light at higher band of frequencies before it breaks up
(D) The filament consumes less electrical power towards the end of the life of the bulb
4.
A human body has a surface area of approximately 1 m2. The normal body temperature is 10 K above the
surrounding room temperature T0. Take the room temperature to be T0 = 300 K. For T0 = 300 K, the value of
sT04 = 460 Wm-2 (where s is the Stefan-Boltzmann constant). Which of the following options is/are correct ?
(A) The amount of energy radiated by the body in 1 second is close to 60 Joules
[IIT-JEE 2017]
(B) If the surrounding temperature reduces by a small amount DT0 << T0, then to maintain the same body
temperature the same (living) human being needs to radiate DW = 4sT03 DT0 more energy per unit time
(C) Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same
body temperature while reducing the energy lost by radiation
(D) If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted
by the body would shift to longer wavelengths
Match the Column
1.
Column–I gives some devices and Column–II gives some processes on which the functioning of these devices
depend. Match the devices in Column–I with the process in Column–II.
[IIT-JEE 2007]
Column I
Column II
(A) Bimetallic strip
(p) Radiation from a hot body
(B) Stem engine
(q) Energy conversion
(C) Incandescent lamp
(r) Melting
(D) Electric fuse
(s) Thermal expansion of solids
Subjective Questions
1.
A solid body X of heat capacity C is kept in an atmosphere whose temperature is T A = 300 K. At time t = 0, the
temperature of X is T0 = 400 K. It cools according to Newton's law of cooling. At time t 1 its temperature is
found to be 350 K. At this time (t1) the body X is connected to a large body Y at atmospheric temperature TA
through a conducting rod of length L, cross–section area A and thermal conductivity K. The heat capacity of Y
is so large that any variation in its temperature may be neglected. The cross–section area A of the connecting rod
is small compared to the surface area of X. Find the temperature of X at time t = 3t 1.
[IIT-JEE 1998]
2.
An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat S of the
container varies with temperature T according to the empirical relation S = A + BT, where A = 100 cal/kg–K
and B = 2 × 10–2 cal/kg–K2. If the final temperature of the container is 27°C, determine the mass of the
container (Latent heat of fusion for water = 8 × 104 cal/kg, specific heat of water = 103 cal/kg–K).
[IIT-JEE 2001]
3.
A 5m long cylindrical steel wire with radius 2 × 10 –3 m is suspended vertically from a rigid support and carries
a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire
ignoring losses. (For the steel wire : Young's modulus = 2.1 × 10 11 Pa; Density = 7860 Kg/m3 ; Specific heat =
420 J/kg–K).
[IIT-JEE 2001]
131
JEE-Physics
4.
The top of an insulated cylindrical container is covered by a disc having emissivity 0.6 and conductivity 0.167 W/
km and thickness 1 cm. The temperature is maintained by circulating oil as shown :
[IIT-JEE 2003]
O il out
O il in
(i) Find the radiation loss to the surroundings in J/m 2 s if temperature of the upper surface of disc is 127°C and
temperature of surroundings is 27°C.
(ii) Also find the temperature of the circulating oil. Neglect the heat loss due to convection.
(Give : s =
17
× 10–8 W–2K–4]
3
5.
A cube of coefficient of linear expansions as is floating in a bath containing a liquid of coefficient of volume
expansion g1. When the temperature is raised by DT, the depth upto which the cube is submerged in the liquid
remains the same. Find the relation between as and g1 showing all the steps.
[IIT-JEE 2004]
6.
One end of a rod of length L and cross–sectional area A is kept in a furnace of temperature T 1. The other end
of the rod is kept at a temperature T2. The thermal conductivity of the material of the rod is K and emissivity of
the rod is e. it is given that T2 = Ts + DT, where DT < < TS, Ts being the temperature of the surroundings. If
DT µ (T1 – Ts), find the proportionality constant that heat is lost only by radiation at the end where the temp, of
the rod is T2.
[IIT-JEE 2004]
Insulated
Furnace
T1
L
T1
T2
Insulated
7.
In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed find the final temperature
of the mixture (in kelvin). Given : Lfusion = 80 cal/g = 336 J/g; Lvaporization = 540 cal/g = 2268 J/g;
Sice = 2100 J/kg, K = 0.5 cal/gK; and Swater = 4200 J/kg, K = 1 cal/gK
[IIT-JEE 2006]
8.
A metal rod AB of length 10x has its one end A in ice at 0°C and the other end B in water at 100 °C. It a point
P on the rod is maintained at 400 °C, then it is found that equal amounts of water and ice evaporate and melt
per unit time. The latent heat of evaporation of water 540 cal/g and latent heat of melting of ice is 80 cal/g. If
the point P is at lx distance from the ice end then find out the value of l. [Neglect any heat loss to the
surrounding.]
[IIT-JEE 2009]
9.
A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by
the metal. The sensor has a scale that displays log2(P/P0), where P0 is a constant. When the metal surface is at
a temperature of 487 °C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains
constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to
2767 °C ?
[IIT-JEE 2016]
132
Thermal Physics
10.
Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept
at temperatures T1 = 300 K and T2 = 100 K, as shown in the figure. The radius of the bigger cylinder is twice
that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are
K1 and K2 respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, then
K1/K2 = _________.
[IIT-JEE 2018]
Insulating material
T1
K1
K2
T2
L
L
11.
A liquid at 30°C is poured very slowly into a Calorimeter that is at temperature of 110°C. The boiling temperature
of the liquid is 80°C. It is found that the first 5 gm of the liquid completely evaporates. After pouring another 80
gm of the liquid the equilibrium temperature is found to be 50°C. The ratio of the Latent heat of the liquid to its
specific heat will be ______ °C.
[Neglect the heat exchange with surrounding]
[IIT-JEE 2019]
* * * * *
133
JEE-Physics
ANSWER KEY
EXERCISE-1
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
C
C
B
A
B
C
D
B
A
C
D
B
C
B
A
Que.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
A
A
D
D
B
D
D
D
B
B
C
A
D
A
D
Que.
31
Ans.
B
EXERCISE-2
Que.
1
2
3
4
5
6
7
8
9
10
Ans.
B,D
C
C
B
B,C,D
A,B
C
B
B
A
l
l
Match the Column:
11.
(A) t (B) r, (C) q, (D) t
12.
(A) p (B) r,s (C) s
13.
(A) p,r (B) s (C) q (D) s
Comprehension Based Quesions
Comp.-1 :
14.
(D)
15.
(D)
Comp.-2:
16.
(A)
17.
(B)
18.
(C)
EXERCISE-3
1. 4L22a2 = L21a1
d
2. a - a t - t
( 1 2 )( 2 1 )
3. (i) 240W (ii) 1.5
4. q1 = 1160C, q2 = 740C
5. 3.739 kg
6. 9000 W
0
7. 5 C
8. (i) –100 0C/m, (ii) 1000J
9. 1927 K
10. (i) 1934 K (ii) 1.5 mm
EXERCISE-4
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
Que.
Ans.
A
16
B
C
17
D
B
18
B
D
19
C
C
20
B
D
A
B
B
A
D
D
A
A
C
EXERCISE-5
l
33
MCQ's (Single Correct answers)
1. (D)
2. (B)
3. (A)
4. (B)
5. (D)
6. (C)
7. (C)
8. (B)
9. (C)
10. (D)
11. (B)
12. (A)
13. (C)
14. (D)
15. (A)
16. (A)
17. (A)
18. (A)
19. (B)
20. (A)
21. (A)
134
Thermal Physics
l
MCQ's (one or more than one correct)
1. (B,D)
2. (A,B,C,D)
l
Match the column :
l
Subjective Questions
3. (C,D)
4. (A,B,C OR C)
1. (A) q,s (B) q, (C) p,q(D) q,r
2KAt1
1. éë300 + 12.5e - CL ùû K
2. 0.495 kg
3. 4.568 × 10–3 0C
4. (i) 595 W/m 2 (ii) 162.60C
5. g l = 2a S
K
6. 4esLT 3 + K
S
7. 273 K
9. 9
10. 4
8. 9
11. 270
*****
135
FLUID
MECHANICS
Recap of Early Classes
Matter W
is normally classified as being in one of three states: solid, liquid, or gas.
A fluid is a collection of molecules that are randomly arranged and held together by weak
cohesive forces and by forces exerted by the walls of a container. Both liquids and gases are
fluids.
In our treatment of the mechanics of fluids, we’ll be applying principles we have already
discussed. First, we consider the mechanics of a fluid at rest, that is fluid statics, and then
study fluids in motion, that is, fluid dynamics.
1.0 FLUID STATICS
1.1 Pressure Exerted by a Liquid (Effect of Gravity)
1.2 Pressure in case of Accelerating Fluid
1.3 Force on Side Wall of Vessel
1.4 Torque on the Side Wall due to Fluid Pressure
1.5 Pascal's Law
1.6 Floatation
1.7 Rotatory – Equilibrium in Floatation
2.0 FLUID DYNAMICS
2.1 Equation of Continuity
2.2 Bernoulli's Theorem
2.3 Venturimeter
2.4 Torricelli's Law of Efflux (Fluid Outflow)
2.5 Illustrations of Bernoulli's Theorem
EXERCISE-1
EXERCISE-2
EXERCISE-3
EXERCISE-4
EXERCISE-5
JEE-Physics
FLUID MECHANICS
1.0 FLUID STATICS
Matter exists in three states solid, liquid and gas. Liquids and gases are referred to as fluids. Any state of matter
that can flow is a fluid. Study of a fluid at rest is called fluid statics or hydrostatics and the study of fluid in
motion is called fluid dynamics of hydrodynamics. Both combined are called fluid mechanics.
The intermolecular force in liquids are comparatively weaker than in solids. Therefore, their shape can be
changed easily. Thus liquids assume the shape of the container. But their volume (or density) cannot be
changed so easily. Liquids are incompressible and have free surface of their own.
The intermolecular forces are weakest in gases, so their shape and size can be changed easily. Gases are
compressible and occupy all the space of the container.
Density (r)
Mass per unit volume is defined as density. So density at a point of a fluid is represented as
r = Lim
DV ®0
SI UNIT :
kg/m3
Dm dm
=
Density is a positive scalar quantity.
DV dV
CGS UNIT : g/cc
Dimensions :
[ML–3]
Relative Density
It is defined as the ratio of the density of the given fluid to the density of pure water at 4°C.
Relative density (R.D.) =
density of given liquid
density of pure water at 4°C
Relative density or specific gravity is a unitless and dimensionless positive scalar physical quantity.
Being a dimensionless/unitless quantity R.D. of a substance is same in SI and CGS system.
Specific Gravity
It is defined as the ratio of the specific weight of the given fluid to the specific weight of pure water at 4°C.
Specific gravity =
r ´ g rl
specific weight of given liquid
= R.D. of the liquid
= l
=
3
specific weight of pure water at 4°C (9.81 kN m ) rw ´ g rw
Thus specific gravity of a liquid is numerically equal to the relative density of that liquid and for calculation
purposes they are used interchangeably.
Illustration 1.
Solution
A hollow metallic sphere has inner and outer radii, respectively, as 5 cm and 10 cm. If the mass of
the sphere is 2.5 kg, find (a) density of the material, (b) relative density of the material of the sphere.
The volume of the material of the sphere is
éæ 10 ö 3 æ 5 ö 3 ù 4
4
æ 4ö
V = ç ÷ p r23 - r13 = ´ 3.14 ´ êç
÷ø - çè 100 ÷ø ú = 3 ´ 3.14 ´ [ 0.001 - 0.000125]
è 3ø
è
3
ëê 100
ûú
(
=
)
4
× 3.14 × 0.000875 m3 = 0.00367 m3
3
(a) Therefore, density of the material of the sphere is r =
(b) Relative density of the material of the sphere rr =
M
2.5
=
kg/m3 = 681.2 kg/m3
V 0.00367
681.2
= 0.6812
1000
Density of a Mixture of substance in the proportion of mass
Let a number of substances of masses M 1, M2 , M3 etc., and densities r1 , r2 , r3 etc. respectively are mixed
together. The total mass of the mixture = M 1 + M2 + M3 +....
136
Fluid Mechanics
M1 M2
M3
M1 + M 2 + M 3 ....
The total volume = r + r + r + .... therefore, the density of the mixture is r =
M1 M2 M 3
1
2
3
+
+
+ ....
r1
r2
r3
For two substances the density of the mixture r =
r1r2 ( M1 + M2 )
r1M2 + r2M1
Illustration 2. Two immiscible of densities 2.5 g/cm3 and 0.8 g/cm3 are taken in the ratio of their masses as 2:3
respectively. Find the average density of the liquid combination.
Solution
æ 2m 3m ö
+
Let masses be 2m g & 3m g, then V = V1 + V2 = ç
cm3
è 2.5 0.8 ÷ø
Total mass = 2m + 3m = 5m g.
Therefore, the average density
r av =
5m
5m
10
5
=
=
=
gm/cm3 = 1.09 gm/cm3
2m 3m
V
2
3
9.1
+
+
2.5 0.8
2.5 0.8
Density of a mixture of substance in the proportion of volume
Suppose that a number of substances of volume V1, V2, V3 etc. and densities r1 , r2 , r3 etc. respectively are
mixed. The total mass of the mixture is
The total volume of the mixture is
= r1V1 + r2 V2 + r3 V3 + ....
= V1 + V2 + V3+.........
Therefore, the density of the mixture is
r=
r1 V1 + r2 V2 + r3 V3
V1 + V2 + V3 + ....
Therefore, for two substances we can write
r=
r1 V1 + r2 V2
V1 + V2
Illustration 3. Two miscible liquids of densities 1.2 gm/cc and 1.4 gm/cc are mixed with a proportion ratio of their
volumes equal to 3:5 : What is the density of resulting liquid?
Solution
V
r1 1 + r2
r1 V1 + r2 V2
(1.2)(3 / 5 ) + 1.4 3.6 + 7
V
r1 =
2
r=
r =
Þ
Þ
V1 + V2
( 3 / 5 ) + 1 = 8 Þ r = 1.325
V1 / V2 + 1
PRESSURE
The pressure P is defined as the magnitude of the normal force acting on a unit surface area.
DF
here DF = normal force on a surface area DA
P=
DA
DA
DF
DF
The pressure is a scalar quantity. This is because hydrostatic pressure is transmitted equally in all directions
when force is applied, which shows that a definite direction is not associated with pressure.
Consequences of Pressure
(i)
Railway tracks are laid on large sized wooden or iron sleepers. This is because the weight (force) of the train is
spread over a large area of the sleeper. This reduces the pressure acting on the ground and hence prevents the
yielding of ground under the weight of the train.
137
JEE-Physics
(ii)
(iii)
A sharp knife is more effective in cutting the objects than a blunt knife. The pressure exerted = Force / area. The
sharp knife transmits force over a small area as compared to the blunt knife. Hence the pressure exerted in case
of sharp knife is more than in case of blunt knife.
A camel walks easily on sand but a man cannot inspite of the fact that a camel is much heavier than man. This
is because the area of camel's feet is large as compared to man's feet. So the pressure exerted by camel on the
sand is very small as compared to the pressure exerted by man. Due to large pressure, sand under the feet of
man yields and hence he cannot walk easily on sand.
TYPES OF PRESSURE
There are three types of pressure
(i) Atmospheric pressure (P0)
(ii) Gauge pressure (Pgauge)
•
Atmospheric pressure :
Force exerted by air column on unit cross–section area of sea level called
atmospheric pressure (Po)
Po =
•
(iii) Absolute pressure (Pabs.)
F
=101.3 kN/m2
A
up to top of
atmosphere
\ Po = 1.013 × 105 N/m2
sea
level
air
column
2
area=1m
Barometer is used to measure atmospheric pressure.
Which was discovered by Torricelli.
Atmospheric pressure varies from place to place and at a particular place from time to time.
Gauge Pressure :
Excess Pressure ( P– Patm) measured with the help of pressure measuring instrument called Gauge pressure.
Patm
manometer
gas
Pabsolute
h
Pgauge = hrg or Pgauge µ h
Gauge pressure is always measured with help of "manometer"
•
Absolute Pressure :
Sum of atmospheric and Gauge pressure is called absolute pressure.
Pabs = Patm + PgaugeÞ Pabs = Po + hrg
The pressure which we measure in our automobile tyres is gauge pressure.
P1
y1
VARIATION OF PRESSURE WITH DEPTH
(i)
Let pressure at L is P1 and pressure at M is P2
Then, P2A = P1 A + rgA (y2 – y1) Þ P2 = P1 + rg(y2 – y1)
Here pressure gradient
L
A
M
A
y2
dP
= rg
dy
P2
(ii)
Pressure is same at two points in the same horizontal level.
As body is in equilibrium, P1A = P2A Þ P1 = P2
Note : Pressure P is independent of shape of container
138
P1
L
M
A
A
P2
Fluid Mechanics
Illustration 4.
Assuming that the atmosphere has a uniform density of (1.3 kg/m3) and an effective height of 10
km, find the force exerted on an area of dimensions 100 m × 80 m at the bottom of the atmosphere .
Solution
F = PA = rgh A = (1.3) (9.8) (104) (100 × 80) = 12.74 × 8× 107 N = 1.0192 × 109 N
1.1
(i)
(ii)
(iii)
Pressure Exerted by a Liquid (Effect of Gravity)
Consider a vessel containing liquid. As the liquid is in equilibrium, so every volume element of the fluid is also
in equilibrium. Consider one volume element in the form of a cylindrical column of liquid of height h and of
area of cross section A. The various forces acting on the cylindrical column of liquid are :
Force, F1 = P1 A acting vertically downward on the top face of the column. P 1 is the pressure of the liquid on
the top face of the column and is known as atmospheric pressure.
Force, F2 = P2 A acting vertically upward at the bottom face of the cylindrical column. P2 is the pressure of the
liquid on the bottom face of the column.
Weight, W = mg of the cylindrical column of the liquid acting vertically downward.
Since the cylindrical column of the liquid is in equilibrium, so the net force acting on
the column is zero. i.e. F1 + W – F2 =0
F1
Þ P1A +mg – P2A = 0 ÞP1A + mg = P2 A
\ P2 = P1 +
mg
...(i)
A
Now, mass of the cylindrical column of the liquid is,
m = volume × density of the liquid = Area of cross section × height × density = Ahr
mg
h
Ahrg
F2
, P2 = P1 + hrg ...(ii)
A
P2 is the absolute pressure at depth h below the free surface of the liquid. Equation (ii), shows that the absolute
pressure at depth h is greater than the atmospheric pressure (P1) by an amount equal to hrg.
Equation (ii) can also be written as (P2–P1) = hrg which is the difference of pressure between two points
separated by a depth h.
\ equation (i) becomes P2 =P1 +
1.2
Pressure in case of Accelerating Fluid
(i)
Liquid placed in elevator :
When elevator accelerates upward with acceleration a 0 then pressure in the fluid,
at depth 'h' may be given by,
a0
P = hr[ g + a0 ]
(ii)
Free surface of liquid in case of horizontal acceleration :
h1
q
h2
1
2
a0
tan q =
h
ma0 a0
=
mg
g
If P1 and P2 are pressures at point 1 & 2 then P1–P2 = rg (h1– h2) = rgltanq = rla0
(iii)
Rotating Vessel
Consider a cylindrical vessel, rotating at constant angular velocity about its axis. If it contains fluid then after
an initial irregular shape, it will rotate with the tank as a rigid body. The acceleration of fluid particles located
at a distance r from the axis of rotation will be equal to w2r, and the direction of the acceleration is toward the
axis of rotation as shown in the figure. The fluid particles will be undergoing circular motion.
Lets consider a small horizontal cylinder of length dr and cross-sectional area A located y below the free surface
of the fluid and r from the axis. This cylinder is accelerating in ground frame with acceleration w2r towards the
axis hence the net horizontal force acting on it should be equal to the product of mass (dm) and acceleration.
139
JEE-Physics
dm = Adrr
P2A – P!A = (Adrr)w2r
If we say that the left face of the cylinder is y below the free surface of the fluid then the right surface is y + dy
below the surface of liquid. Thus P2 – P1 = rgdy
dy rw 2
=
dr
g
Thus solving we get
and, therefore, the equation for surfaces of constant pressure is y =
w2r 2
+ constant
2g
This equation means that these surface of constant pressure are parabolic as shown in figure.
z
axis of
rotation
r
p1
p1
p2
constant p2
pressure p3
lines
p4
ar =rw
p3
p4
w
2 2
2g
r
2
wr
y
x
The pressure varies with the distance from the axis of rotation, but at a fixed radius, the pressure varies
hydrostatically in the vertical direction as shown in figure.
Illustration 5. An open water tanker moving on a horizontal straight road has a
cubical block of cork floating over its surface. If the tanker has an
acceleration of 'a' as shown, the acceleration of the cork w.r.t. container
is (ignore viscosity)
Solution
N
a rel
a
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
ma
mgsinq- macosq
mgcosq+ masin q
mg
marel = mgsinq – macosq but for water surface tan q = a/g Þ arel = 0
Illustration 6. An open rectangular tank 1.5 m wide, 2 m deep and 2m long is half filled with water. It is accelerated
horizontally at 3.27 m/s2 in the direction of its length. Determine the depth of water at each end of
tank. [g = 9.81 m/s2]
Solution
Here tan q =
a 1
= , depth at corner 'A' = 1– 1.5 tan q = 0.5 m
g 3
3m
1m
B
q
3.27 m/s2
A
depth at corner 'B' = 1 + 1.5 tan q = 1.5 m
140
Fluid Mechanics
MEASUREMENT OF ATMOSPHERIC PRESSURE
1.
Mercury Barometer :
To measure the atmospheric pressure experimentally, torricelli invented a
mercury barometer in 1643.
pa =hrg
The pressure exerted by a mercury column of 1mm high is called 1 Torr.
1 Torr = 1 mm of mercury column
Torricelli
vaccum
h
A
Mercury
Trough
OPEN TUBE MANOMETER :
Open-tube manometer is used to measure the pressure gauge. When equilibrium is reached, the pressure at the
bottom of left limb is equal to the pressure at the bottom of right limb.
i.e. p + y1 rg = pa + y2 rg
pa
p – pa = rg (y2 – y1) = rgy
p
p – pa = rg (y2 – y1) = rgy
p+y1r g
p = absolute pressure, p – pa = gauge pressure.
y=y2-y1
y2
y1
p2+y2r g
Thus, knowing y and r (density of liquid), we can measure the gauge pressure.
Illustration 7.
The manometer shown below is used to measure the difference in water level between the two
tanks. Calculate this difference for the conditions indicated
Liquid
(sp gravity = 0.9)
40cm
Water
Solution
pa + h1 rg – 40r1g + 40rg = pa + h2 rg
h2 rg – h1 rg = 40 rg – 40 r1g as r1 = 0.9r
(h2 – h1) rg = 40rg – 36rg
Water
pa
h2
pa
h1
40cm
Water
Water
h2 – h1 = 4 cm
WATER BAROMETER
1.013 ´ 105
rg
The height of the water column in the tube will be 10.3 m. Such a long tube cannot be managed easily, thus
water barometer is not feasible.
Let us suppose water is used in the barometer instead of mercury. hrg = 1.013 × 105 or h =
Illustration 8.
In a given U-tube (open at one-end) find out relation
between p and pa.
Given d2 = 2 × 13.6 gm/cm3 d1 = 13.6 gm/cm3
y=25cm
d2
p
x=26cm
d1
141
JEE-Physics
Solution
Pressure in a liquid at same level is same i.e. at A – A–,
Pa
pa + d2 yg + xd1 g = p
In C.G.S.
Þ
Illustration 9.
pa + 13.6 × 2 × 25 × g + 13.6 × 26 × g = p
pa + 13.6 × g [50 + 26] = p
2pa = p as pa = 13.6 × g × 76
d2
y
P
x
A
Find out pressure at points A and B. Also find angle ‘q’.
A
A
d1
Patm
h
Patm
B
Pressure at A – PA = Patm – r1 gl sin q
Pressure at B PB = Patm + r2 gh q
But PB is also equal to
PB = PA + r3 gl sin q
Hence - Patm + r2 gh = PA + r3 gl sin q
Patm + r2 gh = Patm – r1 gl sin q + r3 gl sinq
Solution
sin q =
r2 h
(r3 - r1 )l .
Illustration 10. In the given figure, the container slides down with acceleration ‘a’ on an
incline of angle ‘q’. Liquid is stationary with respect to container. Find out
(i) Angle made by surface of liquid with horizontal plane.
(ii) Angle if a = g sin q.
Solution
Consider a fluid particle on surface. The forces acting on it are shown in figure.
a
q
ma sinq
ma
ma
a
mg
ma cosq
Normal to
liquid surface
q
Normal to
liquid surface
a
= ma cosq
a
q
liquid surface
mg
liquid surface
(mg-ma sinq)
Resultant force acting on liquid surface, will always normal to it
ma cos q
a cos q
tan a =
=
mg - ma sin q
(g - a sin q)
Thus angle of liquid surface with the horizontal is equal to a = tan–1
æ a cos q ö
g sin q cos q
(ii) If a = g sin q, then a = tan–1 çè g - g sin 2 q ÷ø = tan–1
g cos2 q
= tan–1 (tan q) Þ a = q
Illustration 11. An L shaped glass tube is kept inside a bus that is moving
with constant acceleration. During the motion, the level of
the liquid in the left arm is at 12 cm whereas in the right arm,
it is at 8 cm when the orientation of the tube is as shown.
Assuming that the diameter of the tube is much smaller than
levels of the liquid and neglecting effect of surface tension,
acceleration of the bus find the (g = 10 m/s 2).
h2 - h1
a
4cm
2
=
tan q = =
Solution
g h2 tan 45° + h1 tan 45° 20 cm Þ a = 2 m/s
142
a cos q
(g - a sin q)
12cm
45°
8cm
Fluid Mechanics
1.3
Force on Side Wall of Vessel
Force on the side wall of the vessel can not be directly determined as a different depths pressures are
different. To find this we consider a strip of width dx at a depth x from the surface of the liquid as shown in
figure, and on this strip the force due to the liquid is given as : dF = xrg × bdx
x
dx
h
dF
b
a
This force is acting in the direction normal to the side wall. Net force can be evaluated by integrating
equation
h
F = ò dF = ò xrgbdx
F=
0
rgbh2
2
Average pressure on side wall
The absolute pressure on the side wall cannot be evaluated because at different depths on this wall pressure is
different. The average pressure on the wall can be given as :
F 1 rgbh2 1
=
= rgh
bh 2 bh
2
Equation shows that the average pressure on side vertical wall is half of the net pressure at the bottom of the
vessel.
P
1.4
av
=
Torque on the Side Wall due to Fluid Pressure
As shown in figure, due to the force dF, the side wall experiences a torque about the bottom edge of the side
which is given as
dt = dF ´ ( h - x ) = xrgb dx ( h - x )
This net force is
t = ò dt = ò rgb ( hx - x2 ) dx
h
0
é h3 h3 ù 1
= rgb ê - ú = rgbh 3
3û 6
ë2
Illustration 12. Water and liquid is filled up behind a square wall of side l. Find out
A
h1 =5m
water r
B
square wall
side l=10m
h2=5m
liquid 2r
C
Solution
(a) Pressures at A, B and C
(b) Forces in part AB and BC
(c) Total force and point of application of force. (Neglect atmosphere pressure in every calculation)
(a) As there is no liquid above ‘A’
A
x
So pressure at A, pA = 0
dx
Pressure at B, pB = rgh1
Pressure at C, pC = rgh1 + 2rgh2
B
(b) Force at A = 0
Take a strip of width ‘dx’ at a depth ‘x’ in part AB.
C
143
JEE-Physics
Pressure is equal to rgx.
Force on strip = pressure × area
dF = rgx ldx
h1
Total force upto B: F = ò rgxldx =
0
rgxlh12 1000 ´ 10 ´ 10 ´ 5 ´ 5
=
= 1.25 ´ 106 N
2
2
In part BC for force take a elementary strip of width dx in portion BC. Pressure is equal to
= rgh1 + 2rg(x – h1)
Force on elementary strip = pressure × area dF = [rgh1 + 2rg(x – h1)] l dx
l
F = ò [ rgh1 + 2 rg (x - h1 )] l dx
Total force on part BC
A
h1
x
h1
l
é
é x2
ùù
gh
x
2
g
r
+
r
ê
ê - h1 x ú ú l
1
=
ë2
û û h1
ë
B
dx
él - h
2ù
= rgh1h2 l + 2rgl ê 2 - h1l + h1 ú
ë
û
2
2
1
C
2rgl 2
[l + h12 – 2h1l] = rgh1h2l +rgl (l – h1)2
2
= rgh2l [h1 + h2] = rgh2l2 = 1000 × 10 × 5 × 10 × 10 = 5 × 106 N
(c) Total force = 5 × 106 + 1.25 × 106 = 6.25 × 106 N
Taking torque about A
= rgh1h2l +
Total torque of force in AB = ò dF × x =
h1
ò rgxldx.x
0
h1
é rglx3 ù
rglh13 1000 ´ 10 ´ 10 ´ 125 1.25 ´ 107
=ê
=
=
=
N-m
ú
3
3
3
ë 3 û0
Total torque of force in BC =
ò dF × x
On solving we get = rgh1h2l[h1 +
h2
2h2
] + rgh22l[h1 +
]
2
3
= 1000 × 10 × 5 × 5 × 10 [5 + 2.5] + 1000 × 10 × 25 × 10 [5 +
= 2.5 × 7.5 × 106 +
Total torque =
10
]
3
118.75
62.5
× 106 =
× 106
3
3
11.875 ´ 107 1.25 ´ 107 13.125 ´ 107
+
=
3
3
3
Total torque = total force × distance of point of application of force from top = F.xp
6.25 × 106 xp =
xp = 7m
1.5
•
•
•
13.125 ´ 107
3
Pascal's Law
If the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid
without being diminished in magnitude. Pascal's law is stated in following ways –
The pressure in a fluid at rest is same at all the points if gravity is ignored.
A liquid exerts equal pressures in all directions.
If the pressure in an enclosed fluid is changed at a particular point, the change is transmitted to every
point of the fluid and to the walls of the container without being diminished in magnitude.
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Fluid Mechanics
Applications of pascal's law hydraulic jacks, lifts, presses, brakes, etc
For the hydraulic lift
Pressure applied =
F1
A1
F1
F2
\ Pressure transmitted =
A2
\ Upward force on A2 is F2 =
Ex.
Q
P
F1
F2
=
A1
A2
Q
A1
A2
F1
A2
× A2 =
A1
A1 × F1
A vertical U–tube of uniform cross–section contains mercury in both
arms. A glycerine (relative density = 1.3) column of length 10 cm is
introduced into one of the arms. Oil of density 800 kg m –3 is poured into
the other arm until the upper surface of the oil and glycerine are at the
same horizontal level. Find the length of the column. Density of mercury
is 13.6 × 103 kgm–3
Sol. Pressure at A and B must be same
C
h
10cm
0.1- h
Pressure at A = P0 + 0.1 × (1.3 × 1000) × g
Pressure at B = P0 + h × 800 × g + (0.1 – h) × 13.6 × 1000 g
Þ 0.1 × 1300 = 800 h + (0.1 – h) × 13600
A
B
Þ h = 0.096 m = 9.6 cm
BUOYANCY AND ARCHIMEDE'S PRINCIPLE
•
Buoyant Force : If a body is partially or wholly immersed in a fluid, it experiences an upward force due to the
fluid surrounding it. This phenomenon of force exerted by fluid on the body is called buoyancy and force is
called buoyant force or upthrust.
•
Archimede's Principle : It states that the buoyant force on a body that is partially or totally immersed in a
liquid is equal to the weight of the fluid displaced by it.
Now consider a body immersed in a liquid of density s.
Top surface of the body experiences a downward force
F1
F1 = AP1 = A[h1sg + P0] ...(i)
Lower face of the body will experiences a upward force
F2 = AP2 = A[h2sg + P0] ...(ii)
As h2 > h1. So F2 is greater than F1
So net upward force : F = F2 – F1 = Asg[h2 – h1]
\ F = AsgL = Vsg [Q V = AL]
1.6
(i)
(ii)
L
h1
h2
F2
Floatation
When a body of density (r) and volume (V) is completely immersed in a liquid of density (s), the forces acting
on the body are :
Weight of the body W = Mg = Vrg (directed vertically downwards through C.G. of the body).
Buoyant force or Upthrust Th = Vsg (directed vertically upwards through C.B.).
The apparent weight WApp is equal to W – Th.
The following three cases are possible :
Case I : Density of the body (r) is greater than that of liquid (s)
In this case if r > s then W > Th
So the body will sink to the bottom of the liquid.
WApp = W – Th = Vrg – Vsg = Vrg (1 – s/r) = W (1 – s/r).
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JEE-Physics
Case II : Density of the body is equal to the density of liquid (r = s)
In this case if r = s then W = Th
So the body will float fully submerged in the liquid. It will be in neutral equilibrium.
WApp = W – Th = 0
Case III : Density of the body is lesser than that of liquid (r < s)
In this case if r < s then W < Th
So the body will float partially submerged in the liquid. In this case the body will move up and the volume of
liquid displaced by the body (V in) will be less than the volume of body (V). So as to make Th equal to W
\WApp = W – Th = 0
The above three cases constitute the law of flotation which states that a body will float in a liquid if weight of
the liquid displaced by the immersed part of the body is at least equal to the weight of the body.
1.7
Rotatory – Equilibrium in Floatation
When a floating body is slightly tilted from equilibrium position, the centre of buoyancy B shifts. The vertical
line passing through the new centre of buoyancy B' and initial vertical line meet at a point M called meta –
centre. If the metacentre M is above the centre of gravity the couple due to forces at G (weight of body W) and
at B' (upthrust) tends to bring the body back to its original position (figure) . So for rotational equilibrium of
floating body the meta–centre must always be higher than the centre of gravity of the body.
M
Th
Th
G
M
G
B
(A)
(B)
W
M
B
B’
G
W
B’
(C)
However, if meta–centre goes below centre of gravity, the couple due to forces at G and B' tends to topple the
floating body. This is why a wooden log cannot be made to float vertical in water or a boat is likely to capsize
if the sitting passengers stand on it. In these situations centre of gravity becomes higher than meta centre and
so the body will topple if slightly tilted.
Illustration 13. A cubical block of wood of edge 3 cm floats in water. The lower surface
of the cube just touches the free end of a vertical spring fixed at the
bottom of the pot. Find the maximum weight that can be put on the
block without wetting it. Density of wood = 800 kg/m3 and spring constant
of the spring = 50 N/m. Take g = 10 m/s2
Solution
The specific gravity of the block = 0.8. Hence the height inside water = 3 cm × 0.8 = 2.4 cm. The
height outside water = 3 cm – 2.4 = 0.6 cm. Suppose the maximum weight that can be put
without wetting it is W. The block in this case is completely immersed in the water.
The volume of the displaced water= volume of the block = 27 × 10–6 m3.
Hence, the force of buoyancy = (27 × 10–6 m3) × 1(1000 kg/m3) × (10 m/s2) = 0.27 N.
The spring is compressed by 0.6 cm and hence the upward force exerted by the spring
= 50 N/m × 0.6 cm = 0.3 N.
The force of buoyancy and the spring force taken together balance the weight of the block plus the
weight W put on the block. The weight of the block is
W¢ = (27 × 10–6 m) × (800 kg/m3) × (10 m/s2) = 0.22 N.
Thus, W = 0.27 N + 0.3 N – 0.22 N = 0.35 N.
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Fluid Mechanics
Illustration 14. A wooden plank of length 1 m and uniform cross-section is
F
C
hinged at one end to the bottom of a tank as shown in figure.
The tank is filled with water up to a height of 0.5 m. The
B
l
q
specific gravity of the plank is 0.5. Find the angle q that the
A mg
plank makes with the vertical in the equilibrium position. (Exclude
0
the case q = 0).
Solution
The forces acting on the plank are shown in the figure. The height of water level is l = 0.5 m. The
length of the plank is 1.0 m = 2l. The weight of the plank acts through the centre B of the plank.
We have OB = l. The buoyant force F acts through the point A which is the middle point of the
dipped part OC of the plank.
We have OA =
l
OC
.
=
2cos q
2
Let the mass per unit length of the plank be r. Its weight mg = 2lrg.
æ l ö
The mass of the part OC of the plank = ç
r.
è cos q ÷ø
The mass of water displaced =
1
2lr
l
.
r=
0.5 cos q
cos q
2lrg
.
cos q
Now, for equilibrium, the torque of mg about O should balance the torque of F about O.
The buoyant force F is, therefore, F =
æ 2lr ö æ l ö
So, mg (OB) sinq = F(OA) sinq or, (2lr)l = ç
è cos q ÷ø çè 2 cos q ÷ø
or, cos2q =
•
•
•
•
•
1
1
, or,,q = 45°.
or, cosq =
2
2
Buoyant force act vertically upward through the centre of gravity (C.G.) of the displaced fluid. This point is
called centre of buoyancy (C.B.). Thus centre of buoyancy is the point through which the force of buoyancy is
supposed to act.
Buoyant force or upthrust does not depend upon the characteristics of the body such as its mass, size, density,
etc. But it depends upon the volume of the body inside the liquid.
It depends upon the nature of the fluid as it is proportional to the density of the fluid.
This is the reason that upthrust on a fully submerged body is more in sea water than in pure water
It depends upon the effective acceleration.
If a lift is accelerated downwards with acceleration a (a< g) then Th = Vin s (g – a)
If a lift is accelerated downwards with a = g then
Th = Vin s (g – a) = 0
If a lift is accelerated upward with accelaration a then
Th = Vin s (g + a)
If a body is weighed in air (WA), in water (WW) and in a oil (WO), then
Specific gravity of oil =
WA - WO
loss of weight in oil
= W -W
loss of weight in water
A
W
Illustration 15. A body weighs 160 g in air, 130 g in water and 136 g in oil. What is the specific gravity of oil?
Solution
Specific gravity of oil =
=
loss of weight in oil
loss of weight in water
160 – 136 24 8
=
=
= 0.8
160 – 130 30 10
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JEE-Physics
Illustration 16. An iceberg is floating partially immersed in sea–water. The density of sea–water is 1.03 gm/cm 3 and
that of ice is 0.92 gm/cm3. What is the fraction of the total volume of the iceberg above the level of
sea–water ?
Solution
In case of flotation weight = upthrust i.e.
mg = Vinsg Þ Vrg = Vinsg \ Vin =
Þ fout =
rù
é
r
V so Vout = V – Vin = V ê1 - ú
sû
s
ë
Vout é
rùé
0.92 ù 0.11
= ê1 - ú ê1 ú = 1.03 = 0.106 Þ fout = 10.6 %
s
V
1.03
ë
ûë
û
Illustration 17. A rubber ball of mass 10 gm and volume 15 cm3 is dipped in water to a depth of 10m. Assuming
density of water uniform throughout the depth if it is released from rest. Find (take g = 980 cm/s 2)
(a) the acceleration of the ball, and
(b) the time taken by it to reach the surface.
Solution
The maximum buoyant force on the ball is FB = Vrw g = 15 × 1× 980 dyne = 14700 dyne.
The weight of the ball is mg = 10 × g = 10 × 980 = 9800 dyne
The net upward force, F = (15 × 980 – 10 × 980) dyne = 5 × 980 dyne = 4900 dyne
(a) Therefore, acceleration of the ball upward a =
(b) Time taken by it reach the surface is t =
F 5 ´ 980
= 490 cm/s2 = 4.9 m/s2
=
m
10
2h
=
a
2 ´ 10
s = 2.02 s
4.9
1.
A glass tube shield at both ends is 1m long. It lies horizontally with the middle 10 cm containing Hg. the two
ends of the tube equal in length contain air at 27 9C and pressure 76 cm of Hg. The temperature at one end is
kept 00C and at the other end it is 1270C. Neglect the change in length of Hg column. Then the change in length
on two sides is
(A) 12.3 cm
(B) 10.311 cm
(C) 9.9 cm
(D) 8.489 cm
2.
A thin tube of uniform cross-section is seated at both ends. If lies horizontally. The middle 5 cm containing Hg
and two equal ends contain air at the same pressure P0. When the tube is held at an angle of 600 with the
vertical, the length of the air column above and below the Hg are 46 cm, 44.5 cm. Calculate pressure P 0 in cm
of Hg. Assume temperature of the system constant.
(A) 55 cm of Hg
(B) 65 cm of Hg
(C) 70.4 cm of Hg
(D) 75.4 cm of Hg
3.
To construct a barometer, a tube of length 1 m is filled completely with mercury and is inverted in a mercury cup.
The barometer reading on a particular day is 76 cm. Suppose a 1 m tube is filled with mercury up to 76 cm. It
is inverted in a mercury cup. The height of mercury column in the tube over the surface in the cup will be
(A) zero
(B) 76 cm
(C) > 76 cm
(D) < 76 cm
4.
A light cylindrical tube ‘T’ of length l and radius ‘r’ containing air is inverted
in water (density d1). One end of the tube is open and the other is closed. A
block ‘B’ of density d2 (>d1) is kept on the tube as shown in the figure. The
tube stays in equilibrium in the position shown. Find the volume of ‘B’. The
density of the air is negligible as compared with the density of water. What is
the pressure of the air in the tube assuming that the atmospheric pressure is
P0.
148
Fluid Mechanics
5.
A container of a large uniform cross-sectional area A resting on a horizontal
surface holds two immiscible, non-viscous and incompressible liquids of
densities ' d ' and ' 2 d ' each of height (1/2)H as shown. The smaller density
(
liquid is open to atmosphere. A homogeneous solid cylinder of length L < 1 H
2
)
cross-sectional area (1/5) A is immersed such that it floats with its axis vertical
to the liquid-liquid interface with length (1/4) L in denser liquid. If D is the
density of the solid cylinder then :
(A) D =
3d
2
(B) D =
d
2
(C) D =
2d
3
(D) D =
5d
4
6.
A cylindrical wooden float whose base area S and the height H drifts on the water surface. Density of
wood d and density of water is r. What minimum work must be performed to take the float out of the
water?
7.
A cube (density 0.5 gm/cc) of side 10 cm is floating in water kept in a cylindrical beaker of base area 1500 cm 2.
When a mass m is kept on wooden block the level of water rises in the beaker by 2mm. Find the mass m.
8.
A U-tube of base length “l” filled with same volume of two liquids of
densities r and 2r is moving with an acceleration “a” on the horizontal
plane. If the height difference between the two surfaces (open to atmos
phere) becomes zero, then the height h is given by :
a
(A) 2g l
(C)
9.
a
l
g
(B)
3a
l
2g
(D)
2a
l
3g
The cubical container ABCDEFGH which is completely filled with an
ideal (nonviscous and incompressible) fluid, moves in a gravity free
space with a acceleration of
r
a = a 0 ( î - ĵ + k̂ )
where a0 is a positive constant. Then the only point in the container
where pressure can be minimum is
(A) H
(B) C
(C) E
10.
(D) F
A block is partially immersed in a liquid and the vessel is accelerating upwards with an acceleration “a”.
The block is observed by two observers O 1 and O 2, one at rest and the other accelerating with an
acceleration “a” upward. The total buoyant force on the block is :
(A) same for O1 and O2
(C) greater for O2 than O1
(B) greater for O 1 than O2
(D) data is not sufficient
149
JEE-Physics
2.0 FLUID DYNAMICS
When a fluid moves in such a way that there are relative motions among the fluid particles, the fluid is said to
be flowing.
TYPES OF FLUID FLOW : Fluid flow can be classified as :
•
Steady and Unsteady Flow
Steady flow is defined as that type of flow in which the fluid characteristics like velocity, pressure and density
at a point do not change with time. In an unsteady flow, the velocity, pressure and density at a point in the
flow varies with time.
•
Streamline Flow
In steady flow all the particles passing through a given point follow the same path and hence a unique line of
flow. This line or path is called a streamline. Streamlines do not intersect each other because if they intersect
each other the particle can move in either direction at the point of intersection and flow cannot be steady.
•
Laminar and Turbulent Flow
Laminar flow is the flow in which the fluid particles move along well–defined streamlines which are straight
and parallel. In laminar flow the velocities at different points in the fluid may have different magnitudes, but
there directions are parallel. Thus the particles move in laminar or layers gliding smoothly over the adjacent
layer.
Turbulent flow is an irregular flow in which the particles can move in zig–zag way due to which eddies
formation take place which are responsible for high energy losses.
•
Compressible and Incompressible Flow
In compressible flow the density of fluid varies from point to point i.e. the density is not constant for the fluid
whereas in incompressible flow the density of the fluid remains constant throughout. Liquids are generally
incompressible while gases are compressible.
•
Rotational and Irrotational Flow
Rotational flow is the flow in which the fluid particles while flowing along path–lines also rotate about their own
axis. In irrotational flow particles do not rotate about their axis. So they have no net angular velocity.
2.1
Equation of Continuity
The continuity equation is the mathematical expression of the law of conservation of mass in fluid dynamics.
v1
A1
v2
A2
v2D t
v1D t
In the steady flow the mass of fluid entering into a tube of flow in a particular time interval is equal to the
mass of fluid leaving the tube.
m1 m 2
=
Þ r1A1v1 = r2A2v2 ( Q r1 = r2) Þ A1v1 = A2v2 \ Avv = constant
Dt
Dt
(Here r = density of fluid, v = velocity of fluid, A = Area of cross–section of tube)
Therefore the velocity of liquid is smaller in the wider parts of a tube and larger in the narrower parts.
2.2
Bernoulli's Theorem
Bernoulli's equation is mathematical expression of the law of mechanical energy conservation in fluid dynamics.
Bernoullis theorem is applied to the ideal fluids.
Characteristics of an ideal fluid are :
(i) The fluid is incompressible.
(ii) The fluid is non–viscous.
(iii) The fluid flow is steady.
(iv) The fluid flow is irrotational.
Every point in an ideal fluid flow is associated with three kinds of energies :
150
Fluid Mechanics
Kinetic Energy
If a liquid of mass (m) and volume (V) is flowing with velocity (v) then Kinetic Energy=
energy per unit volume =
1
mv2and kinetic
2
Kinetic Energy 1 m 2 1 2
=
v = rv
2
volume
2 V
Potential Energy
If a liquid of mass (m) and volume (V) is at height (h) from the surface of the earth then its
Potential Energy = mgh and potential energy per unit volume =
m
Potential Energy
= gh = rgh
volume
V
Pressure Energy
If P is the pressure on area A of a liquid and the liquid moves through a distance ( l) due to this pressure then
Pressure energy = Work done = force x displacement = pressure x area x displacement = PA l = PV
[Q Al = volume V]
Pressure energy per unit volume =
Pressure energy
=P
volume
Theorem
According to Bernoulli's Theorem , in case of steady flow of incompressible and non–viscous fluid through a
tube of non–uniform cross–section, the sum of the pressure, the potential energy per unit volume and the
kinetic energy per unit volume is same at every point in the tube, i.e., P + rgh +
C
P2
A2
P1
A1
(i)
v2
B
v1
h2
h1
Consider a liquid flowing steadily through a tube of non–uniform area of cross–section as shown in figure. If
P1 and P2 are the pressures at the two ends of the tube respectively, work done in pushing the volume DV of
incompressible liquid from point B to C through the tube
W = P (DV) = (P1–P2)DV
...(i)
This work is used by the liquid in two ways :
In changing the potential energy of mass Dm (in the volume DV) from
Dmgh1 to Dmgh2 i.e., DU=Dmg (h2–h1)
(ii)
1 2
rv = constant.
2
In changing the kinetic energy from
...(ii)
1
1
1
Dmv12 to Dmv 22 , i.e. DK = Dm v 22 - v 12
2
2
2
(
)
Now as the liquid is non–viscous, by conservation of mechanical energy,
1
2
2
W = DU+DK i.e., ( P1 - P2 ) DV=Dmg ( h2 - h1 ) + Dm v 2 - v 1
2
1
2
2
P1–P2 = rg(h2–h1) + r v 2 - v1 [as r = Dm / DV ]
2
(
(
P1 + rgh1 +
)
)
1
1 2
1
rv1 = P2 + rgh2 + rv 22 Þ P + rgh+ rv2= constant
2
2
2
This equation is the Bernoulli's equation and represents conservation of mechanical energy in case of moving
fluids.
151
JEE-Physics
2.3
Venturimeter
It is a gauge put on a flow pipe to measure the speed of flow of a liquid shown in figure.
r
v1
A1
A
v2
P2
P1
A2
B
h
rm
Let the liquid of density r be flowing through a pipe of area of cross section A 1. Let A2 be the area of cross
section at the throat and a manometer is attached as shown in the figure. Let v 1 and P1 be the velocity of the
flow and pressure at point A, v 2 and P2 be the corresponding quantities at point B.
Using Bernoulli’s theorem :
P1
1
P
1
+ gh1 + v12 = 2 + gh2 + v 22
r
2
r
2
P1
1
P
1
1
+ gh + v 12 = 2 + gh + v 22 (Since h = h = h) or (P –P )= r(v 2 – v 2)
1
2
1
2
1
2
r
r
2
2
2
we get
....(i)
æ A1 ö
According to continuity equation, A1v1 = A2v2 or v 2 = ç A ÷ v1
è 2ø
Substituting the value of v2 in equation (i)
1
éæ A ö 2
ù1
éæ A ö 2
ù
v 12 - v12 ú rv12 êç 1 ÷ - 1ú
êëè 2 ø
úû 2
êëè A 2 ø
úû
( P1 - P2 ) = 2 r = êç A 1 ÷
we have
Since A1 > A2, therefore, P1 > P2 or v 12 =
2(P1 - P2 )
éæ A ö 2
ù
r êç 1 ÷ - 1 ú
êëè A 2 ø
úû
=
2A 22 (P1 - P2 )
r (A12 - A22 )
where (P1 – P2) = rmgh and h is the difference in heights of the liquid levels in the two tubes.
v1 =
2rm gh
éæ A ö 2
ù
r ê ç 1 ÷ - 1ú
êëè A2 ø
úû
The flow rate (R) i.e., the volume of the liquid flowing per second is given by R = v 1A1.
2.4
Torricelli's Law of Efflux (Fluid Outflow)
As shown in the figure since the area of cross–section at A is very large
as compared to that at orifice B, speed at A i.e. vA » 0. Also the two
fluid particles at A and B are at same pressure P0 (atmospheric pressure).
Applying Bernoulli's theorem at A and B.
P0 + rgH +
1
1 2
1
rv A = P0 + rg(H - h) + rv 2B Þ rv 2B = rgh Þ vB =
2
2
2
2gh
Equation is same as that of freely falling body after falling through h height and is known as Torricelli's law.
Writing equation of uniformly accelerated motion in vertical direction
H–h=0+
152
1 2
2(H - h)
1
gt (from sy = uyt + ayt2) Þ t =
2
g
2
Fluid Mechanics
2(H - h)
= 2 h(H - h)
g
Range R will be maximum when R2 is maximum. i.e.,
Horizontal range
R = v xt =
2gh ×
d
d
H
R2 = 0 Þ 4
(Hh–h2) = 0 Þ H–2h = 0, i.e., h =
dh
dh
2
( )
\ Rmaximum = 2
Hé
Hù
H- ú = H
ê
2ë
2û
Illustration 18. A cylindrical tank 1 m in radius rests on a platform 5 m high. Initially the tank is filled with water
upto a height of 5 m. A plug whose area is 10 –4 m2 is removed from an orifice on the side of the tank
at the bottom. Calculate
A
5m
A0
5m
(a) initial speed with which the water flows from the orifice
(b) initial speed with which the water strikes the ground
Solution
(a) Speed of efflux v H =
(2gh) =
2 ´ 10 ´ 5 ; 10 m/s
(b) As initial vertical velocity of water is zero, so its vertical velocity when it hits the ground
vv= 2gh = 2 ´ 10 ´ 5 ; 10 m/s
So the initial speed with which water strikes the ground, v = v 2H + v 2V =10Ö2 = 14.1 m/s
B
container
of area A
Illustration 19.In a given arrangement
(a) Find out velocity of water coming out of ‘C’
(b) Find out pressure at A, B and C.
Solution
(a) Applying Bernoulli’s equation between liquid surface
and point ‘C’.
h
h2
h1
A
h3
liquid r
1 2
1 2
pa + rv 1 = pa – rgh3 + rv 2
2
2
through continuity equation
v 22 =
2gh3
, v2 =
a2
1- 2
A
Av1 = av2 , v1 =
Pressure at C :
av 2
1 2
1 a2 2
Þ r 2 v 2 = – rgh3 + rv 2
2
A
2 A
2gh3
a2
1- 2
A
(b) Pressure at A just outside the tube
For pressure at B :
C
area of cross
section a
B
h2
pA = patm + rgh1
1
pA + 0 + 0 = pB + rgh2 + rvB2
2
pB = PA – rgh2 – 1 r æ 2gh3 ö
2 ç
a2 ÷
ç1 - 2 ÷
è
A ø
pC = patm
h
h1
v1
A
h3
C
153
v2
JEE-Physics
Illustration 20. A tank is filled with a liquid upto a height H. A small hole is made at the bottom of this tank. Let t 1
be the time taken to empty first half of the tank and t 2 time taken to empty rest half of the tank, then
t1
.
t2
Let at some instant of time the level of liquid in the tank is y. Velocity of efflux at this instant of time
find
Solution
v=
æ dV1 ö
2gy . Now, at this instant volume of liquid coming out the hole per second is çè dt ÷ø
æ dV2 ö
Volume of liquid coming down in the tank per second is ç
è dt ÷ø
æ - dy ö
dV1
dV2
=
Þ av=A ç
è dt ÷ø
dt
dt
æ dy ö
\ a 2gy =A çè - dt ÷ø
...(i)
(Here area of cross-section of hole and tank are respectively a and A)
Substituting the proper limits in equation (i),
ò
t1
0
dt = -
=
A
a 2g
ò
H/2
H
2A
H
é yù
y -1/ 2 dy Þ t1 =
a 2g ë û H / 2
2A é
Hù
A
ê Hú
2 úû = a
a 2g ëê
Similarly,
ò
t2
0
dt = -
A
a 2g
ò
0
H/2
H
g
(
2 -1
)
...(ii)
y -1 / 2 dy Þ t2 =
H
g
A
a
...(iii)
t1
From equations (ii) and (iii), t = Ö2–1 = 0.414
2
Illustration 21.A fixed container of height ' H ' with large cross-sectional area ' A ' is completely filled with water.
Two small orifice of cross-sectional area ' a ' are made, one at the bottom and the other on the
vertical side of the container at a distance H/2 from the top of the container. Find the time taken
by the water level to reach a height of H/2 from the bottom of the container.
Solution
v1 =
2 g (h - H / 2) ; v2 =
\ By continuity equation
2gh
æ d hö
A çè - d t ÷ø = a (v1 + v2)
æ d hö
=a
Þ A çè d t ÷ø
{
a
2 g (h - H / 2) + 2 g h
}
H/2
H/2
v2
or -
A
a 2g
H/2
ò
H
dh
h +
h - H/2
v1
h
t
=ò
0
2A
dt Þ t =
3a
(
)
2 -1
a
H
g
Illustration 22.A cylindrical vessel filled with water upto a height of 2 m stands on a horizontal plane. The side
wall of the vessel has a plugged circular hole touching the bottom. Find the minimum diameter of
the hole so that the vessel begins to move on the floor if the plug is removed. The coefficient of
friction between the bottom of the vessel and the plane is 0.4 and total mass of water plus vessel
is 100 kg.
Solution
From Torricelli's theorem, velocity of efflux =
2gh
Momentum per second carried by water stream
= density × volume coming out per second × velocity = r × av × v = rav2
154
Fluid Mechanics
Hence force on cylindrical vessel = ra2gh
Cylinder starts to move when reaction force is just equal to maximum force of friction.
i.e., mMg=ra2gh Þ a =
Area of circular hole =
mM
0.4 ´ 100
=
= 0.01 m2
2rh
2 ´ 103 ´ 2
pd 2
= 0.01 m2 Þ d =
4
0.01 ´ 4
= 0.113m
p
2.5
Illustrations of Bernoulli's Theorem
•
Magnus Effect (Spinning Ball)
Tennis and cricket players usually experience that when a ball is thrown spinning it moves along a curved path.
This is called swing of the ball. This is due to the air which is being dragged round by the spinning ball. When
the ball spins, the layer of the air around it also moves with the ball. So, as shown in figure the resultant velocity
of air increases on the upper side and reduces on the lower side.
force on ball
speed of air
flow increases
\pressure reduced
spin
speed of air
flow decreases\pressure increased
motion of a spin ball
Hence according to Bernoulli's theorem the pressure on the upper side becomes lower than that on the lower
side. This pressure difference exerts a force on the ball due to which it moves along a curved path. This effect
is known as Magnus–effect.
•
Motion of the Ping–Pong Ball
When a ping–pong ball is placed on a vertical stream of water– fountain, it rises
upto a certain height above the nozzle of the fountain and spins about its axis.
The reason for this is that the streams of water rise up from the fountain with very
large velocity so that the air–pressure in them decreases. Therefore, whenever the
ball goes out from the streams, the outer air which is at atmospheric pressure
pushes it back into the streams ( in the region of low pressure). Thus the ball
remains in stable equilibrium on the fountain.
air
If we blow air at one end of a narrow tube, the air emerges from the other end at high speed and so the pressure
falls there. If a ping–pong ball is left free slightly below this end, it does not fall down due to the large pressure
(atmospheric) below the ball. Similarly, if we blow air in between two ping–pong balls suspended by light
threads near each other, the balls come close to each other due to the decrease of air pressure between them.
Same is the reason that when air is blown below a pan of a physical balance the pan is depressed down.
•
Aerofoil
This is a device which is shaped in such a way so that the relative motion between it and a fluid produces a
force perpendicular to the flow. As shown in the figure the shape of the aerofoil section causes the fluid to flow
faster over the top surface then over the bottom i.e. the streamlines are closer above than below the aerofoil.
By Bernoullis theorem the pressure above is reduced and that underneath is increased.
155
JEE-Physics
high speed, reduced pressure
aerofoil
lift
low speed, increased
pressure
principle of an aerofoil
Thus a resultant upward force is created normal to the flow and it is this force which provides most of lift
upward force for an aeroplane. Illustrations of aerofoils are aircraft wings, turbine blades and propellers.
•
Pull–in or Attraction Force by Fast Moving Trains
If we are standing on a platform and a train passes through the platform with very high speed we are pulled
towards the train. This is because as the train comes at high speed, the pressure between us and the train
decreases. Thus the air behind us which is still at atmospheric pressure pushes us towards the train. The reason
behind flying–off of small papers, straws and other light objects towards the train is also the same.
•
Sprayer or Atomizer
This is an instrument used to spray a liquid in the form of small droplets (fine spray). It consists of a vertical
tube whose lower end is dipped in the liquid to be sprayed, filled in a vessel.
spray
Rubber
bulb
air flowing out with
high velocity
The upper end opens in a horizontal tube. At one end of the horizontal tube there is a rubber bulb and at the
other end there is a fine bore (hole). When the rubber bulb is squeezed, air rushes out through the horizontal
tube with very high velocity and thus the pressure reduces (according to Bernoulli's theorem). So the liquid rises
and comes out through narrow end in form of droplets. It is used in spray gun, perfumes, deodorant and etc.
•
•
Bunsen's Burner (Jet)
It is based on the working of jet. It consists of a long tube having a fine nozzle O
at the bottom. The burning gas enters the tube through O and burns in a flame at
the top of the tube.
To produce a non–luminous flame, the air of the atmosphere is mixed with the
gas. Since the nozzle O is fine, the gas enters with a large velocity and so the
pressure inside the tube is lowered than the outer atmospheric pressure. Therefore,
air from outside rushes into the tube through a hole and is mixed with the burning
gas.
Filter Pump (Nozzle)
It is based on the working of nozzle. It consists of a wide tube M N in the upper
part of which there is another tube A. The upper end of A is connected to a
water tank, while its lower end has a fine bore through which water comes out
in the form of a jet. The vessel which is to be evacuated is connected to the tube
M N as shown.
The velocity of the emerging water jet is very large. Therefore, the pressure of
the air near the jet becomes less than the pressure in the vessel.
Hence air from the vessel rushes into the tube M N and is carried out along
with the water jet. Thus partial vacuum is created in the vessel.
156
flame
gas +air
air
nosel
gas
water
M
A
vessel
air
water jet
N
water and air
Fluid Mechanics
•
Blowing–off of Tin Roof Tops in Wind Storm
When wind blows with a high velocity above a tin roof, it causes lowering of pressure above the roof, while the
pressure below the roof is still atmospheric.
Due to this pressure–difference the roof is lifted up.
P
wind
P0
1.
A water tank stands on the roof of a building as shown. Then the value of h for which the distance covered by
the water 'x' is greatest is
(A) 0.5 m
2.
4.
v large
So P<P0
(B) 0.67 m
(C) 1 m
(D) none of these
The velocity of the liquid coming out of a small hole of a large vessel containing two different liquids of
densities 2r and r as shown in figure is
(A) 6gh
(B) 2 gh
(C) 2 2gh
(D) gh
Fountains usually seen in gardens are generated by a wide pipe with an enclosure at one end having many small
holes. Consider one such fountain which is produced by a pipe of internal diameter 2 cm in which water flows
at a rate 3ms-1. The enclosure has 100 holes each of diameter 0.05 cm. The velocity of water coming out of the
holes is (in ms-1)
(A) 0.48
(B) 96
(C) 24
(D) 48
157
JEE-Physics
5.
A fixed container of height ' H ' with large cross-sectional area ' A ' is completely filled with water. Two small
orifice of cross-sectional area ' a ' are made, one at the bottom and the other on the vertical side of the
container at a distance H/2 from the top of the container. Find the time taken by the water level to reach
a height of H/2 from the bottom of the container.
6.
Water flows through a frictionless duct with a cross-section varying
as shown in figure. Pressure p at points along the axis is represented
by:
7.
(A)
(B)
(C)
(D)
A light semi cylindrical gate of radius R is piovted at its mid point O, of the diameter
as shown in the figure holding liquid of density r. The force F required to prevent the
rotation of the gate is equal to
(A) 2pR3rg
8.
(B) 2rgR3l
(C)
2R2 lrg
3
(D) none of these
A slender homogeneous rod of length 2L floats partly immersed in water, being supported
by a string fastened to one of its ends, as shown. The specific gravity of the rod is 0.75.
The length of rod that extends out of water is:
(A) L
(B)
1
L
2
(C)
1
L
4
(D) 3 L
9.
A cylindrical block of area of cross–section A and of material of density r is placed
in a liquid of density one–third of density of block. The block compresses a spring
and compression in the spring is one–third of the length of the block. If acceleration
due to gravity is g, the spring constant of the spring is:
(A) rAg
(B) 2rAg
(C) 2rAg/3
(D) rAg/3
10.
A fire hydrant delivers water of density r at a volume rate L. The water travels
vertically upward through the hydrant and then does 90° turn to emerge horizontally
at speed V. The pipe and nozzle have uniform crosssection throughout. The force
exerted by the water on the corner of the hydrant is
(A) rVL
158
(B) zero
(C) 2rVL
(D) 2rVL
Fluid Mechanics
SOME WORKED OUT ILLUSTRATIONS
Illustration 1.
The pressure of water in a water pipe when tap is opened and closed is respectively 3 × 105 Nm–2 and
3.5 × 105 Nm–2. With open tap, the velocity of water flowing is
(A) 10 m/s
Ans. (A)
Solution
Popen +
(B) 5 m/s
(C) 20 m/s
(D) 15 m/s
2 ( Pclosed - Popen )
1 2
2 ´ ( 3.5 - 3) ´ 105
rv = Pclosed Þ v =
=
= 10 m/s
r
2
103
Illustration 2.
A large cylindrical tank of cross-sectional area 1m2 is filled with water. It has a small
hole at a height of 1m from the bottom. A movable piston of mass 5 kg is fitted on the
top of the tank such that it can slide in the tank freely. A load of 45 kg is applied on the
top of water by piston, as shown in figure. The value of v when piston is 7m
above the bottom is (g = 10 m/s2)
(A) 120 m/s
Ans. (D)
Solution
(B) 10 m/s
(C) 1 m/s
45kg
v
(D) 11 m/s
1 2
Mg
2Mg
2 ´ 50 ´ 10
rv = rgh +
Þ v = 2gh +
= 2 ´ 10 ´ 6 +
= 120 + 1 = 121 = 11m / s
rA
2
A
103 ´ 1
Illustration 3.
An open vessel full of water is falling freely under gravity. There is a small hole in one face of the vessel, as
shown in the figure. The water which comes out from the hole at the instant when hole is at height H above the
ground, strikes the ground at a distance of x from P. Which of the following is correct for the situation described?
(A) The value of x is 2
(B) The value of x is
2hH
3
2h/3
4hH
3
H
g
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
(C) The value of x can't be computed from information provided.
Ground
P
(D) The question is irrevalent as no water comes out from the hole.
Ans. (D)
Solution
As vessel is falling freely under gravity, the pressure at all points within the liquid remains the same as the
atmospheric pressure. If we apply Bernoulli's theorem just inside and outside the hole, then
Pinside +
rv 2inside
rv 2
+ rgeff y = poutside + outside + rgeff y
2
2
vinside = 0 , pinside= poutside = p0
[atmospheric pressure]
Therefore, voutside = 0. i.e., no water comes out.
159
JEE-Physics
Illustration 4.
A cuboid (a × a × 2a) is filled with two immiscible liquids of density 2r & r as
shown in the figure. Neglecting atmospheric pressure, ratio of force on base &
side wall of the cuboid is
(A) 2:3
(B) 1:3
(C) 5:6
(D) 6:5
Ans. (D)
Solution
B
a
A
r
C
2r
a
a
a
Fb = ( 2rgh + rgh) a = 3rgha ;
2
2
5
é rgh ù
Fw = ê
ah + ( 2rgh ) ah = rgha2 [here h=a];
ú
ë 2 û
2
Fb
3
æ 6ö
=
=ç ÷
Fw 5 / 2 è 5 ø
Illustration 5.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height
must the blood container be placed so that blood may just enter the vein ?
[Density of whole blood = 1.06 × 103 kg m–3].
(A) 0.192 m
(B) 0.182 m
(C) 0.172 m
(D) 0.162 m
Ans. (A)
Solution
2000
Pressure P = hrg Þ h =
= 0.192m
1.06 ´ 103 ´ 9.8
Illustration 6.
a
A liquid of density r is filled in a U-tube, whose one end is open & at the other end
a bulb is fitted whose pressure is PA. Now this tube is moved horizontally with
acceleration 'a' as shown in the figure. During motion it is found that liquid in both
column is at same level at equilibrium. If atmospheric pressure is P 0 ,
then value of PA is
(A) ral
(B) rgl
(C) P0–ral
(D) P0 + ral
Ans. (C)
Solution
Consider a point B at the horizontal portion of tube, pressure from both side should be same.
PA
PA
\ P0 + rgh = PA + rgh + ral, PA = P0 – ral
h
h
B
Illustration 7.
The tube shown is of non-uniform cross-section. The cross-section area at A is half of the cross-section area at
B,C and D. A liquid is flowing through in steady state.The liquid exerts on the tube-
C
160
D
Fluid Mechanics
Statement I : A net force towards right.
Statement II : A net force towards left.
Statement III : A net force in some oblique direction.
Statement IV : Zero net force
Statement V: A net clockwise torque.
Statement VI: A net counter-clockwise torque.
Out of these
(A) Only statement I and V are correct
(B) Only statement II and VI are correct
(C) Only statement IV and VI are correct
(D) Only statement III and VI are correct
Ans. (A)
Solution
dSv 2
The force has been exerted by liquid on the tube due to change in momentum
2
2
at the corners i.e., when liquid is taking turn from A to B and from B to C.As
dSv
corss-section area at A is half of that of B and C, so velocity of liquid flow at B
and C is half to that of velocity at A. Let velocity of flow of liquid at A be v and
2
cross section area at A be S, the velocity of flow of liquid at B and C would be
dSv
v
[from continuity equation] and corss section area at B and C would be 2S.
2
2
dSv 2
2
Due to flow of liquid, it is exerting a force per unit time of rSv2 on the tube, where r is the density of liquid, S is cross
section areaand v is velocity of flow of liquid. The force exerted by liquid on tube is shown in the figure. Which
clearly shows that a net force is acting on tube due to flowing liquid towards right and a clockwise torque sets in
Illustration 8.
A hole is made at the bottom of a large vessel open at the top. If water is filled to a height h, it drains out
completely in time t. The time taken by the water column of height 2h to drain completely is
(A) Ö2t
(B) 2t
(C) 2Ö2t
(D) 4t
Ans. (A)
Solution
dh ö
æ
Here çè –A
= (av) = a 2gh
dt ÷ø
h
Þ ò h –1 2 dh =
0
t
A 2h
a
2g ò dt Þ t =
Þ t µ Öh
A
a g
0
Therefore for height 2h t' µ Ö2h Þ t' = Ö2 t
Illustration 9.
A flat plate moves normally with a speed v 1 towards a horizontal jet of water of uniform area of cross-section.
The jet discharges water at the rate of volume V per second at a speed of v 2. The density of water is r. Assume
that water splashes along the surface of the plate at right angles to the original motion. The magnitude of the
force acting on the plate due to jet of water is :–
(A) rVv1
æ Vö
2
(B) r ç v ÷ (v1 +v 2 )
è 2ø
rV
2
(C) v + v v1
1
2
(D) rV(v1+v2)
Ans. (D)
Solution
F=
m
Dp m(v1 + v 2 )
=
= Vr(v1 + v 2 ) where
= V = volume/sec
Dt
Dt
Dt
Illustration 10.
A tank full of water has a small hole at its bottom. Let t 1 be the time taken to empty first one third of the tank
and t2 be the time taken to empty second one third of the tank and t 3 be the time taken to empty rest of the tank
then
(A) t1 = t2 = t3
(B) t1 > t2 > t3
(C) t1 < t2 < t3
(D) t1 > t2 < t3
161
JEE-Physics
Ans. (C)
Solution
As the height decreases, the rate of flow with which the water is coming out decreases.
Illustration 11.
A closed cylinder of length 'l' containing a liquid of variable density r(x) =r0 (1+ax).
Find the net force exerted by the liquid on the axis of rotation. (Take the cylinder to
be massless and A = cross sectional area of cylinder)
é1 1 ù
(A) r0 Aw l ê + al ú
ë2 3 û
é1 2 ù
(B) r0 Aw l ê + al ú
ë2 3 û
2 2
x
dx
l
A
2 2
ù
2 2 é1
(C) r0 Aw l ê + al ú
ë2
û
Ans. (A)
Solution
l
w
4 ù
2 2 é1
(D) r0 Aw l ê + al ú
ë2 3 û
dm = rAdx; dF = (dm)w2 x
l
æ1 1 ö
Þ F = ò w2 xrAdx = w2r0 A ò (1 + ax)x dx = r0 Aw 2 l2 ç + al÷
è2 3 ø
0
0
Illustration 12
A solid right cylinder of length l stands upright at rest on the bottom of a large tub filled with water up to height
h as shown in the figure-I. Density of material of the cylinder equals to that of water. Now the cylinder is pulled
slowly out of water with the help of a thin light inextensible thread as shown in figure-II. Find the work done by
the tension force develop in the thread.
h
l
Figure-I
(A) mgh
Ans. (C)
(B) mgl
Figure-II
(C) 0.5 mgl
(D) mg(0.5l+h)
Solution
Work done by tension force = work done by gravity = 0.5 mgl
Illustration 13.
Some pieces of impurity (density =r) is embedded in ice. This ice is floating in
water (density = rw). When ice melts, level of water will
(A) fall if r > rw
(B) remain unchanged, if r < rw
(C) fall if r < rw
(D) rise if r > rw
Ans. (A,B)
Solution
Level will fall if initially the impurity pieces were floating along will ice and later it sinks.
Level will remain unchaged if initially they were floating and later also they keep floating.
162
Fluid Mechanics
Illustration 14 to 16.
Velocity of efflux in Torricelli's theorem is given by v = 2gh , here h is the height of hole from the top surface,
after that, motion of liquid can be treated as projectile motion.
(ii)
2m
(i)
30°
O
14.
Liquid is filled in a vessel of square base (2m × 2m) upto a height of 2m as shown in figure (i). In figure (ii)
the vessel is tilted from horizontal at 30°. What is the velocity of efflux in this case. Liquid does not spills out?
(A) 3.29 m/s
(B) 4.96 m/s
(C) 5.67 cm
(D) 2.68 m/s
15.
What is its time of fall of liquid on the ground?
(A)
16.
1
2
s
(B)
1
3
s
(C)
1
5
s
(D) Ö2s
At what distance from point O, will be liquid strike on the ground?
(A) 5.24 m
(B) 6.27 m
(C) 4.93 m
(D) 3.95 m
Solution
14. Ans. (B)
The volume of liquid should remain unchanged. Hence, 2 × 2 × 2 =
15.
\ x » 1.42m
Now h = x sin 60° = 1.23 m \ v = 2gh = 2 ´ 10 ´ 1.23 = 4.96m / s
Ans. (C)
v
2m
O
16.
1é
2 ù
x+x+
ê
ú´2
2ë
3û
30°
H = 2 sin 30° = 1m
H
A
\t =
2H
1
=
s
g
5
B
Ans. (D)
OA = 2 cos 30° - 3m Þ AB = vt =
( 4.96 )
5
s
\ OB = 3.95 m
Illustration 17 to 19.
A cylindrical container of length L is full to the brim with a liquid which has mass density r. It is placed on a
weight-scale; the scale reading is w. A light ball which would float on the liquid if allowed to do so, of volume
V and mass m is pushed gently down and held beneath the surface of the liquid with a rigid rod of negligible
volume as shown on the left.
rigid rod
L
17.
What is the mass M of liquid which overflowed while the ball was being pushed into the liquid?
(A) rV
(B) m
(C) m – rV
(D) none of these
163
JEE-Physics
18.
What is the reading of the scale when the ball is fully immersed
(A) w – rVg
(B) w
(C) w + mg – rVg
(D) none of these
19.
If instead of being pushed down by a rod, the ball is held in place by a thin string attached to the bottom of the
container as shown on the right. What is the tension T in the string?
(A) (rV–m)g
(B) rVg
(C) mg
(D) none of these
Solution
17. Ans. (A)
Ball will displace V volume of liquid, whose mass in rV.
18.
Ans. (C)
Weight mg is entered while weight rVg of liquid is overflowed.
19.
Ans. (A)
F
F = T + mg Þ Vrg = T + mg \ T = (Vr – m)g
T
mg
Illustration 20.
An ideal liquid is flowing in a tube as shown in figure. Area of cross-section at points A, B and C are A 1, A2 and
A3 respectively (A1 > A3 > A2). v1, v2, > v3 are the velocities at the points A, B and C respectively.
h1
A
h2
h3
C
B
Column II
(P) Less than h3
(Q) More than h3
(R) Less than v3
(S) More than v3
(T) The maximum value amongst the three velocities.
Ans. (A) Q (B) P (C) R (D) ST
Solution
By using A1v1 = A2v2 Q A2 < A3 < A1 \ v2 > v3 > v1. Now by using P + ½ rv2+ rgh = constant
We have P2 < P3 < P1 so h1 > h3 and h2 < h3
(A)
(B)
(C)
(D)
Column I
h1 is
h2 is
v1 is
v2 is
O'
Illustration 21.
A horizontal oriented tube AB of length 5 m rotates with a constant angular
velocity 0.5 rad/s about a stationary vertical axis OO' passing through the end
A. The tube is filled with ideal fluid. The end of the tube is open, the closed
end B has a very small orifice. The velocity with which the liquid comes out
from the hole (in m/s) is
Ans. 2
Solution
Apply Bernoullie theorem between two ends
5
P0 + rò xw 2 dx = P0 +
3
164
w
B
A
2m
5m
O
2
1 2
rv Þ rw [ 52 - 32 ] = 1 rv 2 Þ (0.5)2 × 16 = v2 Þ v = 2m/s
2
2
2
Fluid Mechanics
Illustration 22.
A ball of density r0 falls from rest from a point P onto the surface of a liquid of density r in time T. It enters the
liquid, stops, moves up, and returns to P in a total time 3T. Neglect viscosity, surface tension and splashing. Find
the ratio of
r
.
r0
Ans. 3
Solution
It strike the surface of liquid with velocity v1 = gT
P
t=0
t=T
2v1
2gT
r
=
Þ
=3
In water (liquid) its time of flight T =
r0
æ r
ö
æ r
ö
1
g
1
g
çè r
÷ø
çè r
÷ø
0
0
Illustration 23.
A cylinder fitted with piston as shown in figure. The cylinder is filled
with water and is taken to a place where there is no gravity. Mass of the A
piston is 50 kg. The system is accelerated with acceleration 0.5 m/sec 2
in positive x–direction. Find the force exerted by fluid on the surface
AB of the cylinder in decanewton. Take area of cross–section of cylinder
to be
B
5m
0.01 m2 and neglect atmospheric pressure (1 decanewton =10N)
Ans. 5
Solution
Force due to piston = 50 × 0.5 = 25 N
Force due to fluid = (rah) A=A (1000 × 5 × 0.5) = 2500 Pa × 0.01 m2 = 25 N
Force on the surface AB = 50N = 5 decanewton
t=2T
3T
t= 2
D
y
x
C
Illustration 24.
Water (density r) is flowing through the uniform tube of cross-sectional area A
with a constant speed v as shown in the figure. Find the magnitude of force exerted
by the water on the curved corner of the tube is (neglect viscous forces)
Solution
P
60°
v
uuur
3
|DP x |= mv sin 60° =
mv
2
uuur
mv
3
|DP y |=
+ mv = mv
2
2
t=3T
A
v
v
uuur
æ 9 3ö
2
2
Þ |DP net |= DPx + DPy = ç + ÷ mv
è 4 4ø
uuur
uuur
æ dm ö
|DP net |= 3 mv Þ |DF net |= 3 çè dt ÷ø . v =
3 r A v2
y
x
dm
æ
ö
çè Since, dm = A (v dt) r Þ dt = A r v ÷ø
ANSWERS
BEGINNER'S BOX-1
1. (D)
6.
Sgd 2H 2
2r
2. (D)
3. (D)
7. m = 300 gm.
lö
æ
4. Pa = P0 + d1g ç h + ÷
3ø
è
8. (B)
9. (A)
5. (D)
10. (A)
BEGINNER'S BOX-2
1. (C)
2. (B)
3. (D)
7. (A)
8. (B)
9. (D)
4. t =
2A
3a
(
)
2 -1
H
g
5. (A)
6. (D)
165
JEE-Physics
1.
A crown made of gold and copper weights 210 g in air and 198 g in water. The weight of gold in crown is:–
[Given : Density of gold = 19.3 g/cm3 and Density of copper = 8.5 g/cm3]
(A) 93 g
(B) 100 g
(C) 150 g
(D) 193 g
2.
Two vessels A and B have the same base area and contain water to the same height, but the mass of water
in A is four times that in B. The ratio of the liquid thrust at the base of A to that at the base of B is :–
B
l
(A) 4 : 1
3.
(B) 2 : 1
(C) 1 : 1
(D) 16 : 1
Water stands upto a height h behind the vertical wall of a dam. What is the net horizontal force pushing the
dam down by the stream, if width of the dam is s ? ( r = density of water) :–
(A) 2hsg
(B)
h2 srg
2
(C)
h2srg
4
(D)
hsrg
4
4.
The side of glass aquarium is 1 m high and 2 m long. When the aquarium is filled to the top with water, what
is the total force against the side–
(A) 980 × 103 N
(B) 9.8 × 103 N
(C) 0.98 × 103 N
(D) 0.098 × 103 N
5.
The gauge pressure of 3 ´ 105 N/m2 must be maintained in the main water pipes of a city. How much work
must be done to pump 50,000 m3 of water at a pressure of 1.0 ´ 105 N/m2–
(A) 1011 J
(B) 1010 J
(C) 109 J
(D) 108 J
6.
The barometric pressure and height on the earth are 105 Pa and 760 mm respectively. If it is taken to moon,
then barometric height will be :–
(A) 76 mm
(B) 126.6 mm
(C) Zero
(D) 760 mm
7.
When a large bubble rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure
is equal to that of column of water height H, then the depth of lake is :–
(A) H
(B) 2H
(C) 7H
(D) 8H
8.
A body floats in a liquid contained in a beaker. The whole system is shown in figure
falling under gravity. The upthrust on the body due to liquid is :–
(A) zero
(B) equal to weight of liquid displaced
(C) equal to weight of the body in air
(D) equal to the weight of the immersed body.
9.
A boat having a length of 3 metre and breadth 2 metre is floating on a lake. The boat sinks by one cm when
a man gets on it. The mass of the man is
(A) 60 kg
(B) 62 kg
(C) 72 kg
(D) 128 kg
10.
A body of volume 100 c.c. is immersed completely in water contained in a jar. The weight of water and the
jar before immersion of the body was 700g wt. After immersion weight of water and jar will be
(A) 700 g wt
(B) 800 g wt
(C) 500 g wt
(D) 100 g wt
166
Fluid Mechanics
11.
The total weight of a piece of wood is 6 kg In the floating state in water its 1/3part remains inside the water.
On this floating solid, what maximum weight is to be put such that the whole of the piece of wood is to be
drowned in the water :–
(A) 12 kg
(B) 10 kg
(C) 14 kg
(D) 15 kg
12.
A sample of metal weights 210 grams in air, 180 grams in water and 120 grams in an unknown liquid. Then:–
(A) the density of metal is 3 g/cm3
(B) the density of metal is 7 g/cm3
(C) density of metal is 4 times the density of the unknown liquid
(D) the metal will float in water
13.
A wooden cube first floats inside water when a 200 g mass is placed on it. When the mass is removed the cube
is 2 cm above water level. The side of cube is :–
(A) 5 cm
(B) 10 cm
(C) 15 cm
(D) 20 cm
14.
A piece of ice with a stone frozen in it on water is kept in a beaker. The level of water when ice completely melts–
(A) Increase
(B) Decrease
(C) Remain the same
(D) None of these
15.
A rectangular block is 5 cm ´ 5cm ´ 10 cm. The block is floating in water with 5 cm side vertical. If it floats
with 10 cm side vertical, what change will occur in the level of water :–
(A) No change
(B) It will rise
(C) It will fall
(D) It may rise or fall depending on the density of block.
16.
A wooden block of volume 1000 cm3 is suspended from a spring balance. Its weight is 12 N in air. It is suspended
in water such that half of the block is below the surface of water. The reading of spring balance is :–
(A) 10 N
(B) 9 N
(C) 8 N
(D) 7 N
17.
A boat carrying a number of large stones is floating in a water tank. What will happen to the water level if the
stones are unloaded into the water :–
(A) Rise
(B) Fall
(C) Remain unchanged
(D) Rise till half the number of stones are unloaded and then begin to fall
18.
An aeroplane of mass 3 × 104 kg and total wing area of 120 m2 is in a level flight at some height. The difference
in pressure between the upper and lower surfaces of its wings in kilo pascal is (g=10m/s2 ) :–
(A) 2.5
(B) 5.0
(C) 10.0
(D) 12.5
19.
One end of a horizontal pipe is closed with the help of a valve and the reading of a barometer attached to
the pipe is 3 × 105 pascal. When the value in the pipe is opened then the reading of barometer falls to
105 pascal. The velocity of water flowing through the pipe will be in m/s :–
(A) 0.2
(B) 2
(C) 20
(D) 200
20.
A tank of height 5 m is full of water. There is a hole of cross sectional area 1 cm2 in its bottom. The initial
volume of water that will come out from this hole per second is :–
(A) 10–3 m3/s
(B) 10–4 m3/s
(C) 10 m3/s
(D) 10–2 m3/s
21.
In the figure below is shown the flow of liquid through a horizontal pipe. Three tubes A, B and C are connected
to the pipe. The radii of the tubes, A, B and C at the junction are respectively 2 cm, 1 cm. and 2 cm. It can
be said that the :–
A
B
C
(A) Height of the liquid in the tube A is maximum
(B) Height of the liquid in the tubes A and B is the same
(C) Height of the liquid in all the three tubes is the same
(D) Height of the liquid in the tubes A and C is the same
167
JEE-Physics
22.
The diagram (figure) shows a venturimeter, through which water is flowing. The speed of water at X is 2 cm/s.
The speed of water at Y (taking g = 1000 cm/s2) is :–
5.1mm
x
(A) 23 cm/s
(B) 32 cm/s
y
(C) 101 cm/s
(D) 1024 cm/s
23.
Water contained in a tank flows through an orifice of a diameter 2 cm, under a constant pressure difference
of 10 cm of water column. The rate of flow of water through the orifice is:–
(A) 44 cc/s
(B) 4.4 cc/s
(C) 444 cc/s
(D) 4400 cc/s
24.
One end of light inelastic string is tied to a helium filled balloon and its other end is tied to bottom of a water
filled container at point O. The container lies on a fixed horizontal surface and is pulled horizontally towards right
with constant horizontal acceleration of magnitude a. Assuming no relative motion of balloon and water with
respect to container, the string will be inclined with vertical line passing through O by an angle. (g is acceleration
due to gravity)
-1
(A) q = tan
25.
a
and string will be on right of vertical line passing through O..
g
-1
(B) q = tan
g
and string will be on right of vertical line pasing through O.
a
-1
(C) q = tan
a
and string will be on left of vertical line passing through O..
g
-1
(D) q = tan
g
and string will be on left of vertical line passing through O..
a
A large open tank has two small holes in its vertical wall as
shown in figure. One is a square hole of side 'L' at a depth '4y'
from the top and the other is a circular hole of radius 'R' at a
depth 'y' from the top. When the tank is completely filled with
water, the quantities of water flowing out per second from both
holes are the same. Then, 'R' is equal to :
(A)
L
2p
(B) 2pL
(C)
2
×L
p
a
right
y
4y
v1
2R
v2
L
(D)
L
2p
26.
A vessel contains oil (density = 0.8 gm/cm3) over mercury (density = 13.6 gm/cm3). A uniform sphere floats with
half its volume immersed in mercury and the other half in oil. The density of the material of sphere in gm/cm 3 is:
(A) 3.3
(B) 6.4
(C) 7.2
(D) 12.8
27.
Following are some statements about buoyant force : (Liquid is of uniform density)
(i) Buoyant force depends upon orientation of the concerned body inside the liquid.
(ii) Buoyant force depends upon the density of the body immersed.
(iii) Buoyant force depends on the fact whether the system is on moon or on the earth.
(iv) Buoyant force depends upon the depth at which the body (fully immersed in the liquid) is placed inside the
liquid.
Of these statements :
(A) Only (i), (ii) and (iv) are correct.
(B) Only (ii) is correct
(C) Only (iii) and (iv) are correct.
(D) (i), (ii) and (iv) are incorrect
168
Fluid Mechanics
28.
29.
Figure shows a metal ball suspended by thread of negligible mass from an
upright cylinder that floats partially submerged in water. The cylinder has
height 6 cm, face area 11 cm2 on the top and bottom and density
0.5 g/cm3. 4 cm of cylinder's height is inside the water surface. If density
of the metal ball is 8 gm/cm3 then its radius is equal to (rw = 1 gm/cm3)
æ 3ö
(A) çè ÷ø
8
1/ 3
æ 6ö
(C) çè ÷ø
8
1/ 3
6 cm
4cm
1/ 3
cm
æ 3ö
(B) çè ÷ø
4
cm
æ 5 ö
(D) ç ÷
è 11 ø
cm
1/ 3
r
cm
Sh
.A
V
very small hole of area S is made at depth 'h'. Water of volume 'V' is filled in the container. The friction is not
sufficient to keep the container at rest. The acceleration of the container initially is
In the figure shown, a light container is kept on a horizontal rough surface of coefficient of friction m =
(A)
V
g
Sh
(C) zero
30.
2
A=11cm
(B) g
(D)
h
Sh
g
V
A fixed container is filled with a liquid of density r up to a height 4m. A horizontal slit of small width but of
area = 0.5m2 is made at a height of 2m form bottom. The speed of top surface of the water level is (area of top
surface of container is 4 m2 and g = 10 m/s2)
(A)
20
m / sec
63
(B)
40
m / sec
63
(C)
80
m / sec
63
(D) None of these
4m
Slit
2m
169
JEE-Physics
1.
The spring balance A read 2 kg with a block m suspended from it. A balance B reads
5 kg when a beaker with liquid is put on the pan of the balance. The two balances
are now so arranged that the hanging mass is inside the liquid in the beaker as shown
in figure. In this situation :–
A
(A) The balance A will read more than 2 kg
m
(B) The balance B will read more than 5 kg
(C) The balance A will read less than 2 kg and B will read more than 5 kg
B
(D) The balance A and B will read 2 kg and 5 kg respectively.
2.
A steel ball is floating in a trough of mercury. If we fill the empty part of the trough with water, what will happen
to the steel ball :–
(A) It will continue in its position
(B) It will move up
(C) It will move down
(D) It will execute vertical oscillations
3.
A balloon filled with air is weighted, so that it barely floats in water as shown in
figure. When it is pushed down so that it gets submerged a short distance in water,
then the balloon :–
(A) Will come up again to its former position
(B) Will remain in the position it is left
(C) Will sink to the bottom
(D) Will emerge out of liquid
4.
A wooden ball of density D is immersed in water of density d to a depth h below the surface of water upto
which the ball will jump out of water is :–
(A)
5.
d
h
D
éd
ù
(B) ê - 1ú h
ëD
û
(C) h
(D) zero
A solid uniform ball having volume V and density r floats at the interface of two
unmixible liquids as shown in figure. The densities of the upper and the lower liquids
are r1 and r2 respectively, such that r1 < r < r2 . What fraction of the volume of
the ball will be in the lower liquid :–
r - r2
(A) r - r
1
2
6.
r1 - r
(C) r - r
1
2
(D)
r1 - r2
r2
The cylindrical tube of a spray pump has a radius R, one end of which has n fine holes, each of radius r. If the
speed of flow of the liquid in the tube is v, the speed of ejection of the liquid through the hole is :–
v éR ù
(A) ê ú
nërû
7.
r1
(B) r - r
1
2
1
v é R ù2
(B) ê ú
nër û
3
v é R ù2
(C) ê ú
nër û
A cylindrical container of radius 'R' and height 'h' is completely filled with a
liquid. Two horizontal L shaped pipes of small cross–section area 'a' are connected
to the cylinder as shown in the figure. Now the two pipes are opened and
fluid starts coming out of the pipes horizontally in opposite directions. Then
the torque due to ejected liquid on the system is–
(A) 4 aghrR
(B) 8 aghrR
(C) 2 aghrR
(D) none of these
170
(D)
v éR ù
n êë r úû
2
2R
h/2
R
h/2
R
Fluid Mechanics
8.
A small uniform tube is bent into a circular tube of radius R and kept in the vertical
plane. Equal volumes of two liquids of densities r and s (r > s) fill half of the
tube as shown. q is the angle which the radius passing through the interface makes
with the vertical. The value of q is :
çè r + s ÷ø
s-rö
÷
(B) q = tan ç
ès+rø
-1 æ r ö
÷
(C) q = tan ç
èr+ sø
-1 æ r ö
÷
(D) q = tan ç
èr-sø
(A) q = tan
9.
10.
-1 æ r - s ö
R
s
-1 æ
q
r
A U–tube of base length 'l' filled with same volume of two liquids of densities
r and 2r is moving with an acceleration 'a' on the horizontal plane. If the height
difference between the two surfaces (open to atmosphere) becomes zero, then
the height h is given by–
(A)
a
l
2g
(B)
3a
l
2g
(C)
a
l
g
(D)
2a
l
3g
h
l
A narrow tube completely filled with a liquid is lying on a series of cylinder
as shown in figure. Assuming no sliding between any surfaces, the value
of acceleration of the cylinders for which liquid will not come out of the
tube from anywhere is given by
(A)
gH
2L
(B)
(C)
2gH
L
(D)
a
open to
atmosphere
L
H
gH
L
a
gH
2L
Match the Column
11.
A cube is floating in a liquid as shown in figure.
Column I
12.
Column II
(A)
If density of liquid decreases then x will
(p)
increase
(B)
If size of cube is increased then x will
(q)
decrease
(C)
If the whole system is accelerated
upwards then x will
(r)
remains same
A solid is immersed completely in a liquid. The coefficients of volume expansion of solid and the liquid are g1 and
g2 (<g1). If temperatures of both are increased, then
Column I
Column II
(A)
Upthrust on the solid will
(p)
increase
(B)
apparent weight of the solid will
(q)
decrease
(C)
Fraction of volume immersed in the
liquid if allowed to float
(r)
remains same
171
JEE-Physics
Comprehension Based questions
Comprehension-1
If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed.
ax
. In case of vertically accelerated
In case of horizontally accelerated liquid (a x), the free surface has the slope
g
liquid (ay) for calculation of pressure, effective g is used. A closed box with horizontal base 6m × 6m and
a height 2m is filled with liquid. It is given a constant horizontal acceleration g/2 and vertical downward
g
acceleration .
2
13.
The angle of the free surface with the horizontal is equal to :
(A) 30°
14.
(B) tan-1
2
3
(C) tan-1
1
3
Length of exposed portion of top of box is equal to:
(A) 2 m
(B) 3 m
(C) 4 m
(D) 45°
(D) 2.5 m
15.
Water pressure at the bottom of centre of box is equal to:
(atmospheric pressure = 105 N/m2, density of water = 1000 kg/m3, g = 10 m/s2)
(A) 1.1 MPa
(B) 0.11 MPa
(C) 0.101 MPa
(D) 0.011 MPa
16.
Maximum value of water pressure in the box is equal to :
(A) 1.4 MPa
(B) 0.14 MPa
(C) 0.104 MPa
17.
(D) 0.014 MPa
What is the value of vertical acceleration of box for given horizontal acceleration (g/2). so that no part of
bottom of box is exposed:
g
g
g
upward
(B)
downward
(C)
upward
(D) Not possible
2
4
4
Comprehension-2
In a U-tube, if different liquids are filled then we can say that pressure at same level of same liquid is same.
(A)
20 cm
10 cm
h
18.
In a U–tube 20 cm of a liquid of density r is on left hand side and 10 cm of another liquid of density 1.5 r
is on right hand side. In between them there is a third liquid of density 2r. What is the value of h.
(A) 5 cm
(B) 2.5 cm
(C) 2 cm
(D) 7.5 cm
19.
If small but equal lengths of liquid –1 and liquid –2 are increased in their corresponding sides then h will.
(A) remain same
(B) increase
(C) decrease
(D) may increase or decrease
Comprehension-3
Newton's laws of motion can be applied to a block in liquid also. Force due to liquid
(e.g., upthrust) are also considered in addition to other forces. A small block of weight
W is kept inside. The block is attached with a string connected to the bottom of the
vessel. Tension in the string is W/2.
20.
21.
2m
The string is cut. Find the time when it reaches the surface of the liquid
1
2
(A)
s
(B) 5 s
s
(C) 3 s
(D)
5
5
If weight of the block is doubled, then tension in the string becomes x times and the time calculated above
becomes y times. Then
(A) x = 2
(C) both (A) and (B) are correct
172
(B) y =
2
(D) both (A) and (B) are wrong
Fluid Mechanics
1.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of
cross–section of the piston carrying the load is 425 cm2. What maximum pressure would the piston have
to bear? (taking g = 10 m/s2)
2.
For the system shown in the figure, the cylinder on the left at L has a mass of 600 kg and a cross sectional
area of 800 cm2. The piston on the right, at S, has cross sectional area 25 cm2 and negligible weight. If the
apparatus is filled with oil ( r = 0.75 gm/cm3) Find the force F required to hold the system in equilibrium.
S
F
600kg
8m
3.
L
A piston of mass M = 3kg and radius R = 4cm has a hole into which a thin pipe of radius
r = 1cm is inserted. The piston can enter a cylinder tightly and without friction, and initially
it is at the bottom of the cylinder. 750 gm of water is now poured into the pipe so that
the piston & pipe are lifted up as shown. Find the height H of water in the cylinder and
height h of water in the pipe.
h
H
An air bubble doubles its volume as it rises from the bottom of a tank to its surface. If the atmospheric pressure
be 76 cm of Hg, what is the depth of the tank ?
5.
Two identical cylindrical vessels with their bases at the same level each contain a liquid of density r. The height
of the liquid in one vessel is h1 and in the other is h2. The area of either base is A. What is the work done
by gravity in equalising the levels when the two vessels are connected ?
6.
A tube of uniform cross–section has two vertical portions connected with
a horizontal thin tube 8 cm long at their lower ends. Enough water to occupy
22 cm of the tube is poured into one branch and enough oil of specific
gravity 0.8 to occupy 22 cm is poured into the other. Find the position of
the common surface E of the two liquids.
7.
A block weight 15 N in air and 12 N when immersed in water. Find the specific gravity of block.
8.
The diagram shows venturimeter through which water is flowing. The speed of water at X is 2 cm/sec. Find
the speed of water at Y (taking g = 1000 cm/sec2).
5.1mm
4.
173
JEE-Physics
9.
8
cm with its crest A 1.8 m above water level as in figure.
p
Find (i) velocity of flow (ii) discharge rate of the flow in m 3/sec (iii) absolute pressure at the crest level A. [Use
P0 = 105 N/m2 & g = 10 m/s2]
A siphon has a uniform circular base of diameter
A
1.8m
3.6m
10.
A syringe containing water is held horizontally with its nozzle at a height h above the ground as shown in
fig. The cross–sectional areas of the piston and the nozzle are A and a respectively. The piston is pushed
with a constant speed v. Find the horizontal range R of the stream of water on the ground.
v
174
Fluid Mechanics
1.
There is a circlar tube in a vertical plane. Two liquids which do not mix and of densities d 1 and d2 are filled in the
tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle a with vertical.
Ratio
(A)
2.
d1
is :
d2
1 + sin a
1 - sin a
(D)
1 + sin a
1 - cos a
(B) 4.8 m
(C) 2.9 m
(D) 6.0 m
(B) 2.9 cm
(C) 1.7 cm
(D) 5.1 cm
(B) 0.1
(C) 2.0
(D) 0.4
(B)
3 2
pv
4
1 2
(C) 2 pv
1 2
(D) 4 pv
Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is 5 cm, the Reynolds
number for the flow is of the order of : (density of water = 1000 kg/m 3, coefficient of viscosity of water = 1mPas)
(A) 106
7.
1 + tan a
1 - tan a
A liquid of density p is coming out of a hose pipe of radius a with horizontal speed v and hits a mesh. 50% of the
liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same
speed. The resultant pressure on the mesh will be:
[JEE(Main) 2019]
(A) PV2
6.
(C)
A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the
liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then
the difference in the heights between the centre and the sides, in cm, will be:
[JEE(Main) 2019]
(A) 1.2
5.
1 + cos a
1 - cos a
Water flows into a large tank with flat bottom at the rate of 10–4 m3s–1. Water is also leaking out of a hole of area
1 cm2 at its bottom. If the height of the water in the tank remains steady, then this height is:[JEE(Main) 2019]
(A) 4 cm
4.
(B)
The top of a water tank is open to air and its water level is maintained. It is giving out 0.74 m 3 water per minute
through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of
water in the tank is close to :
[JEE(Main) 2019]
(A) 9.6 m
3.
[JEE(Main) 2014]
(B) 103
(C) 104
(D) 102 [JEE(Main) 2019]
4
of its volume submerged. When certain amount of an oil
5
is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water
and half in oil. The density of oil relative to that of water is :[JEE(Main) 2019]
A wooden block floating in a bucket of water has
(A) 0.5
(B) 0.7
(C) 0.6
(D) 0.8
175
JEE-Physics
8.
Water from a tap emerges vertically downwards with an initial speed of 1.0 ms –1. The cross-sectional area of the
tap is 10–4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is
streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be : (Take g = 10 ms –2)
[JEE(Main) 2019]
(A) 1 ×
9.
10–5
m2
(B) 5 ×
10–5
m2
(C) 2 ×
10–5
m2
(D) 5 × 10–4 m2
A submarine experiences a pressure of 5.05 × 10 6 Pa at a depth of d1 in a sea. When it goes further to a depth
of d2, it experiences a pressure of 8.08 × 106 Pa. ,Then d2 – d1 is approximately (density of water = 10 3 kg/m3
and acceleration due to gravity = 10 ms–2)
[JEE(Main) 2019]
(A) 500 m
10.
(B) 400 m
(C) 300 m
(D) 600 m
A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight
that can be put on the block without fully submerging it under water? (Take density of water = 103 kg/m3)
[JEE(Main) 2019]
(A) 65.4 kg
176
(B) 87.5 kg
(C) 30.1 kg
(D) 46.3 kg
Fluid Mechanics
MCQ's with One Correct Answer
1.
Water from a tap emerges vertically downwards with an initial speed of 1.0 m/s. The cross–sectional area of tap
is 10–4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is steady, the
cross–sectional area of stream 0.15 m below the tap is :–
[IIT-JEE 1998]
(A) 5.0 × 10–4 m2
2.
(D) 2.0 × 10–5 m2
(B) lower in front side
(C) lower in rear side
(D) lower in upper side
A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the
other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the
quantities of water flowing out per second from the holes are both same. Then, R is equal to :–
(A)
4.
(C) 5.0 × 10–5 m2
A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of
gravity. Then, the pressure in the compartment is :–
[IIT-JEE 1999]
(A) same every where
3.
(B) 1.0 × 10–4 m2
L
2p
(B) 2pL
(C) L
(D)
L
2p
[IIT-JEE 2000]
A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the
remaining cylinder is V and mass M. It is suspended by a string in a liquid of density r, where it stays vertical.
The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder
by the liquid is :–
[IIT-JEE 2001]
h
r
(A) Mg
5.
(B) Mg – Vrg
(C) Mg + pR2hrg
(D) rg (V + pR2h)
A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance l and h are
shown there. After sometime the coin falls into the water. Then :–
[IIT-JEE 2002]
coin
l
(A) l decreases and h increases
(C) both l and h increase
h
(B) l increases and h decreases
(D) both l and h decrease
177
JEE-Physics
6.
Water is filled in a cylindrical container to a height of 3 m. The ratio of the cross–sectional area of the orifice and
the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is (g = 10 m/s 2)
[IIT-JEE 2005]
3m
(A) 50 m2s2
7.
52.5 cm
(B) 50.5 m2s2
(C) 51 m2s2
(D) 52 m2s2
A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest,
the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the
following, state of the lift’s motion is given in List I and the distance where the water jet hits the floor of the lift
is given in List II. Match the statements from List I with those in List II and select the correct answer using the
code given below the lists.
[JEE(Adv.) 2014]
List - I
List-II
P.
1.
d = 1.2 m
2.
d > 1.2 m
3.
d < 1.2 m
4.
No water leaks out of the jar
Lift is accelerating vertically up.
Q. Lift is accelerating vertically down with an
acceleration less than the gravitational
acceleration
R.
Lift is moving vertically up with
constant speed.
S.
Lift is falling freely.
Cobe :
(A) P–2, Q–3, R–2, S–4 (B) P–2, Q–3, R–1, S–4
(C) P–1, Q–1, R–1, S–4 (D) P–2, Q–3, R–1, S–1
MCQ'S (Multiple Correct answers)
1.
A solid sphere of radius R and density r is attached to one end of a mass-less spring of force constant k. The
other end of the spring is connected to another solid sphere of radius R and density 3r. The complete
arrangement is placed in a liquid of density 2r and is allowed to reach equilibrium. The correct statement(s)
is (are) :
[IIT-JEE 2013]
4 pR 3 rg
3k
(C) the light sphere is partially submerged
4 pR 3 rg
3k
(D) the light sphere is completely submerged
(A) the net elongation of the spring is
2.
(B) the net elongation of the spring is
Two spheres P and Q of equal radii have densities r1 and r2, respectively. The spheres are connected by a massless
string and placed in liquids L1 and L2 of densities s1 and s2 and viscosities h1 and h2, respectively. They float in
equilibrium with the sphere P in L1 and sphere Q in L2 and the string being taut (see figure). If sphere P alone in L2
r
r
has terminal velocity VP and Q alone in L1 has terminal velocity VQ , then
[IIT-JEE 2016]
r
| VP | h1
r
=
(A)
| VQ | h2
178
r
| VP | h2
r
=
(B)
| VQ | h1
L1
P
s1
L2
Q
s2
r r
(C) VP . VQ > 0
r r
(D) VP . VQ < 0
Fluid Mechanics
Comprehension Type Questions
Comprehension-1
[IIT-JEE 2006]
A wooden cylinder of diameter 4r, height H and density r/3 is kept on a hole of diameter 2r of a tank, filled with
liquid of density r as shown in the figure.
1.
Now level of the liquid starts decreasing slowly. When the level of liquid is at a height h 1 above the cylinder the
block starts moving up. At what value of h1, will the block rise :–
h1
r
3
H
r
2r
(A)
2.
4h
9
(B)
5h
9
(C)
(D) remains same
The block in the above question is maintained at the position by external means and the level of liquid is
lowered. The height h2 when this external force reduces to zero is :–
r
3
(A)
3.
5h
3
4h
9
(B)
5h
9
h2
(C) remains same
(D)
2h
3
If height h2 of water level is further decreased, then :–
(A) cylinder will not move up and remains at its original position
(B) for h2 = h/3, cylinder again starts moving up
(C) for h2 = h/4, cylinder again starts moving up
(D) for h2 = h/5 cylinder again starts moving up
Comprehension-2
[JEE(Adv.) 2014]
A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross
section is connected to the nozzle. the other end of the tube is in a small liquid container. As the pistion
pushes air throught the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the
spray gun shown, the radii of the pistion and the nozzle are 20 mm and 1 mm respectively. The upper end of
the container is open to the atmosphere.
1.
If the piston is pushed at a speed of 5 mms–1, the air comes out of the nozzle with a speed of
(A) 0.1 ms–1
2.
(B) 1 ms–1
(C) 2 ms–1
(D) 8 ms–1
If the density of air is ra and that of the liquid rl, then for a given piston speed the rate (volume per unit time)
at which the liquid is sprayed will be proportional to
(A)
ra
rl
(B)
ra rl
(C)
rl
ra
(D) rl
179
JEE-Physics
Assertion–Reason
1.
Statement–1 : The stream of water flowing at high speed from a garden hose pipe tends to spread like a
fountain when held vertically up, but tends to narrow down when held vertically down.
[IIT-JEE 2008]
and
Statement–2 : In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.
(A) Statement–1 is True, Statement–2 is True; Statement–2 is a correct explanation for Statement–1
(B) Statement–1 is True, Statement–2 is True; Statement–2 is not a correct explanation for Statement–1
(C) Statement–1 is True, Statement–2 is False
(D) Statement–1 is False, Statement–2 is True
Subjective Type Questions
1.
A wooden stick of length L, radius R and density r has a small metal piece of mass m (of negligible volume)
attached to its one end. Find the minimum value for the mass m (in terms of given parameters) that would make
the stick float vertically in equilibrium in a liquid of density s (> r).
[IIT-JEE 1999]
2.
A uniform solid cylinder of density 0.8 g/cm 3 floats in equilibrium in a combination of two non–mixing liquids
A and B with its axis vertical. The densities of the liquids A and B are 0.7 g/cm 3 and 1.2 g/cm3, respectively. The
height of liquid A is hA = 1.2 cm. The length of the part of the cylinder immersed in liquid B is h B = 0.8 cm.
[IIT-JEE 2002]
Air
A
B
h
hA
hB
(i) Find the total force exerted by liquid A on the cylinder.
(ii) Find h, the length of the part of the cylinder in air.
(iii) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is
then released. Find the acceleration of the cylinder immediately after it is released.
3.
Consider a horizontally oriented syringe containing water located at a height of 1.25 m above the ground. The
diameter of the plunger is 8 mm and the diameter of the nozzle is 2 mm. The plunger is pushed with a constant
speed of 0.25 m/s. Find the horizontal range of water stream on the ground. (Assume liquid is compressible and
non–viscous) (Take g = 10 m/s2).
[IIT-JEE 2004]
D=8mm
D=2mm
1.25m
180
Fluid Mechanics
4.
A U–shaped tube contains a liquid of density r and it is rotated about the line as shown in the figure. Find the
difference in the levels of liquid column.
[IIT-JEE 2005]
w
H
L
Integer Type Questions
1.
A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and
water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is
opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of
water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice. [Take
atmospheric pressure = 1.0 × 105 N.m2, density of water = 1000 kg/m3 and g=10 m/s2. Neglect any effect of
surface tension.]
[IIT-JEE 2009]
181
JEE-Physics
ANSWER KEY
EXERCISE-1
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
D
C
B
B
B
C
C
A
A
B
A
B
B
B
A
Que.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
D
B
A
C
A
D
B
C
A
C
C
D
A
D
B
EXERCISE-2
Que.
1
2
3
4
5
6
7
8
9
10
Ans.
C
B
C
B
C
D
A
A
B
A
l
Match the Column:
11. (A) p, (B) p, (C) r
12. (A) p, (B) q, (C) q
l
Comprehension Based Questions
Comprehension-1 :
13. (D)
14. (C)
Comprehension-2:
18. (B)
19. (C)
Comprehension-3 :
20. (D)
21. (A)
15. (B)
16. (B)
17. (A)
EXERCISE-3
1. 7.06 × 10 5 Pa
é h1 - h2 ù
5. Arg ê
ú
ë 2 û
2. 37.5 N
3. H =
6. BE = 6 cm
7. 5
11
2
m, h = m
32p
p
4. 10.34 m
2
9. (i) 6 2 m/s, (ii) 9.6
8. 32 cm/s
Av
–3
3
4
2
2 × 10 m /s (iii) 4.6 × 10 N/m 10. R = a
2h
g
EXERCISE-4
Que.
Ans.
1
C
2
A
3
D
4
C
5
B
6
C
7
C
8
B
9
C
10
B
EXERCISE-5
l
MCQ's (Single Correct answers)
l
MCQ's (Multiple Correct answers)
l
Comprehension Based Questions
l
1. (C)
2. (B)
6. (A)
7. (C)
1. (A, D)
2. (A, D)
Comprehension-1 :
1. (C)
2. (A)
Comprehension-2 :
1. (C)
2. (A)
Assertion–Reason
1. (A)
182
3. (A)
3. (A)
4. (D)
5. (D)
Fluid Mechanics
l
Subjective Questions
2
1. pR L
4. H =
l
Integer Type Questions
(
rs - r
)
2. (i) zero (ii) 0.25 cm (iii)
g
6
3. 2m
w 2L2
2g
1. 206
*****
183
Important Notes
Important Notes