-16
Technology gives rise to new physics; similarly
physics gives rise to new technologies.
• Very short range (10 m ) weak force
• Not weak as gravitational force but much
weaker than the strong nuclear force and
electromagnetic forces.
Chapter - 1
Principal Thrust
To derive the properties of bigger and more
complex system from the properties of its
constituent into simpler parts.
Scop
Classical Physics deals with macroscopic
Study of the basic laws of nature Excitemeeand
nt of phenomenon and includes subject like
Physics
and manifestation in different
Mechanics, Electrodynamics, Optics and
natural phenomenon
Thermodynamics. Quantum Physics deals with
microscopic phenomena at the minute scales
of atoms, molecules and nucleus.
Physical World
Fundamental Forces of Nature
• Very short range, strongest force
• Attractive in nature
• Range – 10-15 m
• Strongest among all fundamental forces.
To explain diverse physical phenomena
in terms of concepts and laws.
Systematic observations, controlled
experiments, qualitative and quantitative
reasoning, mathematical modelling, prediction
and verification or falsification of theories.
• Any subject is said to be science when
it is studied in sequence pattern. The
experimenting, exploring and predicting
from what we see around us.
• Force of mutual attraction between any two
objects by virtue of their masses
• Plays a key role in large scale.
• The phenomena of the universe, such as
formation of and evolution of stars, galaxies
and galactic clusters are by gravitational
force governed.
• Force between the charged particles
• Both attractive and repulsive
• It is 1036 times stronger than gravitational force
between two protons, for any fixed distance.
Mind map : learning made simple
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
1




mean
=
1
( ∆a
δa =
∆a mean
× 100
amean
n
+ ∆ a2 + ...+ ∆ an
Percentage Errror
∆amean =
∆a
= mean
amean
n
a1 + a2 + ...... + an
Relative Errror
a
∆an = an – amean
Absolute Errror
)
Types
±∆ Z= ± ∆ A ± ∆ B
All non-zero digits
All zero between two non-zero digits
If the number is less than 1, the zero(s)
on the right of decimal point but to the
left of the non-digit zero are not significant.
Terminal or trailing zero (s) in a number
without a decimal point are
not significant
e.g  In 020342.010
Here 20342.01 are significant
Comparison with a certain
internationally accepted reference
standard is called unit
Reference standard use to
measure physical quantities.
Measurement
Units and
Measurements
∆Z  ∆ A   ∆B 
=
+

Z  A   B 
ived
Z = A ×B / C
q
r
∆Z
 ∆A 
 ∆B   ∆C 
= p
 + q
 + r

Z
 A 
 B   C 
If
p
Universal acceptance
Non-perishable
Well defined
Does not change with time
Quantities plane angle and
solid angle have radian and
steradian as units
(Internationally accepted)
S.I - System International
MKS– Metre, Kilogram, Second
FPS – Foot, Pound, Second
CGS – Centimeter, Gram, Second
Set of fundamental and derived units

The units which are derived from
fundamental units are called
ts
derived units.
Physical
quantities like speed

acceleration have derived units
like m/s, m/s2
uni
Mass, length, time, temperature,
electric current, luminous intensity
and amount of substance have fundamental
units. Kilogram, metre, second, Kelvin,
ampere, candela and mol respectively.
Closeness of the measured
value to the true value
ity
nt
a
u
er
w
po
If Z=A/B
Der
ts
ni
Chapter - 2
Pr
op
er
tie
s
Limit of resolution of
quantity measured.
Types
Me
a
rai
sed sure
d
to
q
th
e
Combination of Errors
in
Error = True Value – Measured Value


Expression which shows how and which of the
base quantities represent the dimensions of a
physical quantity is called dimension formula.
e.g:- Force =  MLT –2 


Mind map : learning made simple
Unit
Fund
am
e
nta
lU
m
ste
Sy
its
un
f
o
Multiplication
or Division
2
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI

x = v0t + at
(ii)
In opposite direction, it will be Difference
In same direction, it will be Sum
vavg

Rate of change of position
of an object w.r.t time in
given direction .Vector Quantity
Av
er
ag
e
=
∆x
∆t
Total displacement
=
Total time taken
ity
oc
l
Ve
(ii)

 
vAB = vAn − vB



vBnA = vBB − vA
a ( 2n-1)
2
Types
Motion in a
Straight Line
t
Rate of change
of

velocity, a = v
Chapter - 3
orm

∆x
vinst = lim
∆t → 0 ∆t

= dx
dt



Av
e
The shotest distance
traversed by any
object in called
displacement
 It in a vector quantity.

Total length of the path traversed
by any object is called its distance
 It is scalar quantity.

When the magnitude or the direction
of velocity changes w.r.t. time.
Equal distances are traversed in
equal amount of time.
v 1t 1 + v2 t2 + ..... + vn tn
t 1 + t 2 + ..... + t n
When object traversed different
speeds in different time of
intervals.
s1 s2 s3
s
+ + + ...... n
v1 v2 v3
vn
When object traversed different
distances with different speeds.
Distance traversed/time taken .
Scalar quantity
ment
Displace
e
tanc
Dis
n
o ti o n
ti o
Mo
Non - uni form m
if
Un

ed
pe
eS
g
ra
(i)
(iv) xnth = v0+
2
2
(iii) v = v0 + 2ax
v = v+
at
0
(i)
Free
Fall
ns of
quatio
tic e
a
n motion
nem cceleratio
Ki
a
y
l
rm
i fo
un
Earth’s gravity (g= 9.8m/s2).
On neglecting air resistance, it is a
case of motion with uniform
acceleration. e.g Apple falling
from a tree.
An object falling because of
Total time interval

∆v
=
∆t
Total change in velocity
Accel eration
aavg =
Ve locity


∆v
dv
a inst = lim
=
∆t → 0 ∆t
dt
Mind map : learning made simple
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
3
2
Horizontal range,R = u sin2θ
g
Total time of flight,Tf = 2u sinθ
g
Maximum height,Hmax = u2sin2θ
2g
Vectors having
common starting
point.
If, A = – B
Vectors having
same magnitude
but opposite direction
eg. A is a negative of B
Vectors having
same direction
and magnitude
gnitu
It has magnitude as
one or unity
A
^
A =
A
y=x tanθ–
Equ
Pla
ne
tile
jec
n
o
tio
Pr
mo
Mo
tio
ni
na
Motion
in a Plane
x=(ucosθ)t;
y=(u sinθ) t - 1 gt2
2
Motion of an object that
is in flight after being thrown
or projected.
r
Vec
to
→
P
tile
ation of path of projec
g
x2
2u2cos2
→
P
Law of Parallelogram
Law of Triangle
→ → →
R=P+Q
→
→
→
→
R
R
Q
Q
Addition of
Vectors
at any instant
po
ax = 0,ay = g C o m
→
→
|V
& AB|=| VBA|
→
→
VAB = –VBA
→→
→ →
V BA = VB –VA
→
→ →
V AB =VA –VB
ac=v2/r = rw2 = 4π2 rv2
A body in a circular motion acted
upon by an acceleration directed
towards centre of the circular
motion.
Angular velocity ,ω= θ/t
Angular acceleration,d=∆ω/∆t
eg- merry go around.
When an object follows a circular path at
a constant speed , the motion of the
object is called uniform circular motion.
→
a = axi + ayj; ax=dvx/dt & ay= dvy/dt
→
|a|=√ax2+ay2
→
v =vx i + vy j ,
→
magnitude|v | = √vx2+vy2
→
i + y^
j
Position vector , r = x^
→
Displacement vector , ∆ r = ∆ x^
j
i +∆y^
→ →
λ × A=λA
u=
ucosθ,u=
u sinθ
x
y
Motion of Body under
2 dimensional frame.
→→ → →
A–B=A+(–B)
Chapter - 4
ne
It has zero magnitude
and orbitary direction
Mind map : learning made simple
ele r a ti o n
f acc
Centripeta
l
n ts
o
4
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
Acce
lera
tio
n



n
io
ct
mv
R
Opposes impending
relative motion
FS = µsR
Ce
ntr
n
ctio
Fri
ip e t a l f o r c e
Force which opposes
the relative motion
of a body
Centrifugal force is
equal and opposite
to centripetal force.
F=
2
on est
Kinetic Fric
tion
,
e g car
A push or pull which
changes or tends to
change state of rest
or of uniform
motion of a body.
Oppose actual
rolling motion
µ R < µk < µs
tion
fric
g
llin
Ro
Oppose actual relative
motion
Fk = µkR
e
Laws of Motion
Types
In
rt
ia
of m
w
o tion
h
c o e n m o ving
me
s i n r e st
A particle is said to
be in equilibrium
when net external
force on the
particle is zero.
 
 
F1 + F2 + ..... + Fn = 0
M
om
en
t
ton’s law of
New
tion
of mo
Resistance to
change its state of
rest or motion.
Ne
Product of mass and
velocity of the body
p = mv
II l a w
mAuB + mBuB = mAvA + mBvB
The total momentum of an
isolated system of interacting
particles is conserved
dt
Impulse = force × time
I = change in momentum
I = mv −
mu
I = m ∆v
Total change in momentum
To every action there is always
an
equal
& opposite reaction.
FAB = FBA
Action & reaction act on two
different bodies
where ‘a’ = acceleration
The rate of change of momentum
of a body is directly proportional
to the applied force and takes place
in the direction in which the force
dp
acts.
∝ F ⇒ F = ma
oti o n
o n’s
of m
wt
law
’s
on
rtia
Ine
Every body continues
to be in its state of rest or
uniform motion unless
acted upon by a non-zero
n external force.
o
i
t
Mo
Also called law of Inertia.
of
Chapter - 5
lse
pu
Im
Maximum value of static friction
n
ca
ro
n
Force which makes a body
move along a circular path
with a uniform speed.
a car on bank
ed r
oad
Moti
on
of
a
When friction force is
taken into consideration
½

µ + tan θ 
v max =  Rg s

 1 − µ s tan θ 
When no friction force is
considered
v max = Rg tan θ
of
on
oti
M
Vmax = µS Rg
a
d
roa
el
v
le
)
(P
Limitin
g
fri
cti
o
st
Fo r c e
n
io
ot
m
of
St
a
t
ic
fri
re
um
of conservation
Law
of
momentum
r
r in i
eg, when a c a m ot
co m e s i n
of
In e rti a
aw
II l
’s I
n
o
t
New
article
of a p
ium
ibr
uil
Eq
Ne
w
t
Mind map : learning made simple
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
5
When force and displacement
are perpendicular to
each other. W = zero
when θ = π
2
When force (F) and displacement (S)
are in opposite direction.
W will be (-ve) when π <0<π
2
When force (F) and displacement (S)
are in same direction. π
W will be+ve when 0<θ<
2
An instance of one
moving body striking
against another
Types
Z
o
er
rk
wo
s
Type
• Total energy and linear
momentum conserved.
• Kinetic energy not conserved.
( )
Time ( t )
Work w
Power (
P)
Work, Energy and
Power
1hp=746watt
1kWH=3.6 × 106J

 ds  
Pinst . = F ⋅ = F ⋅ v
dt
P =
s2
1
1
 
W = ∫s Fdscosθ = ∫s F ⋅ ds
s2
Work done by a variable force
 
W = Fscos θ = F⋅ s
Work is said to be
done when force produces
displacement.
Work done by constant
force
y
rg
Types of mechanical energy
Capacity of doing
work
Various forms
1
1
mv 2 – mu2 = W
2
2
Change in Kinetic Energy =
Work done by net force on
the body i.e;
By virtue of position
Ep = mgh
k= spring constant P ote n
1
U = kx 2
2
1
p2
E K = mv 2 =
2
2m
By virtue of velocity
Nuclear Energy
Electrical Energy
Chemical Energy
Heat Energy
Energy can neither be created nor
destroyed; it can only be
transformed from one form to
another.
Equivalence of mass and Energy
E=∆mc2
Formula
Rate of doing work
Kine
tic
E
n
erg
y
• Total mechanical/kinetic
energy is conserved.
• Momentum is conserved.
En
e
Chapter - 6
ti a
lE
ne
r
Mind map : learning made simple
ri n g
f Sp
(W)
Work
gy
o
6
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
mo
me
nt
of
fo
rc
e
Vcm =
n
i i
i=1
mi
n
i =1
I = ∑mi ri 2
Inertia of rotational
motion, M.I.,
Systems of Particles and
Rotational Motion
Vcm =
n
i=1
i
∑m
i =1
n
∑m v
i i
The point where the whole
mass of the system is
supposed to be concentrated
e
ody
Rigid b
tion
us of gyra
Radi
ax
i =1
n
acm = ∑
Acc
ele
rat
ion
mas
so
f th of c
e
es
ys
te
i
mi ai
∑m
ass for rigid bodie
s
tre of m
Cen
el
body
ped
a
h
rs
ula
of p
reg
Theorem
e
som
of
a
i
t
ter
f in
o
t
en
om
M
ll
ara
I = I cm + mr 2
Icm = M.I.about the parallel
axis through the centre
of mass
is independent of the state
i.e., rest or motion of the body.
A body with perfectly definite
and unchanging shape.
n
r12 + r2 2 + ........ + rn 2
I
K=
m
K=
mI 2
12
2
I one edge = mI
3
(2) Ring I cm = mR2
2
Idiameter = mR
2
2
(3) Disc I cm = mR
2
2
Idiameter = mR
4
(4) Solid sphere
Idiameter = 2 mR2
5
(1) Rod I cm =
Chapter - 7
of
re
t
n
m
∑
i =1
n
∑m r
mass
e of
r
t
cen
tem
of
n
sys
o
e
i
it
th
of
Angular m
omen
tum
Torque or c
oup
le o
r
Position of centre of mass
of an object changes in
translatory motion
but remains unchanged
in rotatory motion.
Position of centre of mass
depends upon shape,
size, distribution of mass
of the body.
L= Ιω = mvr = r × mv
L= r × p = rpsinθ
→ → → ^
τ = r × F = r Fsinθ n^
τrotational = Ια
perpendicular axes x, y
and z respectively
Th
eo
Equ
rem
atio
of p
ns
of
erpe
ax
ndicular
ro
es
Theorems of moment
tat
io
of Inertia
na
Po
s
Moment
Inertia
e of
Centr
mass
ion
ot
lm
Velocity of
Center mass
of the system
s
Iz = Ix + I y
I x , I y & I z moments of inertia about
Mind map : learning made simple
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
7
2m
L
2
3
= Constant
Gm
r2
−Gm
work done
=
mass
r
U =
r
−GMm
· Gravitational Potential Energy
Vg =
· Gravitational potential
I=
· Gravitational field Intensity
(iii) Law of Periods: T ∝a
where, a = length of
semi-major axis
dt
=
F=
Gm1 m2
, G = 6.67 ×10−11 Nm2 kg −2
r2
(i)
(ii)
(iii)
(iv)
el
ne
ss
(g
)
vorb=
gR =
2 gR
(iii) with rotation of earth about its own axis
g’=g– Rω 2 cos 2 λ ,
0
At poles, λ = 90 , g maximum
0
At equator, λ = 0 , g minimum
i.e. g decreases with depth
g′ = g  1− d  ,
 RE 
2h

g ′= g 1 −
,
 RE 
i.e. g decreases with height
R2e
G Me
with height
g =
Acceleration possessed by an
object during free fall
due to earth’s gravitational pull,
During free fall under gravity
inside a spacecraft or satellite
a body is weightless i.e., effective
weight becomes zero.
2
ve
2 GM / R
1 Mm
GMm  − GMm
=
+
=– G
2r   r 
2
r
E = K.E +P.E
Total Energy of a satellite
Orbitals speed
(ii) with depth
(i)
gra
vit
y
Weakest force in Nature.
Central as well as conservative.
Always attractive in nature.
Applicable for all bodies irrespective
of their shape, size and position.
fo
rc e
Acceleratio
n du
e to
Cha
ract
eris
tics
of
gr
av
ita
tio
na
l
w of
Ne a w
l
to
Gravitation
When two mass bodies are separted by a distance,
they experience an attractive force
which is directly proportional to the product of
their masses and inversaly proportional to the
square of their separation.
l
na
tio
a
t
tial
ten
o
dp
an
d
l
Fie
motion
w- of planetary
er ’s La
Kepl
Sat
lit
Heavenly object that revolves
around a planet
=
v esc =
Earth ’s v esc = 11.2km/s
Escape speed
Chapter - 8
ss
dA
(i) Law of Orbit: Every
planet revolves around
the Sun in an elliptical
orbit and Sun is at its
one focus.
(ii) Law of Area: The radius
vector drawn from the
Sun to a planet sweeps
out equal areas in equal
intervals of time.
Geostationary satellite
Polar
satellite
Height from earth’s surface = 880km
Time period = 84min
Orbital velocity = 7.92km/s
e
Height from surface of earth =36000km
Radius of orbit = 42,400km
Time periods = 24 hours
Orbital velocity = 3.1km/s
Mind map : learning made simple
tl e
Gr
av
i
W
h
eig
n’
G s un
ra i
v
it a v e r s a
tio l
n
8
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
Deforming force applied normal Fn
= A
Area
Force Changing Volume FV
=
A
Area
pe
longitudinal stress F / A
=
longitudinal strain l / ∆l
F ∆l Mg∆l
=
π r 2l
Al
Y =
Y =
Young’s modulus of elasticity
Property of material by
virtue of which it regains it
original shape & size after the
removal of deforming force
Stress
9 1 3
9 Bη
= + or Y =
Y B η
η +3B
(iv)
E
s Modulus
Young’
Restoring force per unit area
i.e., stress= AF
s
(iii) Shearing area or tangential stress
Tangential Force
F
=
= t
A
Area
=
(ii) Volumetric stress
=
(i) Longitudinal stress
σ=
(iii)
3B − 2 η
2η + 6B
Relation between Y, B, η & σ
(i)
Y=3B (1 − 2 σ)
Y=2η (1 + σ)
(ii)
Mechanical Properties
of Solids
Elasticity
Elastic potential energy in a stretched
1
wire= ×stress × strain × volume of the wire
2
Ty
∆P 1
B=−
; = compressibility
∆V B
V
hydraulic stress
B=
volume strain
Bulk modulus or volume modulus of elasticity
Bulk Modulus
Types of Modulus of Elasticity
Within the elastic limit, stress is
directly proportional to strain.
i.e., stress ∝ strain
Hooke’s
law
V
change in length ∆l
=
original length
lo
o
in volume ∆V
=
ε = change
original volume
V
=
Shearing strain=angular displacement of the
plane perpendicular to the fixed surface = θ
Volumetric strain
Longitudinal strain
T
η=
η=
σs
F
F
= A =
εs θ
Aθ
tangential stress
shearing strain
Rigidity or shear modulus of elasticity
Ratio of change in
configuration to original
configuration
change in configuration
Strain =
original configuration
Modulus of Rigid
ity
Strain
(iii)
(ii)
(i)
Laternal strain (β)
∆d / d
=
Longitudinal strain (α ) ∆l / l
Value of σ lies between 0 and 0.5
Poision’s ratio(σ) =
Chapter - 9
es
rgy
ne
e
l
tia
ten
o
cP
sti
la
yp
Mind map : learning made simple
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
9
2
1
2
2
2hρm g
ρ (a 2 – a 2
)
of continuity
m=a1v1ρ1=a2v2ρ2
for an incompressible liquid,
ρ1=ρ2 then a1v1=a2v2
or av=constant
 Pascal’s law : The pressure exerted
at any point on an enclosed liquid is
transmitted equally in all direction.
Hydraulic brakes and hydraulic lifts
are based on Pacal’s law.
 Equation
Fl
Ven
turim
eter
Applications
ow
of
fl u
( )
∆E
∆A
=
That can flow like liquids and gases
Mechanical Properties
of fluids
Form u
Vi
P)
solu
s
te P res
ur
e(
Total or actual pressure at a point.
Absolute pressure= atmospheric pressure
+ gauge pressure
G
a
lae
it y
Difference between the absolute pressure at a point
and the atmospheric pressure.
Pg=absolute pressure(P) – atmospheric pressure(Pa)
density of substance
density of water at 4oC
Density of water at 4°C i.e.,
maximum density of water=1.0×103 kg/m3
Volume(v)
Density(ρ)= Mass(m)
F
tension S=
l
work done in increasing area W
 Surface Energy=
=
increase in surfacearea
∆A
2S
cosθ
 Capillary rise or fall, h=
r ρg
 Excess Pressure inside a drop (liquid)
2S
Pexcess =
R
 Excess Pressure inside a bubble (soap)
4S
Pexcess =
R
Stroke’s law F=6 ηvr
Opposing force between different layers of
fluid in relative motion
Viscous drag F= −ηA dv
dx
η=coefficient of viscosity
Lift of an aircraft wing.
Sprayer or atomizer
Blowing off the roofs during windstorm.
 Surface
o
o
o
Chapter - 10
Relative Density or specific gravity=
)
(Pg
re
su
s
re
eP
g
u
Pressure (atm) exerted by the atmosphere.
At sea level, 1 atm=pressure exerted by 0.76m
of Hg=hρg=0.76×13.6×103×9.8=1.013×105 Nm-2
=101.3kPa
Pa)
re(
u
s
es
Pr
ic
r
he
Pressure(P) = thrust(F) = ∆lim
A→ 0
dF
area(A)
dA
Pressure exerted by a liquid column of height h, (p)=hρg
ids
s
s co
For an incompressible non-viscous streamline,irrotational flow of fluid,
1
P + ρ v 2 + ρ gh = constant
2
uids
of Fl
s
w
La
1
2
Streamline : In liquid flow
when
the Velocity
is less than critical velocity, each particle of the
liquid passing through a point travels along
the same path and same velocity as the
preceding particles.
Turbulent : When velocity of liquid flow is
greater than critical velocity and particles
follow zig-zag path.
Q = a 1a2
Velocity of efflux of liquid through an orifice
V = 2 gh
To ric ell’s
Device used to measure the rate of flow of liquid.
Volume of liquid flowing per second
Mind map : learning made simple
Law
Ab
At
m
os
p
Bernoulli’s
principle
10
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
1
or λ m =
b
4
4
)
Stefan’s
Boltzmann law
Wien
disp
lac
em
en
t la
w
Kr
ich
of
Tempera
t
ure
α=
L – l0
∆l
=
l0 ∆Τ
l0(T – T0)
a
n
l co
Thermal Properties
of Matter
T h er m
ng
oli
co
f
o
law
n’s
o
t
New
Types (In solids)
Increase in dimensions due
to increase in temperature
d
Pri
nci
ple
of
A – A0
∆A
dA
β=
=
=
A ∆ T A dT A0(T–T0)
Increase in area of a solid on heating.
Coefficient of superficial expansion
or
a:b:γ=1:2:3
γ=2β=3a
Speci
fic H
ea
t
∆T
∆Q
calo
rime
try
s ∆Q
=
n n∆ T
V0 ∆T
dV
V0 dT
γ = ∆V =
=
V – V0
V0(T – T0)
Increase in volume of a solid on heating.
Coefficient of cubical expansion.
∆F
 ∆l 
= Y   , Y = Young’s modulus
A
 l 
Heat lost = Heat Gain
Molar specific heat capacity, c =
Heat capacity s =
s
Q
Specific heat capacity C = m= m∆T
A form of energy, transferred between
two systems by virtue of temperature
difference.
t
5
For water, latent heat of fusion, L f = 3.33× 10 J / kg
5
Latent heat of vapourisation, Lv = 22.6 × 10 J / kg
Cubical
or vo
lum
e ex
pan
Re
lat
sio
ion
n
t
ea
H
Chapter - 11
Heat required to change the state of unit mass substance
without changing its temperature , L = Q
m
Q. x
(K) =
A( T2 − T1 ) t
For small temperature difference between
a body and its surroundings, the loss of
dQ
heat is given by −
= k ( T2 − T1 )
dt
Increase of length of a solid on heating.
Coefficient of linear expansion
TC − 0
T − 32
TK − 273.15
–0
= F
=
–
=R
100 − 0 212 − 32 373.15 − 273.15 80 – 0
Relation among different temperature scales
Degree of hotness or coldness of a body
measuring device = Thermometer
(i) Conduction : heat transfer through molecular collisions
without any actual motion of matter.
(ii) Convection : heat transfer by actual motion of matter
within the medium. Land breeze, sea breeze, trade
winds based on natural convection are some examples.
(iii) Radiation : method of heat transfer requiring
no material medium.
E = σ T −T0
(
For a perfectly black body
σ = 5.672 Js −1 m−2 k−4
E∝T 4 ; E = σ T 4
T
T
−3
b(wine’s constant) = 2.9 ×10 mK
λm
law
Th
e
r
m
Expan al
sion
eλ
aλ = Eλ = constant
al
erm
Th aws
L
Superficial
or area
Expansion
it y
tiv
uc
f ’s
La
ten
t
H
ea
At any given temperature
Mind map : learning made simple
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
11
Two systems are in thermal
equilibrium with each other
if they have the same
temperature.
Heat Engine
V1
∫
V2
nRT
dv
V
W
Q
T
Efficiency η= =1– 2 = 1– 2
QT1
Q
T1
Converts continuously heat into
mechanical energy in cyclic process.
W =
P
γ
T
n- γ
=Constant;
PV = Constant
γ −1
TV = Constant
γ
=γ
Cv
Cp
Refrigerator and heat pump
V1 V2
=
T1 T2
= Constant
P1 P2
=
T1 T2
V
CP
=γ
CV
Mayer’s Equation
CP −C
=R
v V
β =
W
Q2
=
=
Q1 - Q 2
Q2
2
T1 - T
T2
Refrigerator is a heat engine working in the
reverse direction. Coefficient of performance,
The pressure (P) & temperature (T) of the
environment can differ from those of the
system only infinitesimally.
In this process, system
returns to initial state
for a cyclic process
∆U = 0 (zero)
Pressure = Constant
T
Isochoric
Process
Volume = Constant
P
= Constant
T


Infinity slow process such that systems remain
in thermal & mechanical equilibrium with
the surroundings throughout.
process
i-static
Quas
Cycli
c Pr
oce
ss
cess
Pro
c
i
ar
ob
A thermally insulated
system neither gains
nor loses heat
Is
Branch of Science which deals with concepts of
heat & temperature and their interconversion by
thermodynamic process
Thermodynamics
capacity Rela
tion
ific heat
Spec
Property
Temperature = Constant
PV= Constant = n RT
Efficiency of Carnot’s engine
Q
T
η= 1 – 2 = 1 – 2
Q1
T2
An Ideal engine works on a reversible
cycle of four operations in succession.
(i) isothermal expansion ,
(ii) adiabatic expansion,
(iii) isothermal compression,
(iv)adiabatic compression.
It is the statement of law of conservation of
It is impossible for an engine working between
Energy; ∆Q = ∆U + ∆W
= ∆U+P∆U (Here, ∆U=P∆U) a cyclic process to extract heat from a reservoir
and convert completely into work.
Chapter - 12
If two systems A&B are in thermal equilibrium
ics
with a third system C, then A&B are in thermal
am
n
equilibrium with each other.
w
y
od
t la y
rm
Zer
irs
e
d
F
h
o
oth
Measure of molecular disorder of a system.
ft
The
Law
rm
wo
δQ
of
rmo
the econd la
Change in entropy δ S =
dyna
S
mics
Laws
T
Thermo
dyna
mic
state v
ariab
le
n of
a
m
ics
(i) Extensive – indicate the size of the
system. e.g. U, volume, total mass.
(ii) Intensive – do not indicate size of the
system. e.g. , pressure & temperature
ot’ Engine
Carn
ersible & Irreversibl
Rev
e Pr
oce
Any process made to proceed in the
ss
reverse direction by changing its
conditions is called Reversible
Process.
Any process which cannot
be retraced in the reverse direction
exactly is called Irreversible Process.
quilibrium
rmal E
The
Mind map : learning made simple
py (S)
Entro
I
s
o
t
her
ma
l
P
ro
ce
ss
Adiabati
c
P
roc
ess
12
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
=
3PV
3P
3 RT
=
=
ρ
M
M
Cv
Cp
5
3
=γ=
(4 + f)
(3 + f )
De g
re
Energy associated with each degree of freedom per molecule =
1
kT
2
Pa
th
d om
Fr
ee
fre e
M
ea
n
e of
(f)
and Ki ne ti c Ene
y
erg
En
Hence, f is the degree of freedom
for polyatomic gases,
for diatomic
=γ=
7
=γ=
Cv
5
Cp
Cv
Theor
y of Gases
Cp
b
rted
λ=
)
2πnd 2
1
ssure (
P
rgy (E)
Re l ati on be twee n Pre
s
xe
ee
ur
s
y ga
=
3
2
PV
1 2
P = ρV rms
3
Total pressure of a mixture
of non - reacting gases,
P = P1 + P2+...... +Pn
2 π d2 P
kBT
For polyatomic gas:
(i)at room temperature, f = 6
(ii)at high temperature, f = 8
For monoatomic gas: f = 3
For diatomic gas :
(i) at room temperature, f = 5
(ii) at high temperature, f = 7
E=
e
Dal ton’s Law of Parti al Pre ssur
1
V
Chapter - 13
(If T = constant)
PV = constant
or P1V1=P2V2
P∝
es
Pr
ion
rtit
a
ip
qu
for monoatomic
Kinetic Theory
Ass
ump
tion of Kinetic
• All the molecules of a gas are identical.
• The molecules of different gases are different.
• The molecules of gases are in a state of random
motion.
• The collisions of gases molecules are perfectly
elastic.
ea
cH
i
f
i
c
Spe
e
ed
y
cit
pa
a
tC
Sp
eed
E
of
Specific Heat Capacity
for an ideal gas, Cp– Cv = R
vmp =
are
Sp
bl e
ob a
r
p
t
s qu
s
Mo
Root
me a
n
w
La
2 RT  2 
=
= 0.816 vrms
 3  vrms
M




va v = 8 RT =  8  vrms = 0.92 v rms
π M  3π 
v + v2 + v3 + ... + vN
va v = 1
N
rms
v
w
La
e’s
l
y
Bo
An ideal gas satisfies equation PV = nRT at
all pressure and temperature n = no. of
moles, R = NAkB universal gas constant
Behaviour
Under the same condition of temperature
and pressure equal volumes of all gases
contain equal no. of molecules.
i.e. N1= N2
Gu
yL
us
sac
’s
La
w
V ∝ T (If P = constant)
V = constant
T
V1 V 2
or
=
T1
T2
Charles’s
Law
of Gases
P ∝ T (If V = constant)
P = constant
T P P
or 1 = 2
T1
T2
Mind map : learning made simple
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
13
Frequency
F
= spring constant
x
Damping force Fd= –bv, b = damping constant
-(b/2m)t
x = xme
Amplitude decreases with time as
x = xme-bt/2m sin(ωt+φ)
m
lace
Disp
e
g
sprin
to a
e
u
ns d
atio
l
l
i
c
Os
ns
io
at
n
SH
O
sc
illa
tio
ns
2
2
Fo
2 2
d
2 2
d
{m (ω − ω ) +ω b }
−vo
tanφ =
ωd xo
Amplitude, A =
Displacement, x (t)=Acos (ωd +φ)
1
2
Every oscillatory motion is periodic,
but every periodic motion need not be
oscillatory.
To and fro motion repeatedly about
a fixed point in a definite interval of time.
Driving force, F (t) = Fo cosωd t


Oscillations
i
nt
E
Velocity in SHM
A= Amplitude i.e.,maximum displacement of particles.
d
Displacement of damped oscillator
Phase
Simple Harmonic Motion is
The smallest interval of time T
)
im e
the simplest form of
p eri o d ( T
after which the motion
oscillatory motion
is repeated
Time varying
quantity ( ωt + φ )
ce
Oscillation of a body whose amplitude goes on decreasing with time.
Here, k =
m
k
I
l
=2 π
mgl
g
Time period T =2 π
Time period T = 2 π
The number of oscillations
per second
i.e.,
1
ω
f= =
T 2π

x(t)= A(cosωt+φ)
Chapter - 14(a)
M
Phase constant or Phase angle (φ)
It depends upon velocity ( v) and
displacement of particle at t = 0
Mind map : learning made simple
Fo r
D
a
mp
ed
a=
dv
=−ω 2A cos (ωt +φ)
dt
a max= ω 2 A
dx
= – ωA sin(ωt+φ )
d (t )
vmax = ωA
v=
=
When frequency of external force
is equal to natural frequency
of oscillator.
∴ω= k
m
F= −k ( x) , k = mω2
Force law equation for SHM
1
kA2 cos 2 ( ωt+ φ) + sin2 ( ωt +φ) 
2
Potential Energy
1
P.E =
mω 2 A2 cos2 (ωt +φ)
2
Total energy, T.E.=K.E.+ P.E.
Kinetic Energy
1
K.E = mω2A2 sin2 (ωt + ϕ)
2
HM
in S
y
rg
ne
O
sc
ill
SHM
14
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
2l
m
v
2l
at
io n
(iii)  In open organ pipe, higher harmonics, both
odd & even.
 In closed organ pipe, higher harmonics, odd
only.
4l
(b) Closed at one end ν = v
Fundamental freq. or I harmonic ν =
(ii) In organ pipe:
(a) Open at both ends,
(i) In string: Fundamental Frequency, ν = 1 T
Stationary waves
Waves associated with particles
like-electrons, protons, neutrons,
atoms, molecules etc.
tter
Ma
po
When two or more waves are
Principle of super
propagating with different
s
displacements then the net
o f w a ve
displacement of collective wave
is given as Y = y1 + y2 + y3+.....+yn
ves
Wa
on
siti
ar y Waves
1
Frequency
Wave number, v =
Types
Do not require any material
medium for their propagation
e.g., light waves
Mech
anic
al W
ave
s
Waves
1
λ
)
V =
er
s
in a
it u d
Long
s
wave
l
√
Individual particles of
the medium oscillate
along the direction of
propagation of wave.
Individual particles of the
medium oscillate perpendicular
to the direction of propagation
of wave
T = tension & M = mass/length
√ √ √
Speed of a longitudinal wave speed of sound, V = B
ρ
Y
P
γP
=
for air, γ = 7/5
ρ = ρ = ρ
Require material medium
for their propagation
v
ns
Tra
√ mT ,
Speed of transverse wave in a stretched string
Difference in frequencies of two
superposing waves, νbeat= ν1∼ ν2
(
( t)) = A sin (–kx + t) ωφ
y (x,
y (x,t) = Asin 2 π t – x
T λ
( ) amplitude
Velocity
v0=Aω2
Acceleration Amplitude a0 = Aω2
Essential properties
for propagation
 Elasticity
 Inertia
 Minimum friction
la
re e
t
iv
en
m ress
ce
a
g
l
o
ts
sp
pr
Di
Bea
wavelength
1
=
1
= ν
Angular Frequency (ω) = 2 πν
v = νλ
Time period (T) =
t
n
io leng
t
a
l e
Re wav
ν ± νs ± νm
n1 = n 
ν ± ν0 ± νm 


Here, v = speed of
sound,
v0 = observer speed
vs = source speed
vm = medium speed
General Formula,
Apparent Frequency
Electromagnetic Waves
e
la n
tio
w n in
av
ap
e
Properties
b/
w
h wa
(λ ve
)
& v elo
fre
c it y
q
u
e n c ( v),
y (ν)
Chapter - 14(b)
e
St
es
po
av
Do
ct
fe
Ef
r
le
w
Mind map : learning made simple
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
15
sin ( δmin + A ) / 2
Ray Optics
MP =–
fo
& L = fo + fe
fe
When image is at D
f 
f 
MP = − o  1 + e 
fe–
D
 f

& L = f o+ e

f
D
+
e

Magnifying Power (MP)
When image is at ∞
Telescope
D
v
D L 
MP = −  1 +  =  1 + 
fe 
u
fe  fo 
n
bi
m
o
C
through
light
f
o
ion
ers
Prism
sp
i
D
Optical
Instr
um
ent
Tota
l In
ter
na
lR
ef
le
ct
Astronomical
Magnifying Power (MP)
When image is at ∞
vD
MP = −   & L = u + fe
u  fe 
When image is at D
P=
1
f
1 1 1
= +
f f1 f2
P = P1 + P2
sin A / 2
δ = A (µ −1)
µ =
δ=i+e−A
A = r1 + r2
• Compound microscope
• Astronomical microscope
• Newtonian Telescope
The complete reflection of a light
ray reaching an interface with a
less dense medium when the
angle of incidence exceeds the
critical angle.
&
ws
La
R
2

1  n2
=  − 1
f  n1

 1
1 
−


R
R
2 
 1
Mirror formula ⇒
1 1 1
= +
f v u
I −v
f
f −v
(2) Linear magnification, m = =
=
−
O u
f−u
f
(1) Focal length f =
The separation of visible light into its different
colours.
δred < δviolet because ηred < ηviolet
(ii) Angular dispersion : θ = δv – δr=(ηv – ηr )A
δ − δr
(iii) Dispersive power : ω = v
δr
When beam of light goes from one medium to another
medium, it deviates from its path, such deviation of
light in called refraction of light.
Laws :
(i) Incident ray, refracted ray & normal lie in the same
plane.
sin i
(ii) Snell’s Law, refractive index (η) = sin r
Sphe
rica
lM
irr
ors
Fo
rm
ul
a
Chapter - 15
When beam of incident light bounces in the same
medium.
Laws :
(i) Incident ray, reflected ray and normal lie in
the same plane.
(ii) Angle of incidence (∠ i) = Angle of reflection (∠ r)
Dispersion of light
n
io
ct
a
r
f
Re
f
l
ec
tio
n
&
La
w
s
Mind map : learning made simple
at
io
C
n
o
of
m
an
th
po
in
d
u
p
n
l
en
dM
ow
se
er
ic
s
r
o
sco
pe
n
io
Re
16
OSWAAL CBSE Chapterwise Mind Maps, PHYSICS, Class - XI
er
ak
M
s
n’
Le