Target Publications Pvt. Ltd.
Physics
Board Answer paper: October 2014
Physics
SECTION – I
Q.1. Attempt any THREE:
i.
a.
The components of forces acting on a vehicle moving on a curved banked road are
represented in diagram as follows:
N
N cos θ
θ
N sin θ
F cos θ
θ
G
C
F
A
h
θ
F sin θ
B
W = mg
b.
AC : inclined road surface
AB : horizontal surface
BC : height of road surface
G : centre of gravity of vehicle
W : (mg) weight of vehicle
N : normal reaction exerted on vehicle.
θ : angle of banking
Equation for maximum safety speed for the vehicle moving on the curved banked road
[1]
⎡ µ + tan θ ⎤
rg ⎢ s
⎥
⎣1 − µs tan θ ⎦
where, r is radius of curved road.
µs is coefficient of friction between road and tyres,
θ is angle of banking.
[1]
Significance of the terms
i.
The maximum safety speed of a vehicle on a curved road depends upon friction
between tyres and roads.
ii.
It depends on the angle through which road is banked. Also absence of term ‘m’
indicates, it is dependent of mass of the vehicle.
[1]
Maxwell’s distribution of molecular speed:
a.
When a given mass of gas is considered as a system, each molecule travels with
different velocity.
b.
For fixed pressure, volume and temperature conditions, collisions change direction and
speed of molecules.
c.
The distribution of speed is constant when a state of equilibrium is attained by system.
d.
Maxwell was first to give distribution of speeds among the gas molecules at a given
temperature.
[½]
is vmax =
c.
ii.
1
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Target Publications Pvt. Ltd.
e.
f.
g.
Board Answer Paper: October 2014
If dN(v) represents number of molecules between the speeds v and v + dv, dN(v)
remains almost constant at equilibrium and is proportional to dv.
In a system of N molecules, if probability that a molecule has speed between v and
v + dv is ηvdv, then, dN(v) = Nηvdv
The graph of ηv against v is shown in the figure.
[½]
ηv(dv)
ηv
2.5
1 dv 1.5
2
v
Maxwell distribution
The fraction of molecules with speed between v and v + dv equals the area of the strip
dN(v)
= ηvdv
shaded. Thus
N
Total area corresponds to the fraction of molecules with speed between 0 and ∞ and
hence equals 1.
0.5
h.
i.
iii.
6
−11
2
2
[1]
[1]
6
h = 1800 km = 1.8 × 10 m, G = 6.67 × 10 N m /kg , R = 6400 km = 6.4 × 10 m,
m = 800 kg, M = 6 × 1024 kg
Total Energy (T.E.) = ?, Binding Energy (B.E.) = ?
T.E. = −
GMm
2(R + h)
[½]
T.E. = −
6.67 × 10−11 × 6 × 1024 × 800
2 ( 6.4 + 1.8 ) × 106
[½]
32016 × 107
2 × 8.2
T.E. = −1.953 × 1010 J
The total energy of the artificial satellite is –1.953 × 1010 J.
B.E. = −T.E.
B.E. = 1.953 × 1010 J
The binding energy of the artificial satellite is 1.953 × 1010 J.
=−
∴
iv.
∴
∴
∴
∴
2
70
70
m, λ2 =
m
153
157
Frequency (n) = ?, Velocity of sound (v) = ?
As v = nλ
For two notes in air, n1λ1 = n2λ2
70
70
n1 ×
= n2 ×
157
153
λ1 > λ2
n2 > n1
From the given condition in the question,
n2 = n + 8 and n1 = n − 8
70
70
= (n − 8)
(n + 8)
157
153
153 (n + 8) = 157 (n− 8)
[1]
[1]
λ1 =
[½]
[½]
Target Publications Pvt. Ltd.
∴
Physics
n = 620 Hz
[1]
The frequency of tuning fork is 620 Hz
Using formula, v = nλ
70
157
70
= (620 + 8)
157
70
v = 628 ×
157
v = (n + 8)
∴
= 280 m/s
The velocity of sound in air is 280 m/s.
[1]
Q.2. Attempt any SIX:
i.
Different stages (cases) of projection for artificial satellite:
Earth
Hyperbola
(v > ve)
→
vc
R
circle
(v = vc)
Parabola
(v = ve)
Ellipse
(vc< v < ve)
ii.
iii.
(drawing and labelling)
(1 + 1)
Statement: The angular momentum of a body remains constant, if resultant external torque
acting on the body is zero.
Example:
a.
A ballet dancer makes use of the law of conservation of angular momentum to vary her
angular speed.
b.
Torque acting on her body is zero. By law of conservation of angular momentum,
L = constant
1
Iω = constant ⇒ ω ∝
I
c.
When she suddenly folds her arms and brings the stretched leg close to the body, her
angular velocity increases on account of decrease in moment of inertia.
This helps in rotating safely on her legs.
When a liquid is in contact with a solid, the angle between tangent drawn to the free surface
of the liquid and the surface of solid at the point of contact measured inside the liquid is
called angle of contact.
Characteristics:
a.
The angle of contact is constant for a given liquid-solid pair.
b.
The value of angle of contact depends upon nature of liquid and solid in contact.
c.
It depends upon the medium which exists above the free liquid surface.
d.
The angle of contact changes due to impurity or temperature.
(Any two characteristics)
(½ × 2)
3
[2]
[1]
[1]
[1]
[1]
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Target Publications Pvt. Ltd.
Board Answer Paper: October 2014
Ferry’s perfectly black body:
iv.
Double walled hollow
metal sphere
Incident radiation
Conical
projection
a.
b.
Aperture
Lamp black
Evacuated space
It consists of a double walled hollow sphere, having a small aperture in it.
The inner surface of the sphere is coated with lamp black. The outer surface is nickel
polished.
There is a conical projection diametrically opposite to the aperture.
The space between the two walls is evacuated to avoid loss of heat by conduction and
convection.
c.
d.
[1]
[1]
m = 5 kg, r = 0.8 m, g = 9.8 m/s2
Minimum velocity at the highest point (vH) = ?
Minimum velocity at midway point (vM) = ?
At highest point,
v.
VH =
=
rg
0.8× 9.8
= 7.84
= 2.8 m/s
The minimum velocity at the highest point is 2.8 m/s.
At midway point,
VM =
=
3rg
∴
[½]
[½]
3 × 2.8
= 1.732 × 2.8
= 4.849
≈ 4.85 m/s
The minimum velocity at the midway point is 4.85 m/s.
vi.
[½]
[½]
vmax = 0.16 m/s, amax =0.64 m/s2
Period (T) = ?
vmax = Aω and amax = Aω2
a max
2π
=ω=
T
v max
[1]
2π× v max
a max
2 × 3.142 × 0.16
=
0.64
= 1.571 s
Period of particle is 1.571 s
[½]
T=
4
[½]
Target Publications Pvt. Ltd.
vii.
h1 = 3.2 cm, A2 =
Physics
A1
, h2 = ?
2
1
r
h1r1 = h2r2
πr 2
πr22 = 1
2
r
⇒ r2 = 1
2
3.2 × r1
h2 =
r1 2
As, h ∝
∴
∴
∴
[½]
[½]
= 3.2 × 2 = 4.525 cm
The water will rise to a height of 4.525 cm.
[1]
viii. l = 36 cm = 0.36 m, n = 280 Hz, T = 24.5 N
linear density (m) = ?
1 T
n =
2l m
T
m = 2 2
4l n
24.5
=
4 × 0.362 × 2802
[½]
[½]
= {antilog[log(24.5) − log 4 − 2log(0.36) − 2log(280)]}
= {antilog[1.3892 − 0.6021 − 2(1.5563) − 2(2.4472)]}
= {antilog[1.3892 − 0.6021 − (−0.8874) − 4.8944]}
{
}
= antilog[4.7801]
−4
∴
= 6.027 × 10 kg/m
Linear density of material of wire is 6.027 × 10−4 kg/m
[1]
Q.3. Select and write the most appropriate answer from the given alternatives for each
sub-questions:
i.
(C)
M
Linear density ρ =
L
∴
M = ρL
Wire of length L is bent into a coil of radius R
L
∴
R=
2π
M.I. of coil through any tangent in the plane of the coil
3
= MR2
2
=
3
⎛ L ⎞
(ρL) × ⎜ ⎟
2
⎝ 2π ⎠
=
3ρL3
8π 2
2
[1]
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Target Publications Pvt. Ltd.
Board Answer Paper: October 2014
ii.
(A)
[1]
iii.
(B)
[1]
iv.
(D)
Elongation ∝
1
r2
(Elongation) thin
(n / 2) 2
= n2
=
(1 / 2) 2
(Elongation) thick
[1]
v.
(B)
[1]
vi.
(C)
[1]
viii. (D)
[1]
Q.4. A.
Sr.
Forced vibrations
Resonance
No.
i.
It is produced by an external periodic It is produced by an external periodic force
force of any frequency.
whose frequency is equal to the natural
frequency of the body.
ii.
The frequency of vibrations is The frequency of vibrations is same as the
different from the natural frequency natural frequency of the body.
of the body.
iii. The amplitude of vibration is small.
The amplitude of vibration is very large.
iv. Vibrations
stop as soon as the Vibrations continue for some time even
external force is removed.
after the external force is removed.
v.
Faint sound is produced.
Loud sound is produced.
(Any two of the above points)
(1 × 2)
[2]
Mode of vibration of a stretched string in second harmonic:
A
A
N
N
N
l = λ1
[1]
Mode of vibration of a stretched string in third harmonic:
N
A
N
A
N
A
N
3λ
l= 2
2
B.
6
A = 0.5 m × 0.5 m = 0.25 m2
h = 1 cm = 10−2 m, x = 0.015 mm = 15 × 10−6 m
η = 4.5 × 1010 N/m2
Strain (θ) = ?, Shearing force (F) = ?
x
θ=
h
[1]
[½]
Target Publications Pvt. Ltd.
Physics
15 ×10−6
10−2
= 1.5 × 10−3
Shearing strain is 1.5 × 10−3
Shearing force F = ηAθ
= 4.5 × 1010 × 0.25 × 1.5 × 10−3
= 1.688 × 107 N
Shearing force is 1.688 × 107 N.
=
[½]
[½]
[½]
[1]
OR
The physical quantity which describes the state of oscillation of a particle performing S.H.M
at any instant is called phase of S.H.M.
In the equation of S.H.M,
x = A sin (ωt + α)
where, (ωt + α) is the phase or phase angle of S.H.M
For a particle starting from extreme position:
Displacement time graph:
Time
(t)
0
T/4
T/2
3T/4
T
[1]
Displacement
(x)
A
0
−A
0
A
Phase
(ωt)
0
π/2
π
3π/2
2π
Graph:
Displacement
Q.4. A.
+A
0
−A
T
4
T
2
3T
4
T
5T
4
3T
2
t
[1]
Velocity-time graph:
Time
(t)
0
T/4
T/2
3T/4
T
Phase
(ωt)
0
π/2
π
3π/2
2π
Velocity
(v)
0
− Aω
0
Aω
0
7
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Target Publications Pvt. Ltd.
Board Answer Paper: October 2014
Graph:
Velocity
+Aω
0
−Aω
T
4
T
2
3T
4
T
3T
2
5T
4
t
[1]
Acceleration-time graph:
Time
(t)
0
T/4
T/2
3T/4
Phase
(ωt)
0
π/2
π
3π/2
Acceleration
(a)
− Aω2
0
Aω2
0
T
2π
− Aω2
Acceleration
Graph:
+Aω2
0
−Aω2
B.
∴
∴
8
T
4
T
2
3T
4
T
5T
4
3T
2
t
T = 20 N m, n = 60 revolutions in 60 s = 1 r.p.s., t = 1 min = 60 s
Moment of inertia (I) = ?
Angular displacement θ = ωt = 2πnt
= 2π × 1 × 60
= 120 π rad
1
θ = ω0t + α t 2
2
1
120 π = 0 + α × 602
2
240 π
α=
602
π
rad/s2
=
15
τ = Iα
τ
I =
α
20
=
π / 15
= 95.48 kg m2
Moment of inertia of the body is 95.48 kg m2
[1]
[½]
[½]
[½]
[½]
[½]
[½]
Target Publications Pvt. Ltd.
Physics
SECTION − II
Q.5. Attempt any THREE:
i.
∴
∴
∴
λ = 4800 Å = 4.8 × 10−7 m, d = 3 mm = 3 × 10−3 m, D = 15 + 85 = 100 cm = 1 m
Distance between 4th bright band on one side and 4th dark band on the other side of the central
bright band, (X4 + X′4) = ?
For nth bright band,
n λD
Xn =
d
4λ D
X4 =
d
For nth dark band,
λD
X′n = (2n – 1)
2d
λD 7 λD
=
2d
2 d
4 λD 7 λ D
+
X4 + X′4 =
d
2 d
[½]
[½]
X′4 = (2 × 4 − 1)
[½]
4 × 4.8 × 10−7 × 1 7 4.8 × 10−7 × 1
+ ×
3 × 10−3
2
3 × 10−3
= 6.4 × 10−4 m + 5.6 × 10−4 m
= 12 × 10−4 m
The distance between 4th bright band on one side and 4th dark band on the other side of
the central bright band is 12 × 10−4 m.
=
∴
ii.
[1]
[½]
Given balanced network of capacitors with
CAB = CBC = CAD = CBD = 5 µF, CDC = (10 + X) µF
The value of X = ?
Resultant capacitance between A and C (Ceq) = ?
C1 C2
for branch DC of the given network we get,
a.
Using formula, CS =
C1 + C 2
10 × X
….(i)
10 + X
Now using formula,
C AB C AD
=
(For balance condition)
CBC CDC
CS =
∴
∴
∴
∴
∴
5
5
=
5 ⎛ 10 × X ⎞
⎜
⎟
⎝ 10 + X ⎠
(10 + X)
1=
2X
2 X = 10 + X
X = 10 µF
The value of X is 10 µF.
[1]
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Target Publications Pvt. Ltd.
b.
Board Answer Paper: October 2014
As the bridge is balanced, no current flows through branch BD. Hence the network can
be reduced as follows:
5 µF B 5 µF
A
C
5 µF D 10 µF
+ –
5V
∴
iii.
Here, 5 µF and 5 µF are in series in the branch ABC.
Using formula (i),
5×5
= 2.5 µF
CS =
5+5
Also, 5 µF and 10 µF are in series in the branch ADC.
5 × 10 50
C′′S =
=
= 2.5 µF
10 + 10 20
Now, C′S and C′′S are in parallel.
∴
CAC = CP = C′S + C′′S = 2.5 µF + 2.5 µF
∴
= 5 µF
The resultant capacitance between A and C is 5 µF.
a.
[½]
[½]
[1]
Suppose that rectangular coil PQRS is kept in uniform magnetic field of induction ‘B’.
Let ‘n’ be the number of turns of the coil with ‘l’ as its length and ‘b’ as its breadth.
The current ‘I’ is passed through it in anticlockwise direction.
→
I
P
B
S
F2
F1
Q
l
b
R
[½]
Torque acting on a rectangular coil
b.
c.
d.
e.
∴
∴
10
→
The forces on QR and SP are equal to zero because they are parallel to B .
Force on PQ,
(directed normally outwards)
F1 = nIBl
Force on RS,
(directed normally inwards)
F2 = nIBl
Two forces F1 and F2 are equal in magnitude but opposite in direction and act at
different points. Hence these forces constitute a torque (τ) and rotate the coil.
Magnitude of torque is given by,
τ = Magnitude of one of the forces × perpendicular distance between these parallel forces
τ = (nBIl) (b) = nBI (lb)
τ = nBIA
where, A = lb = area of rectangular coil PQRS.
[½]
Target Publications Pvt. Ltd.
f.
g.
Physics
This torque deflects the coil hence it is called deflecting torque.
It is given by,
τd = nBIA
….(i)
This torque causes the pointer attached to the coil to deflect and move on a graduated
scale.
As the coil is deflected, the phosphor bronze wire is twisted. This twist in phosphor
bronze wire provides restoring or controlling torque.
[½]
S
N
P
Radial magnetic field
h.
∴
i.
∴
∴
C ⎞
I = ⎛⎜
⎟ θ
nAB
j.
Equation (iii) represents the current flowing through M.C.G.
For a given M.C.G., n, B, A are constant and for given suspension fibre
C = constant.
∴
∴
a.
⎝
⎠
.…(iii)
[½]
[½]
C
= k = constant
nAB
I = kθ
I∝θ
Hence, current flowing through galvanometer is proportional to deflection produced in
it.
[½]
Energy band structure in conductor:
1.
Conductors are those substances which allow the passage of electric current through
them.
2.
In a conductor, the valence and conduction energy band overlap each other. Due to
this overlapping, a slight P.D across the conductor forces electrons to constitute
electric current.
3.
In case of sodium, valence band may be completely filled with an extremely
small energy gap between them as shown in figure (i).
Energy
∴
iv.
This restoring torque is directly proportional to the deflection of the coil.
i.e. τr ∝ θ
τr = Cθ
….(ii)
where, the constant of proportionality C is called twist constant or restoring torque per
unit twist.
For equilibrium of the coil, τd = τr
nBIA = Cθ
Partially filled
Conduction band
Very small
energy gap
Completely filled
valence band
[½]
Figure (i) Valence band and conduction
band in Sodium
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In case of zinc, the valence band is completely filled and the conduction band is
empty but the two overlap each other as shown in figure (ii).
Empty conduction
band
Overlapping region
Energy
4.
Board Answer Paper: October 2014
Completely filled
valence band
Figure (ii) Valence band and conduction
band in Zinc
5.
b.
In both the situations, it can be assumed that there is a single energy band which
is partially filled.
Therefore on applying even a small electric field, metals conduct electricity.
Thus the electrical behaviour of conductors can be satisfactorily explained by the
band energy theory of materials.
[1]
Energy band structure in semiconductor:
Energy
Empty conduction band
1.
2.
3.
4.
Eg < 3 eV
Eg Small forbidden gap
Completely filled
valence band
[½]
In terms of energy bands, we can define semiconductors as those substances,
which at room temperature have partially filled conduction band and partially
filled valence band with a very narrow energy gap between them. It is less than
3 eV.
For Ge, energy gap Eg = 0.7 eV and for Silicon Eg =1.2 eV.
At absolute zero, there are no electrons in their conduction band and their valence
band is completely filled i.e., at absolute zero, semiconductors like Ge and Si
behave like insulators.
As the temperature is increased, the energy gap is decreased. Some of the electrons go
to the conduction band and thus the conductivity increases with temperature.
[1]
Q.6. Attempt any SIX:
i.
A Plane of vibration
P
D
S
Plane polarised light
• •
Unpolarised Q
light
R
B
C
ABCD is plane of vibration,
PQRS is plane of polarisation
ii.
12
Plane of
polarisation
(1 + 1)
(Diagram and labelling)
Conditions for steady interference pattern:
a.
The two sources of light must be coherent.
b.
The sources of light must be monochromatic.
c.
The two light sources must be equally bright i.e. they must emit lights of equal
amplitude and intensity.
d.
The two light sources must be narrow.
[2]
[½]
[½]
[½]
[½]
Target Publications Pvt. Ltd.
iii.
Physics
Expression for magnetic dipole moment:
a.
Consider an electron of mass me and charge e revolves in a circular orbit of radius r
around the positive nucleus in anticlockwise direction, leading to a clockwise
current.
→
L
+e
→
r
v
e−
U.C.M of electron in an atom
b.
c.
The angular momentum of an electron due to its orbital motion is given by,
L0 = mevr ….(i)
For the sense of orbital motion of electron shown in the figure, the angular momentum
→
d.
e.
vector L acts along normal to the plane of the electron orbit and in upward direction.
Suppose that the period of orbital motion of the electron is T. Then the electron crosses
any point on its orbit after every T seconds or 1/T times in one second.
Magnitude of circulating current is given by,
⎛1⎞
I = e⎜ ⎟
⎝T⎠
2πr
v
ev
⎛ 1 ⎞
I = e⎜
⎟=
2
πr
2
π
r
/
v
⎝
⎠
But, T =
∴
f.
[½]
[½]
The magnetic dipole moment associated with circulating current is given by,
ev
× π r2
M0 = IA =
2π r
[½]
[∵ Area of current loop, A = πr2]
∴
evr
….(ii)
2
This is the required expression.
M0 =
[½]
iv.
Information
source
Noise
User of
information
Message
signal
Transmitter
Message
signal
Transmitted Communication Received
Signal
Channel
Signal
Receiver
Block diagram of a generalised communication system
(Diagram and labelling)
(1 + 1)
13
[2]
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Target Publications Pvt. Ltd.
v.
(λr)air = 6400 Å, (λr)glass = 4000 Å,
(λv)air = 4400 Å
Wavelength (λv)glass = ?
Using the formula,
λa
for red and violet colours we get,
aµg =
λg
( λ r )air
and
( λ r )glass
( λ v )air
air ( µ v )glass =
( λ v )glass
( λ r )air ( λ v )air
=
( λ r )glass ( λ v )glass
( λ r )glass 4400× 4000
(λv)glass = (λv)air ×
=
6400
( λ r )air
air
∴
∴
Board Answer Paper: October 2014
( µ r )glass =
∴
∴
(λv)glass = 2750 Å
The wavelength of violet light in glass is 2750 Å.
vi.
l = 5 cm = 5 × 10−2 m, b = 2.5 cm = 2.5 × 10−2 m, h = 1.25 cm = 1.25 × 10−2 m
Mnet = 3 Am2, MZ = ?
M
Now, MZ = net
V
M net
=
l ×b×h
3
=
−2
5 × 10 × 2.5 × 10−2 × 1.25 × 10−2
= 0.192 × 106
= 1.92 × 105 A/m
The intensity of magnetization is 1.92 × 105 A/m.
∴
vii.
14
[½]
[½]
[½]
[½]
[½]
[½]
[1]
−3
L = 125 mH = 125 × 10 H,
C = 50 µF = 50 × 10−6 F
fr = ?
We have,
1
fr =
2π LC
1
=
2 × 3.14 × 125 × 10−3 × 50 × 10−6
1
=
6.28 × 625 × 10−8
1
=
6.28 × 25 × 10−4
104
=
6.28 × 25
400
=
6.28
[½]
[½]
Target Publications Pvt. Ltd.
∴
∴
Physics
= antilog [log (400) – log (6.28)]
= antilog [2.6021 – 0.7980]
= antilog [1.8041]
fr = 63.69 Hz
The resonant frequency in the A.C. circuit is 63.69 Hz.
[1]
c
3 × 108
=
= 108 m/s, h = 6.63 × 10−34 Js,
3
3
m = 9.11 × 10−28 g = 9.11 × 10−31 kg, λ = ?
h
λ=
mv
6.63 × 10−34
=
9.11 × 10−31 × 108
66.3
=
× 10−35 × 1031 × 10−8
9.11
66.3
=
× 10−12
9.11
λ = {antilog [ log(66.3) − log(9.11) ]} × 10−12
viii. v =
∴
[½]
[½]
= {antilog [1.8215 − 0.9595]} × 10−12
= {antilog [ 0.8620]} × 10−12
∴
= 7.278 × 10−12
The de Broglie wavelength of electron is 7.278 × 10−12 m.
[1]
Q.7. Select and write the most appropriate answer from the given alternatives for each
sub-question:
i.
(B)
ii.
(B)
[1]
[1]
d 4
l = 1.5 m, d = 4 cm, r = = = 2 cm = 2 × 10−2 m,
2 2
n = 10 per metre, I = 5 A, µ0 = 4π × 10−7 Wb/Am
Now, B = µ0nI
= 4π × 10−7 × 10 × 5
= 2π × 10−5 T
iii.
(C)
[1]
iv.
(C)
[1]
∴
∴
∴
∴
hc
We have, E =
λ
hc
hc
E1 =
and E2 =
λ1
λ2
E1 – E2 =
hc hc
−
= hc
λ1 λ 2
⎛1
1 ⎞
⎜ − ⎟
⎝ λ1 λ 2 ⎠
….[∵ E1 > E2]
⎛ λ − λ1 ⎞
E1 – E2 = hc ⎜ 2
⎟
⎝ λ1 λ 2 ⎠
(E − E 2 ) λ1λ 2
h= 1
c (λ 2 − λ1 )
15
-
Target Publications Pvt. Ltd.
Board Answer Paper: October 2014
v.
(C)
[1]
vi.
(D)
[1]
vii.
(A)
[1]
E=
∴
Q.8. A.
V
F ma
but E = =
d
q
q
ma V
qV
=
⇒a=
q
d
dm
a.
The phenomenon of production of an induced e.m.f in a coil due to change in current in
the same coil is called self induction.
It is also called as back-e.m.f.
b.
L
−
+
E
K
c.
Consider a coil connected with battery E, plug key K and inductor L carrying current of
magnitude I as shown in figure.
Since magnetic flux linked with the coil is directly proportional to the current,
∴
φ∝I
∴
φ = LI
.…(i)
where, L = constant called coefficient of self induction or self inductance of the coil,
which depends upon the number of turns, shape, area of the coil and material of the
core.
Induced e.m.f in the coil is given by,
d.
e=−
[½]
dφ
dt
e=−L
dI
dt
….(ii)
−ve sign in equation (ii) shows that self induced e.m.f opposes the rate of change of
current.
dI
dI
=L
dt
dt
∴
|e| = − L
∴
Magnitude of self induced e.m.f is given by, e = L
dI
dt
This is required induced e.m.f.
e.
16
The phenomenon of production of induced e.m.f in one coil due to change of current in
the neighbouring coil is called mutual induction.
The e.m.f so induced is called mutually induced e.m.f.
[½]
Target Publications Pvt. Ltd.
f.
Physics
+
E
−
P
S
G
K
Consider primary coil P and secondary coil S fitted with galvanometer G and placed
very close to each other as shown in figure. The coil P is connected in series with the
source of e.m.f (battery) and key K.
g.
When tap key K is pressed, current IP passes through the coil P.
Magnetic flux φS linked with secondary coil S at any instant is directly proportional to
current IP through primary coil P at that instant.
∴
φS ∝ IP
∴
φS = M IP ….(i)
where M is constant called coefficient of mutual induction or mutual inductance of the coil.
h.
e.m.f induced in S at any instant is given by,
d φS
eS = −
dt
d
= − (MIP) [From equation (i)]
dt
dI
∴
eS = − M P
dt
∴
Magnitude of induced e.m.f is given by,
MdIP
MdIP
eS = −
=
dt
dt
eS
M=
∴
dI P / dt
Co-efficient of self induction of a coil is defined as the e.m.f induced in the coil per unit rate
of change of current in the same coil.
Coefficient of mutual induction is defined as the e.m.f induced in the secondary coil per unit
rate of change of current in the primary coil (neighbouring coil).
SI Unit of co-efficient of self induction is henry (H) in SI system or volt A−1 s.
Dimensions of self induction are [L2M1T−2A−2]
B.
∴
∴
∴
∴
∴
L = 4 m, R1 = 5 Ω, E = 2 V, r = 1 Ω, K = 10−3 V/cm = 10−1 V/m, X = ?
V
P.D.
= applied
….(i)
Potential gradient =
length
L
IR
=
L
E
⎛
⎞R
=⎜
….(ii)
⎟
⎝R+r+X⎠ L
where, X = series resistance
2
⎛5⎞
10−1 =
….[From (i) and (ii)]
⎜ ⎟
5 +1+ X ⎝ 4 ⎠
5
6 + X = 20 ×
4
6 + X = 25
X = 25 – 6 = 19 Ω
A resistance of 19 Ω should be connected in series.
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[1]
17
-
Target Publications Pvt. Ltd.
Board Answer Paper: October 2014
OR
Q.8. A.
Expression for total energy of electron in the nth orbit (En)
i.
Kinetic energy (K.E.):
Let, m = mass of electron
rn = radius of nth orbit of Bohr’s hydrogen atom
vn = velocity of electron
−e =
+e =
charge of electron
charge on the nucleus
vn
+e
rn
e−
According to Bohr’s first postulate,
mv2n
rn
=
e2
1
× 2
4πε0
rn
where, ε0 is permittivity of free space.
∴
mv 2n =
1 e2
4πε0 rn
….(i)
The revolving electron in the circular orbit has linear speed and hence it possesses
kinetic energy.
It is given by,
K.E =
1
mv 2n
2
∴
K.E =
1
2
∴
K.E =
e2
8πε 0 rn
ii.
Potential energy (P.E.):
⎛ 1 e2 ⎞
. ⎟ [From equation (i)]
⎜
⎝ 4πε0 rn ⎠
.…(ii)
[1]
Potential energy of electron is given by, P.E = V(−e)
where, V = electric potential at any point due to charge on nucleus
− e = charge on electron.
18
∴
P.E =
1 e
× (−e)
4πε0 rn
∴
P.E =
−e 2
1
×
rn
4πε0
∴
P.E = −
e2
4 πε 0 rn
….(iii)
[½]
Target Publications Pvt. Ltd.
iii.
Physics
Total energy (En):
The total energy of the electron in any orbit is its sum of P.E and K.E.
∴
T.E = K.E + P.E
⎛ e2
⎞ ⎛
e2 ⎞
=⎜
⎟ +⎜−
⎟
⎝ 8πε0 rn ⎠ ⎝ 4πε 0 rn ⎠
[From equations (ii) and (iii)]
1 e2
.
8πε 0 rn
∴
En = −
∴
En = −
iv.
⎛ ε h2 ⎞
But, rn = ⎜ 0 2 ⎟ n 2
⎝ πme ⎠
e2
8πε0 rn
….(iv)
[½]
Substituting for rn in equation (iv),
∴
En = −
1
e2
8πε 0 ⎛ ε 0 h 2 ⎞ 2
n
⎜
2 ⎟
⎝ πme ⎠
=−
∴
1
e2 πme2
×
8πε0 ε0 h 2 n 2
⎛ me4 ⎞ 1
En = − ⎜ 2 2 ⎟ 2
⎝ 8ε0 h ⎠ n
….(v)
This is required expression for energy of electron in nth orbit of Bohr’s hydrogen atom.
v.
The negative sign in equation (v) shows that the electron is bound to the nucleus by an
attractive force and hence energy must be supplied to the electron in order to make it
free from the influence of the nucleus.
vi.
Now, as m, e, ε0, and h in equation (v) are constant,
∴
me 4
= constant
8ε0 2 h 2
[½]
Using equation (v), we get,
⎛ 1 ⎞
En = constant ⎜ 2 ⎟
⎝n ⎠
∴
En ∝
1
n2
Hence, the total energy of electron in a Bohr’s orbit is inversely proportional to the
square of the principal quantum number.
[½]
The amount of energy required to separate all the nucleons from the nucleus is called
binding energy of the nucleus.
[1]
19
-
Target Publications Pvt. Ltd.
B.
Board Answer Paper: October 2014
λ0 = 230 nm = 230 × 10−9 m, λ = 180 nm = 180 × 10−9 m,
h = 6.63 × 10−34 J s, c = 3 × 108 m/s, K.E. (in joule and eV) = ?
Now,
K.E.max = h (ν − ν0)
⎛c c ⎞
=h ⎜ − ⎟
⎝ λ λ0 ⎠
⎛1 1 ⎞
⎛λ −λ⎞
= hc ⎜ − ⎟ = hc ⎜ 0
⎟
⎝ λ λ0 ⎠
⎝ λ λ0 ⎠
[½]
1
1
⎛
⎞
= 6.63 × 10−34 × 3 × 108 ⎜
−
−9
−9 ⎟
230 × 10 ⎠
⎝ 180 × 10
[½]
6.63 × 3 × 10−26 ⎛ 23 − 18 ⎞
⎜
⎟
10−8
⎝ 23 × 18 ⎠
6.63 × 3 × 5
× 10−18
=
23 × 18
=
= {antilog [ log(6.63) + log(3) + log(5) − log(23) − log(18)]} × 10−18
= {antilog [ log(0.8215) + 0.4771 + 0.6990 − 1.3617 − 1.2553]} × 10−18
= {antilog [1.9976 − 2.6170]} × 10−18
{
}
= antilog ⎡⎣ 1.3806 ⎤⎦ × 10−18
∴
∴
∴
20
= 0.2402 × 10−18
K.E.max = 2.402 × 10−19 J
The maximum kinetic energy of the ejected electron is 2.402 × 10−19 J.
2.402 × 10−19
eV
=
1.6 × 10−19
= antilog [log (2.402) − log (1.6)]
= antilog [0.3806 − 0.2041]
= antilog [0.1765]
= 1.5 eV
The maximum kinetic energy of the ejected electron is 1.5 eV.
[1]
[1]