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CHHANDAK SIR PHYSICS CLASSES
Full marks-20
Answer any four questions:
1.(a) A thin spherical shell of metal has a radius of 0.25m and carries a charge of 0.2μC. Calculate the electric
intensity at a point (i) inside the shell, (ii) just outside the shell (iii) 3.0m from the centre of the shell.
[3]
(b) An Al wire of diameter 0.24cm is connected in series to a Cu wire of diameter 0.16cm.The wire carry an
electric current of 10A. Find (1) current density in Al wire,(2) drift velocity of electrons in Cu wire. Given n=
8.4x1028.
[2]
2.(a) Derive an expression for electric potential due to a point charge.
[3]
(b) A circular conducting loop of radius a and resistance R is placed with the plane perpendicular to a time
varying magnetic field B=B0sin𝜔t. Obtain an expression for the current induced in the loop.
[2]
3.(a) A potentiometer is being used for the determination of the internal resistance of a cell of emf E= 1.5V. The
balanced point of the cell in open circuit is at 76.3cm length of potentiometer wire. When a resistor of 9.5Ω is
connected across the cell, the balance point shifts to 64.8cm length of the wire. Find the internal resistance of
the cell.
[3]
(b) An infinite line charge produces an electric field of 9.0x 104NC-1 at a distance of 2cm.What is the liner charge
density?
[2]
4.(a)Obtain self inductance of a solenoid.
[3]
(b) A parallel plate capacitor is charged to a certain potential difference. When a 3mm thick slab is slipped
between the capacitor plates, then to maintain the same p.d. between the plates, the plate separation is to be
increased by 2.4mm. Find the dielectric constant of the slab.
[2]
5.(a) A d.c. electric motor operating at 30V input is drawing a current of 10A. The efficiency of the motor is 40%.
Calculate (1) Mechanical power output, (2) Energy lost at heat in 10sec.
[2]
(b) A voltmeter can measure up to 25V and its resistance is 1000Ω. What will you do to measure up to 250V? [3]
6.(a) With the help of a diagram, explain the principle and working of a cyclotron. Show that cyclotron frequency
does not depend on the speed of the particles.
[4]
(b) Ina moving coil galvanometer, what is meant by a radial magnetic field?
[1]
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SOLUTION OF SECTION A
1.(a) (i)The electric intensity at a appoint inside a charged shell is zero.
(ii) The intensity at a point just outside the shell is
1
𝑞
E=
2
4𝜋𝜀0 𝑅
Here q= 0.2 x 10-6C ,R=0.25m,
∴ E=(9x109) x
0.2 x 10−6
(0.25)2
= 2.88x104 N/C
(iii) The intensity at 3m from the centre of the shell is,
𝐸=
E= (9x109) x
1 𝑞
4𝜋𝜀0 𝑟 2
0.2 x 10−6
(3.0)2
= 200N/C
0.24
(b) (1) Given: Radius =
= 0.12cm = 0.12x10-2m
2
2
A= πr = πx (0.12x10-2)2 = 4.5x100-6m2
10
∴ Current density j= i/A =
= 2.2x106A/m2
4.5x106
(2) Area of cross section of Cu wire, A= πr 2
A= π(0.8x10-2)2 = 2x10-6m2
10
6
2
j=i/A =
−6 = 5x10 A/m
drift velocity vd =
2x10
𝑗
𝑛𝑒
5x10−6
vd = (8𝑥1028)𝑥
(1.6𝑥10−19 )
= 3.7x10-4m/sec
2.(a) Let +q C charge be situated at point O. Let at point P electric potential is to be determined. Force acting ob q 0
is given by,
1 𝑞𝑞0
F=
2
4𝜋𝜀0 𝑥
Work done in carrying the test charge from A to B against the force F is given by,
W= F x displacement
+q
P
1 𝑞𝑞0
=
O
2 x (-dx)
B
4𝜋𝜀0 𝑥
Work done in carrying the test charge +q0 from
infinity to P is,
𝑟
𝑟 −𝑑𝑥
1 𝑞𝑞0
𝑞𝑞
W =∫∞ −
𝑑𝑥 = 0 ∫∞ 2
4𝜋𝜀 𝑥 2
4𝜋𝜀
𝑥
0
0
=
𝑞𝑞0
V=
1
r
x
1
1
[ − ]
=
And V= W/q0 =
F
[ ]
4𝜋𝜀0 𝑥
𝑞𝑞0 1
=
+q 0
A
4𝜋𝜀0 𝑟
1 𝑞𝑞0
4𝜋𝜀0
∞
𝑟
1 𝑞𝑞0
4𝜋𝜀0 𝑟
𝑞0
𝑞
4𝜋𝜀0 𝑟
(b) The instantaneous magnetic flux is given by
𝜑=BA
= B0sin𝜔t x 𝓍a2
= B0𝜔cos𝜔𝑡 𝓍a2
So, i=
|𝑒|
𝑅
=
𝑥𝑞𝑎2 𝐵0 𝜔 cos 𝜔
𝑅
3. (a) Let K is the potential gradient along the potentiometer wire.
Then, E=K x 76.3
And V=K x 64.8
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∴
𝐸
=
𝑉
𝐾 𝑥 76.3
𝐾 𝑥 64.8
=
76.3
64.8
𝐸
76.3
𝑉
64.8
− 1) =1.7Ω
We know r=R( − 1)= 9.5(
(b) E=
1
𝜆
2𝜋𝜀0 𝑟
𝜆
9 x 104 = 9 x 109 x2 x
0.02
∴ λ = 0.1μc/m
4. (a) Let a long air is cored solenoid of length l and area of cross section A has N turns. Let I is the current through
the solenoid. Magnetic field is given by,
𝑁
B=μ0 i
𝑙
𝑁
Magnetic flux through each turn is given by, 𝜙𝐵 = 𝐵𝐴 = μ0 iA
𝑙
∴ Total magnetic flux through the solenoid is :
𝑁
N 𝜙𝐵 = 𝑁μ0 iA=
𝜇0 𝑖𝑁2 𝐴
𝑙
N𝜙
𝑙
𝜇 𝑁2 𝐴
The self inductance of the solenoid is: L= 𝐵 = 0
𝑖
𝑙
Since L is in Henry, it is clear from this equation that we may take the unit for μ 0 as Henry per meter also.
If the solenoid be wound on a wire of constant permittivity μ, then the total magnetic flux through the solenoid is
𝜇𝑁2 𝑖𝐴
𝑙
and the coefficient of the self-inductance becomes
𝜇𝑁2 𝐴
L=
𝑙
(b) Let E0 be the electric field in the entire space between the plates of the capacitor, distant d apart before the slab
is slipped. Then the p.d. between the plates is given by
V0 =E0 x d…….(1)
Suppose on slipping a slab of dielectric constant K and thickness t between the plates, the distance is to be
increased by d’in order to maintain the same p.d. between the plates. Then
𝐸
V0 =E0(d+d’-t)+ 𝑜 t………(2)
𝐾
𝐸
From equation (1) and (2) ,we have E0(d+d’-t)+ 𝑜 t= E0d
∴ K=
𝑡
𝑡−𝑑
′ =
3.0𝑚𝑚
3.00𝑚𝑚−2.4𝑚𝑚
=
3.0
0.6
𝐾
=5
5(a) Power input is given by Va x i=30 x 10=300watt
Efficiency of motor is 𝜂 = 40% =0.4
Power output= 𝜂 x power input = 0.4 x 300 =120W
Since 60% power is lost; so power loss = 0.6 x power input
= 0.6 x 300 = 180W= 180J/s
Energy lost in 10sec= 180 x 10= 1800J.
(b) Let G (=1000Ω) be the resistance of the voltmeter. The current giving full scale deflection in the voltmeter
(reading 25 V) is given by,
25
ig =
= 0.025A
1000𝛺
The required new range of the voltmeter is V’=250V. It will read up to 250 V is an appropriate resistance R is
connected in series with it. Now we have
ig =
R=
𝑉′
𝐺+𝑅
𝑉′
𝑖𝑔
− 𝐺=9000Ω
7. (a) It consists of two horizontal D-shaped hollow metal segments D1 and D2(called the dees) with a small gap
between them. An alternating potential difference of the order to 10 5 V at a frequency of 10 to 15 megacycles
per second is applied across the dees.An intense magnetic field → of about 1.6 T is set up perpendicular to the
𝐵
plane of dees by a large electromagnet. The whole space inside the dees is evacuated to a pressure of about
10-6mm of Hg.
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An ion source is located at the center S in the gap between the dees. It consists of a small chamber containing
a heated filament and a gas such as hydrogen (for protons) or deuterium (for deuterons). The thermions given
out by the filament produce positive ions by ionization of the gas. The ions come out through a small hole in
the ion source and are available to be accelerated.
Working: suppose that an ion of mass m and charge +q emerges from the ion source at an instant when D 2 is at
a negative potential. It will be accelerated towards D2 by the electric field in the gap between the dees enter D 2
with a velocity v(say).
Now under the action of the magnetic field which is perpendicular to t6he plane of dees, the ion
adopts a circular path with constant speed v and of radius r given by
𝑚𝑣
r=
𝑞𝐵
where B is the a magnetic field. The time t required by the ion to complete a semicircular is given by
𝜋𝑟
𝑚𝜋
t= =
𝑣
𝑞𝐵
Let us assume that the frequency of the applied p.d. has been so adjusted that during the one half cycle of
p.d. the ion complete a semi-circle. Then the ion will emerge from D2 into the gap at the instant when D1 is at a
negative potential. The ion is therefore further accelerated while crossing the gap between D 1 and D2 and enters
D1. On accounts of its increased velocity its semi circular path in D 1 is now of greater radius. The time of
passage t through D1 however is still the same. The process is repeated after every half cycle of p.d. and the ion
gains in speed each time it passes from one dee to other. Finally the ion becomes enough energetic to reach the
outer edge of one dee where it is pulled out of the system by a negatively charged deflector plate.
(b) The uniform magnetic field which keeps the plane of the coil always parallel to the direction of the magnetic
field, that is, the angle between the plane of the coil and the magnetic field is zero in all the orientation of the
coil. Further since filed lines always lie in the plane of the loop and hence torque is independent of the
orientation of the loop.
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