Download AIEEE PHYSICS PAPER 2006 Model Solutions

AIEEE PHYSICS PAPER 2006
Model Solutions
1. An electric dipole is placed at an angle of 30° to a non-uniform electric
field. The dipole will experience
(1) a torque as well as a translational force
(2) a torque only
(3) a translational force only in the direction of the field
(4) a translational force only in a direction normal to the direction of the
field
Solution: The correct choice is (1) .
2. A whistle producing sound waves of frequencies 9500 Hz and above is
approaching a stationary person with speed v ms–1. The velocity of sound
in air is 300 ms–1. If the person can hear frequencies upto a maximum of
10,000 Hz, the maximum value of v upto which he can hear the whistle is
15
(2) 30 ms–1
(3) 15 2 ms–1 (4)
ms–1
(1) 15 ms–1
2
Solution: The frequency of sound as heard by the stationary person is
Ê v ˆ
n¢ = n Á
Ë v – us ˜¯
Ê 300 ˆ
10,000 = 9500 Á
Ë 300 – us ˜¯
\
which gives us =15 ms–1. Hence the correct choice is (1).
3. In a region, steady and uniform electric and magnetic fields are present.
These two fields are parallel to each other. A charged particle is released
from rest in this region. The path of the particle will be a
(1) ellipse
(2) circle
(3) helix
(4) straight line
Solution: Consider a particle of charge q in a region of parallel and uniform electric field E and magnetic field B as shown in the adjoining figure. The electric field exerts a force
Fe = q E
in the direction of the field. As a result the particle is accelerated in the
direction of the field. If v is the velocity of the particle at an instant of
time then at that instant the force experienced by the particle due to the
magnetic field is
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M.64 AIEEE Physics in 30 Days
Fm = q (v ¥ B)
E
q
vv
B
Since v is parallel to B, Fm = 0. Hence the particle is moving in a straight
line in the direction of the electric field if it carries a positive charge and
opposite to the direction of the electric field if it carries a negative charge.
Hence the correct choice is (4).
4. A mass of M kg is suspended by a weightless string. The horizontal
force that is required to displace it until the string makes an angle of 45°
with the initial vertical direction is
Mg
(2) Mg ( 2 – 1)
(1)
2
(3) Mg
(
)
2 +1
(4) Mg
2
Solution: Refer to the adjoining figure. Let l be the length of the string.
The mass is moved from B to C by a horizontal force F until q = 45°. The
work done against the force of gravity is
A
Wg = Mgh = Mg DB
q
l
= Mg (AB – AD)
= Mg (l – l cos q)
D
C
h
= Mg l (1 – cos q)
This work is done against the force of gravity.
Hence, by sign convention, it is negative. Thus
Wg = – Mg l (1 – cos q)
B
F
Mg
The work done by the applied horizontal force F is
Fa = F ¥ horizontal distance moved
= F ¥ CD = F l sin q
\
Total work done is
W = Wg + Wa
= – Mg l (1 – cos q) + F l sin q
Since the mass is at rest at positions B and C, the change in kinetic energy
is zero. From work–energy principle, work done = change in kinetic
energy, i.e.
– Mg l (1 – cos q) + F l sin q = 0
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Model Solutions —AIEEE Physics Paper 2006
or
F=
M.65
Mg (1 – cos q )
Mg (1 – cos 45°)
=
sin q
sin 45∞
1 ˆ
Ê
Mg Á1 –
˜
Ë
2¯
=
= Mg
1/ 2
(
)
2 –1
Hence the correct choice is (2).
5. A particle located at x = 0 at time t = 0, starts moving along the positive
x-direction with a velocity v that varies as v = a x , where a is a constant. The displacement x of the particle varies with time t as
(2) t3
(3) t2
(4) t
(1) t1/2
Solution: Given v = a x . Since v =
dx
=a
dt
dx
, we have
dt
x
dx
= a dt
x
or
Integrating, we have
Ú
x
0
t
dx
= a Ú dt
0
x
a 2t 2
which gives – 2 x = a t or x =
. Hence x is proportional to t2. Thus
4
the correct choice is (3).
6. A particle of mass 100 g is thrown vertically upwards with a speed of 5
m/s. The work done by the force of gravity during the time the particle
goes up (take g = 10 ms–2)
(1) 1.25 J
(2) 0.5 J
(3) – 0.5 J
(4) – 1.25 J
Solution: The vertical height to which the particle rises is
h=
u2
(5)2
=
= 1.25 m
2g
2 ¥ 10
Work done Wg = mgh = (100 ¥ 10–3) ¥ 10 ¥ 1.25
= 1.25 J
Since the work Wg is done against the force of gravity, it is negative.
Hence the correct choice is (4), i.e. Wg = – 1.25 J.
7. Needless N1, N2 and N3 are made of a ferromagnetic, a paramagnetic
and a diamagnetic substance respectively. A magnet when brought close
to them will
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M.66 AIEEE Physics in 30 Days
(1) attract N1 strongly, but repel N2 and N3 weakly
(2) attract all three of them
(3) attract N1 and N2 strongly but repel N3
(4) attract N1 strongly, N2 weakly and repel N3 weakly
Solution: The correct choice is (4).
8. A material ‘B’ has twice the specific resistance of ‘A’. A circular wire
made of ‘B’ has twice the diameter of a wire made of ‘A’. Then for the
two
wires
to
have
the
same
resistance,
the
ratio
lB/lA of their respective lengths must be
(1)
1
4
(2) 2
(3) 1
(4)
1
2
2
2
Ê
ˆ
Ê
ˆ
RA Á p d A ˜
RB Á p d B ˜
Ë 4¯
Ë 4¯
rA =
and rB =
lA
lB
Solution:
2
rA
R Êd ˆ
l
= A ¥Á A˜ ¥ B
d
Ë
¯
rB
RB
lA
B
\
2
2
r
lB
R Êd ˆ
1
2
= A ¥ B ¥ Á B ˜ = ¥ ÊÁ ˆ˜ ¥ 1 = 2
r B RA Ë d A ¯
lA
2 Ë 1¯
or
Hence the correct choice is (2).
9. A bomb of mass 16 kg at rest explodes into two pieces of masses 4 kg
and 12 kg. The velocity of the 12 kg mass is 4 ms–1. The kinetic energy of
the other mass is
(1) 192 J
(2) 96 J
(3) 144 J
(4) 288 J
Solution: Let v be the speed of the 4 kg mass just after the explosion.
Since the bomb was at rest, its momentum is zero. From the conservation
of momentum we have
4 ¥ v – 12 ¥ 4 = 0
which gives
v = 12 ms–1. Therefore
1
¥ 4 ¥ (12)2 = 288 J
2
Hence the correct choice is (4).
10. The Kirchhoff’s first law (Si = 0) and second law (SiR = SE), where the
symbols have their usual meanings, are respectively based on
(1) conservation of momentum, conservation of charge
(2) conservation of charge, conservation of energy
(3) conservation of charge, conservation of momentum
(4) conservation of energy, conservation of charge
KE of 4 kg mass =
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Model Solutions —AIEEE Physics Paper 2006
M.67
Solution: Kirchhoff’s first law is a statement of conservation of charge.
It states that the sum of the currents entering a junction is equal to the
sum of the currents leaving it, i.e. the charge must be removed from the
junction at the same rate at which it arrives at it.
The product iR is the potential difference across a resistor R. Since
potential difference is defined as the amount of work done (or energy
spent) in taking a unit charge from one end of the resistor to the other,
Kirchhoff’s second law is a statement of conservation of energy. Hence
the correct choice is (2).
11. If the ratio of the concentration of electrons to that of holes in a semicon7
7
and the ratio of currents is , then what is the ratio of their
ductor is
5
4
drift velocities?
4
5
4
5
(2)
(3)
(4)
(1)
7
8
5
4
Solution: If ne and nh are the concentrations (i.e. number per unit volume) of electrons and holes, ve and vh their respective drift velocities,
then the currents due to the flow of electrons and holes are
Ie = ne eA ve
(1)
(2)
and
Ih = nh eA vh
where e = electronic charge and A = cross-sectional area of the semiconductor. Dividing (1) by (2), we get
ve
ÊI ˆ Ên ˆ
7 5
5
= Á e˜ ¥Á h˜ =
¥ =
Ë I h ¯ Ë ne ¯
vh
4 7
4
Hence the correct choice is (1).
12. The energy spectrum of b-particles [number N(E) as a function of benergy E] emitted from a radioactive source is
(1)
(2)
(3)
(4)
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M.68 AIEEE Physics in 30 Days
Solution: In b-decay, a neutron in the nucleus decays into a proton with
the emission of an electron and an uncharged particle called antineutrino.
The energy released in this process is shared between the electron and
antineutrino. When many nuclei undergo b-decay, all the emitted electrons do not have the same energy. Some electrons have a very low energy and some have a very high energy. In fact, when the antineutrino has
a high energy, the electron has a low energy and vice versa. Hence the
energy spectrum of electrons is as shown in choice (1).
13. A solid which is not transparent to visible light and whose conductivity
increases with temperature is formed by
(1) Vander Waals binding
(2) Metallic binding
(3) Ionic binding
(4) Covalent binding
Solution: The correct choice is (4).
14. A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s.
If the catching process is completed in 0.1 s, the force of the blow exerted
by the ball on the hand of the player is equal to
(1) 30 N
(2) 300 N
(3) 150 N
(4) 3 N
–3
–1
Solution: Given m = 150 ¥ 10 kg, v = 20 ms and t = 0.1 s. Let F be the
force of the impact. Now, impulse = force ¥ time of impact = F ¥ t. Also
impulse = change in momentum = mv. Equating them we have F ¥ t = mv or
(150 ¥ 10 –3 ) ¥ 20 = 30 N
m¥v
=
t
0.1
Hence the correct choice is (1).
15. Which of the following units denotes the dimensions ML2/Q2, where Q
denotes the electric charge?
(2) Weber (Wb) (3) Wb/m2
(4) Henry (H)
(1) H/m2
Solution: Henry is a unit of inductance. The energy stored in an inductor
is given by
F=
or
U=
1
LI 2
2
L=
2U
I2
\ Dimensions of inductance (L) =
=
dimension of energy
dimensions of (current)2
ML2 T –2
ML2 T –2
=
2
I2
(QT–1 )
Qˆ
Ê
ÁËQ I = ˜¯
T
= ML2 Q–2, which is choice (4).
All other choices have dimensions different from ML2 Q–2.
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Model Solutions —AIEEE Physics Paper 2006
M.69
16. When 37 Li nuclei are bombarded by protons, and the resultant nuclei are
8
4
Be , the emitted particles will be
(1) gamma photons
(2) neutrons
(3) alpha particle
(4) beta particle
Solution: The nuclear reaction is represented as
7
3
Li + 11H Æ 84 Be + AZ X
Since the mass and atomic numbers are conserved, we have
7+1=8+A
and
3+1=4+Z
which give A = 0 and Z = 0, which is gamma particle called photon.
Hence the correct choice is (1).
17. In a common base mode of a transistor, the collector current is 5.488 mA
for an emitter current of 5.60 mA. The value of the base current amplification factor (2) will be
(1) 51
(2) 48
(3) 49
(4) 50
Solution: Current gainb =
=
IC
IC
=
IB
I E – IC
5.488
= 49
5.60 – 5.488
Hence the correct choice is (3).
18. A ball of mass 0.2 kg is thrown vertically upwards by applying a force
by hand. If the hand moves 0.2 m while applying the force and the ball
goes upto 2 m height further, find the magnitude of the force. Consider g
= 10 m/s2.
(1) 20 N
(2) 22 N
(3) 4 N
(4) 16 N
Solution: A velocity of the ball just after it is released from the hand is
v=
2 gh =
2 ¥ 10 ¥ 2 =
40 ms–1
Let a be the acceleration imparted to the ball during the time the hand
was moving. During this time, the distance moved is s = 0.2 m. The value
of a is given by
v2 – u2 = 2as
or
40 – 0 = 2 ¥ a ¥ 0.2
which gives a = 100 ms–2. If F is the magnitude of the applied force, then F
– mg = ma or F = m(g + a) = 0.2 ¥ (10 + 100) = 22 N, which is choice (2).
19. The maximum velocity of a particle, executing simple harmonic motion
with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is
(1) 0.1 s
(2) 100 s
(3) 0.01 s
(4) 10 s
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M.70 AIEEE Physics in 30 Days
Solution:
vmax = Aw =
A ¥ 2p
T
22
(7 ¥ 10 –3 ) ¥ 2 ¥
A ¥ 2p
7 = 0.01 s
T=
=
vmax
4.4
or
Hence the correct choice is (3).
20. Assuming the Sun to be a spherical body of radius R at a temperature of
T K, evaluate the total radiant power, incident on Earth, at a distance r
from the Sun.
r02 R 2s T 4
4p r02 R 2sT 4
p r02 R 2s T 4
R 2sT 4
(2)
(3)
(4)
4p r 2
r2
r2
r2
where r0 is the radius of the Earth and s is Stefan’s constant.
Solution: Surface area of the sun = 4p R2. From Stefan’s law, the power
radiated by the sun is
Ps = s AT4 = 4p R2 s T4
(1)
This power is radiated by the sun in all directions. Therefore, power received per unit area on earth’s surface is Ps/4p r2. Hence, the total power
incident on the earth is
Pe =
=
Ps
¥ surface area of earth
4p r 2
4p R 2s T 4
¥ 4p r20
4p r 2
4p r02 R 2 s T 4
, which is choice (3).
r2
21. Consider a two particle system with particles having masses m 1 and m 2.
If the first particle is pushed towards the center of mass through a distance d, by what distance should the second particle be moved, so as to
keep the center of mass at the same position?
=
(1)
m1
d
m2
(2) d
(3)
m2
d
m1
(4)
m1
d
m2 + m2
Solution: If x1 and x2 are the positions of masses m 1 and m 2, the position
of the centre of mass is given by
m x + m2 x2
xcm = 1 1
m1 + m2
If x1 changes by Dx1 and x2 changes by Dx2, the change in xcm will be
Dxcm =
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m1 D x1 + m2 D x2
m1 + m2
(1)
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Model Solutions —AIEEE Physics Paper 2006
M.71
Given Dxcm = 0 and Dx1 = d. Using these values in Eq. (1), we get m 1 d +
m2 Dx2 = 0 or
Dx2 = –
\Distance moved by m 2 =
m1 d
m2
m1 d
, which is choice (1).
m2
22. A string is stretched between fixed points separated by 75.0 cm. It is
observed to have resonant frequencies of 420 Hz and 315 Hz. There are
no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
(1) 1050 Hz
(2) 10.5 Hz
(3) 105 Hz
(4) 1.05 Hz
Solution: Let nm = 420 Hz be the mth harmonic and nn = 315 Hz be the
nth harmonic. Then
and
420 =
m T
2L m
(1)
315 =
n T
2L m
(2)
where m is the mass per unit length of the string. Dividing (1) and (2), we
have
420
m
4
m
=
or
=
315
n
3
n
Hence the lowest values of m and n are m = 4 and n = 3, i.e. 420 Hz is the
4th harmonic and 315 Hz is the 3rd harmonic.
420 =
\
4 T
1 T
420
or
=
= 105
2L m
2L m
4
Now, the lowest resonant frequency is the fundamental frequency, which
is given by
n1 =
1 T
= 105 Hz
2L m
Hence the correct choice is (3).
23. The flux linked with a coil at any instant t is given by
f = 10t2 – 50t + 250
The induced emf at t = 3 s is
(1) 10 V
(2) 190 V
Solution: Induced emf is E = –
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71
(3) – 190 V
(4) – 10 V
df
dt
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M.72 AIEEE Physics in 30 Days
=–
d
(10t2 – 50t + 250)
dt
= – 20t + 50
E (at t = 3s) = – 20 ¥ 3 + 50 = – 10 V,
\
which is choice (4).
24. The threshold frequency for a metallic surface corresponds to an energy
of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in
(1) visible region
(2) X-ray region
(3) ultra-violet region
(4) infra-red region
Solution: Given eV0 = 5 eV and hn0 = 6.2 eV where V0 is the stopping
potential and n0 is the threshold frequency.
Now,
eV0 = hn – hn0
where
E = hn is the energy of the incident photon. Thus
E = eV0 + hn0 = 5 + 6.2 = 11.2 eV
= 11.2 ¥ 1.6 ¥ 10–19 J
Now
E = hn =
l=
hc
. Hence
l
(6.63 ¥ 10 –34 ) ¥ (3 ¥10 8)
hc
=
11.2 ¥1.6 ¥10 –19
E
; 1.11 ¥ 10–7 m = 1110 Å,
which is in the ultraviolet region of the electromagnetic spectrum. Hence
the correct choice is (3).
25. A thermocouple is made from two metals, Antimony and Bismuth. If
one junction of the couple is kept hot and the other is kept cold, then, an
electric current will
(1) not flow through the thermocouple
(2) flow from Antimony to Bismuth at the cold junction
(3) flow from Antimony to Bismuth at the hot junction
(4) flow from Bismuth to Antimony at the cold junction
Solution: At the hot junction the current flows from the metal which occurs earlier in the thermoelectric series to the metal which occurs later in
the series. At the cold junction the current flows from the metal that occurs
later in the series to the one which occurs earlier. Since Bismuth occurs
earlier than Antimony in the series, at the cold junction, the current will
flow from Antimony to Bismuth. Hence the correct choice is (2).
26. In the adjoining circuit, the current I drawn from the 5 volt source will be
(1) 0.67 A
(2) 0.17 A
(3) 0.33 A
(4) 0.5 A
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Model Solutions —AIEEE Physics Paper 2006
M.73
10 W
5W
10 W
20 W
10 W
I
+
–
5 Volt
Solution: The given circuit can be redrawn as follows:
B
10 W
5W
10 W
A
10 W
C
20 W
D
+
–
5 Volt
It is a balanced Wheatstone’s bridge. The resistor of 10 W between
points B and D is not effective as no current flows through it and hence it
can be ignored. The total resistance is a parallel combination of 5 + 10 =
15 W and 10 + 20 = 30 W which is
Reff =
Current I =
\
15 ¥ 30
= 10 W
15 + 30
5 volt
V
=
= 0.5 A
10 ohm
R eff
Hence the correct choice is (4).
27. The time taken by a photoelectron to come out after the photon strikes is
approximately
(2) 10–1 s
(3) 10–4 s
(4) 10–10 s
(1) 10–16 s
Solution : The correct choice is (4).
28. In a Wheatstone’s bridge, three resistances P, Q and R are connected in
the three arms and the fourth arm is formed by two resistances S1 and S2
connected in parallel. The condition for the bridge to be balanced will be
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M.74 AIEEE Physics in 30 Days
(1)
P R (S1 + S 2 )
=
Q
2 S1S2
(2)
P
R
=
Q S1 + S2
(3)
P
2R
=
Q S1 + S2
(4)
P R (S1 + S2 )
=
Q
S1S2
R
Q
S
P
R
P
Solution: The circuit diagram of Wheatstone’s bridge is follows:
S1
Q
S2
Here
S=
S1S 2
S1 + S2
The balance condition is
P
R (S1 + S2 )
R
=
=
Q
S1S2
S
Hence the correct choice is (4).
29. The refractive index of glass is 1.520 for red light and 1.525 for blue
light. Let D1 and D2 be angles of minimum deviation for red and blue
light respectively in a prism of this glass. Then,
(1) D1 can be less than or greater than D2 depending upon the angle of
prism
(2) D1 > D2
(3) D1 < D2
(4) D1 = D2
Solution: The refractive index of a prism is given by
m=
1
( A + D)
2
sin ( A / 2)
sin
(1)
where A = angle of the prism and D = angle of minimum deviation. It
follows from Eq. (1) that the greater the refractive index, the greater is
the angle of minimum deviation. Since the refractive index for blue light
is greater than that for red light, the angle of minimum deviation (D2) for
blue light will be greater than that (i.e. angle D1) for red light. Hence the
correct choice is (3).
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Model Solutions —AIEEE Physics Paper 2006
M.75
1
mv2 bombards a heavy nuclear target of
2
charge Ze. Then the distance of closest approach for the alpha nucleus
will be proportional to
(2) 1/Z3
(3) v 2
(4) 1/m
(1) 1/v4
Solution: The charge of an alpha particle is q = 2e. From the conservation of energy, we have
30. An alpha nucleus of energy
1 2Ze2
1 q (Ze )
1
mv2 =
=
4p Œ0 r0
4p Œ0 r0
2
where r0 is the distance of closest approach, which is given by
r0 =
1 Ze2
◊
p e 0 mv 2
1
. Hence the correct choice is (4).
m
31. In a series resonant LCR circuit, the voltage across R is 100 volts and R
= 1 kW with C = 2 mF. The resonant frequency w is 200 rad/s. At resonance the voltage across L is
(1) 250 V
(2) 4 ¥ 10–3 V (3) 2.5 ¥ 10–2 V (4) 40 V
Solution: The resonant angular frequency is given by
r0 µ
\
w=
which gives
L=
1
LC
1
1
=
2
–6
Cw
(2 ¥ 10 ) (200) 2
100
= 12.5 H
8
At resonance, impedance Z = R. Therefore, the current in the circuit is
=
I=
\
100V
V
V
=
=
= 0.1 A
1000
W
Z
R
Voltage across inductor (VL) = IXL = IwL
= 0.1 ¥ 200 ¥ 12.5
= 250 V
Hence the correct choice is (1).
32. The resistance of bulb filament is 100 W at a temperature of 100°C. If its
temperature coefficient of resistance be 0.005 per °C, its resistance will
become 200 W at a temperature of
(1) 500°C
(2) 200°C
(3) 300°C
(4) 400°C
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M.76 AIEEE Physics in 30 Days
Solution: R2 = R1 [1 + a (t2 – t1)]
which gives
t2 – t1 =
200 – 100
R2 – R1
=
100 ¥ 0.005
R1 a
= 200°C
\ Final temperature t2 = t1 + 200°C = 100°C + 200°C = 300°C.
Hence the correct choice is (3)
33. Two insulating plates are both uniformly charged in such a way that the
potential difference between them is V2 – V1 = 20 V. (i.e. plate 2 is at a
higher potential). The plates are separated by d = 0.1 m and can be treated
as infinitely large. An electron is released from rest on the inner surface
of plate is 1. What is its speed when it hits plate 2?
(e = 1.6 ¥ 10–19 C, m e = 9.11 ¥ 10–31 kg). See the following figure
(2) 3.2 ¥ 10–18 ms–1
(1) 1.87 ¥ 106 ms–1
6
–1
(3) 2.65 ¥ 10 ms
(4) 7.02 ¥ 1012 ms–1
Y
0.01 m
O
1
X
2
Solution: Potential difference between plates 1 and 2 is
V = V2 – V1 = 20 V
Energy gained by the electron is eV. If v is its velocity with which it hits
plate 2, then
1
mv2 = eV
2
or
v=
2eV
m
1/ 2
È 2 ¥ (1.6 ¥ 10 –19 ) ¥ 20 ˘
= Í
˙
9.11 ¥ 10 –31
Î
˚
6
–1
= 2.65 ¥ 10 ms , which is choice (3).
34. In an AC generator, a coil with N turns, all of the same area A and total
resistance R, rotates with frequency w in a magnetic field B. The maximum value of emf generated in the coil is
(1) NABR
(2) w NAB
(3) w NABR
(4) NAB
Solution: The induced emf is given by
E = w NAB sin w t
\
Emax = w NAB. Hence the correct choice is (2).
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35. Starting from the origin a body oscillates simple harmonically with a
period of 2s. After what time will its kinetic energy by 75% of the total
energy?
1
1
1
1
s
(2)
s
(3)
s
(4)
s
3
6
12
4
Solution: Since the body starts from the origin, its displacement x = 0 at
time t = 0. Hence the phase constant f = 0. Now, the displacement of a
body in simple harmonic motion is given by (since f = 0)
x = A sin (w t + f) = A sin wt
(1)
(1)
Kinetic energy
Total energy
Ek =
E=
1
mw2 (A2 – x2)
2
1
mw2 A2
2
\
A2 – x2
Ek
=
A2
E
Given
Ek
3
= 75% = . Hence
4
E
A2 – x2
3
=
A2
4
which gives
x=
A
. Using this is Eq. (1), we have
2
A
= A sin (wt)
2
which gives
sin wt =
Therefore,
p
p
1
2p t
or wt =
or
=
6
6
2
T
1
T
2s
=
= s, which is choice (3).
6
12
12
36. The anode voltage of a photocell is kept fixed. The wavelength l of the
light falling on the cathode is gradually changed. The plate current I of
the photocell varies as follows:
t=
(2)
(1)
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M.78 AIEEE Physics in 30 Days
(3)
(4)
Solution: The maximum kinetic energy of the photo-electrons is given
by
Kmax = hn – W0
hc
– W0
(1)
l
As l increases, Kmax decreases. Therefore, fewer photoelectrons will
reach the plate as l increases. Hence the current through the photocell
decreases as l increases. Therefore, the correct graph is (4).
37. If the lattice constant of this semiconductor is decreased, then which of
the following is correct?
or
Kmax =
Ec
conduction band width
band gap
valence band width
Eg
Ev
(1) Ec and Ev decrease, but Eg increases (2) Ec, Ev and Eg all
decrease
(4) Ec and Ev increase but Eg
(3) Ec, Ev and Eg all increase
decreases
Solution: The correct choice is (4).
38. If the binding energy per nucleon in 37 Li and 42 He nuclei are 5.60 MeV
and 7.06 MeV respectively, then in the reaction
p + 37 Li Æ 242 He
energy of proton must be
(1) 1.46 MeV
(2) 39.2 MeV
(3) 28.24 MeV
(4) 17.28 MeV
Solution: Let Q MeV be the energy of the proton. Then, we have
Q + energy of one 37 Li nucleus = energy of two 42 He nuclei
Q + 7 ¥ 5.6 MeV = 2 ¥ (4 ¥ 7.06 MeV)
Q + 39.6 MeV = 56.48 MeV
\
or
which gives Q = 17.28 MeV. Hence the correct choice is (4).
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39. A thin circular ring of mass m and radius R is rotating about its axis with
a constant angular velocity w. Two objects each of mass M are attached
gently to the opposite ends of a diameter of the ring. The ring now rotates
with an angular velocity w¢ =
wm
wm
(2)
(1)
(m + M )
( m + 2M )
w (m – 2 M )
w (m + 2 M )
(3)
(4)
(m + 2 M )
m
Solution: The angular momentum of the ring before the two objects are
attached is
L = mR2w
After the two objects are attached, the angular momentum becomes
L¢ = (m + 2M)R2 w¢
Since no external torque acts, the angular momentum is conserved, i.e. L
= L¢ or
mR2w = (m + 2M)R2w¢
or
w¢ =
mw
, which is choice (2).
( m + 2M )
40. A force – F kˆ acts on O, the origin of the coordinate system shown in the
figure. The torque about the point (1, –1) is
(1) F (iˆ + ˆj )
(2) – F (iˆ – ˆj )
(3) F (iˆ – ˆj )
(4) – F (iˆ + ˆj )
y
O
x
z
r = x iˆ + y ˆj + z kˆ
= 1 ¥ iˆ + (–1) ¥ ˆj + 0 ¥ kˆ = iˆ – ˆj
Solution:
(
)
F = – F kˆ
Torque = r ¥ F
= ( iˆ – ˆj ) ¥ ( – F kˆ )
= F ( – iˆ ¥ kˆ + ˆj ¥ kˆ )
= F ( ˆj + iˆ )
Hence the correct choice is (1).
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M.80 AIEEE Physics in 30 Days
41. The ‘rad’ is the correct unit used to report the measurement of
(1) the biological effect of radiation
(2) the rate of decay of a radioactive source
(3) the ability of a beam of gamma ray photons to produce ions in a target
(4) the energy delivered by radiation to a target
Solution: The correct choice is (1).
42. The potential energy of 1 kg particle free to move along the x-axis is
given by
Ê x4 x2 ˆ
V(x) = Á – ˜ joule
Ë 4
2¯
The total mechanical energy of the particle is 2 J. Then, the maximum
speed (in m/s) is
(1)
1
2
(2) 2
Solution:
V=
(3)
3
2
(4)
2
x4 x2
–
4
2
dV
= x3 – x
dx
dV
= 0, i.e. if x3 – x = 0 or x(x2 – 1) = 0
V is maximum or minimum if
dx
which gives x = 0, 1 and –1. Now
\
d 2V
= 3x2 – 1
dx2
d 2V
d 2V
V is maximum if
is negative and V is minimum if
is positive.
2
dx
dx2
d 2V
d 2V
For x = 0,
=
–
1.
Hence
for
x
=
0,
V
is
maximum.
For
x
=
±
1,
dx2
dx2
= 2. Hence for x = ± 1, V is minimum. Now
x4 x2
–
4
2
Therefore, the minimum potential energy is (put x = ± 1)
V=
Vmin =
1
( ± 1) 4 ( ± 1) 2
1 1
= – = – joule
–
4
4 2
4
2
1
1
mv2max = v2max (∵ m = 1 kg).
2
2
Given, total energy E = 2 joule. From E = V + K, we have
Maximum kinetic energy is Kmax =
1 1
2 = – + v2max
4 2
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Model Solutions —AIEEE Physics Paper 2006
M.81
3
ms–1, which is choice (3).
2
43. Two spherical conductors A and B of radii 1 mm and 2 mm are separated
by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of
the magnitude of the electric fields at the surfaces of sphere A and B is
(1) 2 : 1
(2) 1 : 4
(3) 4 : 1
(4) 1 : 2
Solution: Let Q1 and Q2 be the charges on spheres A and B and R1 and R2
their respective radii. Their potentials are
which gives
vmax =
1 Q1
1 Q2
◊
and V2 =
◊
4p Œ0 R1
4p Œ0 R2
When equilibrium is attained, V1 = V2, i.e.
V1 =
Q
1
1 Q2
◊ 1 =
◊
4p Œ0 R1
4p Œ0 R2
Q1
R
= 1
Q2
R2
The electric fields on the surfaces of the spheres are
or
E1 =
(1)
Q
1
1 Q2
◊ 12 and E2 =
◊
4p Œ0 R1
4p Œ0 R22
2
E1
R
Q ÊR ˆ
2
\
= 1 ◊Á 2˜ = 2 =
[use Eq. (1)]
E2
R1
Q2 Ë R1 ¯
1
Hence the correct choice is (1).
44. A wire elongates by l mm when a load W is hung from it. If the wire goes
over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (in mm)
(1) zero
(2) l/2
(3) l
(4) 2l
Solution: Let L be the length of wire AB and a be the area of crosssection of the wire. The Young’s modulus is given by
W /a
WL
=
(1)
l/L
al
For a given wire, Y and a are fixed. Then for a given load, it follows from
(1) that L/l = constant, i.e. l µ L. The extension is proportional to the
original length of the wire. In the second case (see the adjoining figure),
L
AC = CB =
since the load on both sides of the pulley is the same.
2
Since the extension is proportional to length, the extension of wire AC of
length L/2 will be l/2. Similarly, the extension of wire BC will be l/2.
l l
Hence the total extension is + = l, which in choice (3).
2 2
Y=
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M.82 AIEEE Physics in 30 Days
45. A inductor (L = 100 mH), a resistor (R = 100 W) and a battery (E = 100 V)
are initially connected in series as shown in the figure. After a long time
the battery is disconnected after short circuiting the points A and B. The
current in the circuit 1 ms after the short circuit is
(1) 0.1 A
(2) 1 A
(3) 1/e A
(4) e A
Solution: The decay of current in an LR circuit is given by
E –Rt/L
e
R
Substituting the values of E, R, L and t we get
I=
1
ampere
e
Hence the correct choice is (3).
46. An electric bulb is rated 220 volt – 100 watt. The power consumed by it
when operated on 110 volt will be
(1) 25 watt
(2) 50 watt
(3) 75 watt
(4) 40 watt
I=
Solution:
(220) 2
V2
V2
. Hence R =
=
= 484 W
100
R
P
V ¢ = 110 volt, the power will be
P=
When operated at
2
(110)2
V¢
=
= 25 watt
484
R
Hence the correct choice is (1).
P¢ =
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M.83
47. A coin is placed on a horizontal platform which undergoes vertical
simple harmonic motion of angular frequency w. The amplitude of oscillation is gradually increased. The coin will lose contact with the platform
for the first time
(1) for an amplitude of g2/w2
(2) at the highest position of the platform
(3) at the mean position of the platform
(4) for an amplitude of g/w2
Solution: The coin will lose contact with the platform if the amplitude A
of oscillation is such that at the extreme position, the maximum acceleration equals the acceleration due to gravity, i.e. if
w2A = g
or
A = g/w2, which is choice (4).
48. The terminal speed of a sphere of gold (density 19.5 ¥ 103 kg m–3) in a
viscous liquid (density 1.5 ¥ 103 kg m–3) is 0.2 ms–1. What is the terminal
of a sphere of silver (density 10.5 ¥ 103 kg m–3) of the same size in the
same liquid?
(1) 0.1 ms–1
(2) 0.2 ms–1
(3) 0.4 ms–1
(4) 0.133 ms–1
Solution: The terminal speed of a sphere of radius r and density s in a
liquid of density r and viscosity h is given by
where
vt =
2 r2g
◊
(s – r) = K(s – r)
9 h
K=
2 r2g
◊
is a constant for this problem.
9 h
For the gold sphere,
0.2 = K (19.5 – 1.5) = 18K
(1)
For the silver sphere,
vt = K (10.5 – 1.5) = 9K (2)
From (1) and (2), we get vt = 0.1 ms–1. Hence the correct choice is (1).
49. The rms value of the electric field of the light coming from the Sun is 720
N/C. The average total energy density of the electromagnetic wave is
(2) 3.3 ¥ 10–3 J/m3
(1) 81.35 ¥ 10–12 J/m3
–6
3
(3) 4.58 ¥ 10 J/m
(4) 6.37 ¥ 10–9 J/m3
Solution: The total energy density of an electromagnetic wave is
u = ue + um
2
B
1
e0 E2 and um =
.
m0
2
2
The electric field of the wave is given by
E = E0 sin wt
where E0 is the peak value of E and w is the angular frequency. The
average energy density associated with electric field is
where
ue =
<ue> =
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M.84 AIEEE Physics in 30 Days
where
<sin2 wt> =
Hence
<ue> =
T
1
1
sin 2 (wt ) dt =
Ú
T0
2
1
e0 E20
4
Also um
<um > =
Hence
<um> =
Total energy density
B2
1
1
=
(E/c)2 =
e0 E2
2m 0
2m0
2
1 ˆ
Ê
C=
ÁËQ
m e ˜¯
1
e0 E20
4
1
1
<u> = <ue> + <um > =
e0 E20 +
e0 E20
4
4
1
=
e0 E20
2
E0 ˆ
Ê
= e0 E2rms
ÁËQ Erms =
˜
2¯
= (8.85 ¥ 10–12) ¥ (720)2
= 4.58 ¥ 10–6 Jm–3
Hence the correct choice is (3).
50. Two rigid boxes containing different ideal gases ae placed on a table.
Box A contains one mole of nitrogen at temperature To, while Box B
contains one mole of helium at temperature (7/3)To. The boxes are then
put into thermal contact with each other, and heat flows between them
until the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the final temperature of the gases, Tf, in terms of To is
(1) Tf =
3
To
2
(2) Tf =
5
To
2
3
7
To
(4) Tf =
To
7
3
Solution: The change in internal energy of an ideal gas is given by
DU = n Cv DT = nCv (Tf – Ti)
where
n = number of moles and Ti = initial temperature.
(3) Tf =
For a diatomic gas (such as nitrogen) Cv =
5
R and for a monoatomic
2
3
R.
2
5R
x (Tf – To)
DU for oxygen = 1 ¥
2
gas (such as helium) Cv =
\
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M.85
7To ˆ
3R Ê
ÁË Tf –
˜
3 ¯
2
Since no heat is given to or taken from the surrounding, the total internal
energy of the system cannot change. Therefore
and
DU for helium = 1 ¥
1¥
7To ˆ
5R
3R Ê
¥ (Tf – To) + 1 ¥
ÁË Tf –
˜ =0
3 ¯
2
2
3
To, which is choice (1).
2
51. The work of 146 kJ is performed in order to compress one kilo mole of a
gas adiabatically and in this process the temperature of the gas increases
by 7°C. The gas is (R = 8.3 J mol–1 K–1)
(1) mixture of monoatomic and diatomic
(2) monoatomic
(3) diatomic
(4) triatomic
Solution: An an adiabatic process, the work is given by
which gives Tf =
W=
nR (T1 – T2 )
g –1
1000 ¥ 8.3 ¥ 7
(∵ DT = 7°C = 7K)
(g – 1)
which gives g = 1 + 0.4 = 1.4. Hence the gas is diatomic. So the correct
choice is (3).
52. In the following, which one of the diodes is reverse biased?
146 ¥ 103 =
or
(1)
(2)
(3)
(4)
Solution: If the p-side is at a lower potential than the n-side of a junction
diode, then the diode is reverse biased. The correct choice is (2). The pside is at zero potential (because it is connected to earth) and the nside at potential + 5V.
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M.86 AIEEE Physics in 30 Days
53. A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its center is 6.28 ¥ 10–2 Weber/m2. Another long solenoid
has 100 turns per cm and it carries a current i/3. The value of the magnetic field at its center is
(2) 1.05 ¥ 10–4 Weber/m2
(1) 1.05 ¥ 10–3 Weber/m2
–2
2
(3) 1.05 ¥ 10 Weber/m
(4) 1.05 ¥ 10–5 Weber/m2
Solution: For the first solenoid
B1 = m0 n1 i1
For the second solenoid
B2 = m0 n2 i2
Dividing we get,
B2
n
i
100 i / 3
1
= 2¥ 2 =
=
¥
B1
n1 i1
200
i
6
1
1
B1 =
¥ 6.28 ¥ 10–2 ; 1.05 ¥ 10–2 Wb/m2
6
6
Hence the correct choice is (3).
54. Four point masses, each of value m, are placed at the corners of square
ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD is
or
B2 =
(2) ml2
(3) 2 ml2
(1) 3 ml2
Solution: Refer to the following figure.
(4)
3 ml2
The moment of inertia of the system about axis XY is
I = m at A ¥ 0 + m at B ¥ (BE)2 + m at C ¥ (AC)2 + m at D ¥ (DF)2
2
Ê l ˆ
=m¥0+m¥ Á
+m¥
Ë 2 ˜¯
(
2 l)
2
Ê l ˆ
+m¥ Á
Ë 2 ˜¯
2
ml 2
ml 2
+ 2ml2 +
= 3 ml2
2
2
Hence the correct choice is (1).
=0+
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55. The circuit has two oppositely connected ideal diodes in parallel. What is
the current flowing in the circuit?
(1) 2.31 A
(2) 1.33 A
(3) 1.71 A
(4) 2.00 A
Solution: Diode D2 is forward biased but diode D1 is reverse biased.
Hence no current flows through the branch containing D1 and the circuit
may be redrawn as follows
\ Current in the circuit is
I=
12 V
12 V
=
= 2A
4W + 2W
6W
Hence the correct choice is (4).
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