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1)
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 = α .
The displacement of the particle varies with time as
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1 )
t3
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2 )
t2
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3 )
t
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4 )
t1/2
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| see the answer see the solution |
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2)
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
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1 )
Mg( -1)
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2 )
Mg(+1)
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3 )
Mg
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4 )
Mg/
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3)
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 is
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1 )
-0.5 J
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2 )
-1.25 J
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3 )
1.25 J
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4 )
0.5 J
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| see the answer see the solution |
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4)
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
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1 )
0.01 s
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2 )
10 s
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3 )
0.1 s
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4 )
100 s
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| see the answer see the solution |
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5)
A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity
of the 12 kg mass is 4 ms-1. The kinetic energy of the other mass is
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1 )
96 J
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2 )
144 J
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3 )
288 J
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4 )
192 J
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| see the answer see the solution |
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6)
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
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1 )
30 ms-1
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2 )
15 ms-1
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3 )
15/ ms-1
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4 )
15 ms-1
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| see the answer see the solution |
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7)
Starting from the origin a body oscillates simple harmonically
with a period of 2 s. After what time
will its kinetic energy be 75% of the total energy?
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1 )
1/6 s
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2 )
1/4 s
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3 )
1/3 s
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4 )
1/12 s
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| see the answer see the solution |
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8)
A string is stretched between fixed points separated by 75 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
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1 )
10.5 Hz
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2 )
105 Hz
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3 )
1.05 Hz
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4 )
1050 Hz
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| see the answer see the solution |
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9)
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.
where r0 is the radius of the Earth and σ is Stefan's constant.
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1 )
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2 )
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3 )
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4 )
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| see the answer see the solution |
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10)
Which of the following units denotes the dimensions
ML2/Q2, where Q denotes the electric charge ?
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1 )
Wb/m2
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2 )
Henry (H)
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3 )
H/m2
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4 )
Weber (Wb)
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| see the answer see the solution |
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11)
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.
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1 )
4 N
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2 )
16 N
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3 )
20 N
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4 )
22 N
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| see the answer see the solution |
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12)
Consider a two particle system with a particles having
masses m1 and m2. If the first particle is pushed
towards the centre of mass through a distance d,
by what distance should the second particle be
moved, so as to keep the centre of mass at the same position?
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1 )
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2 )
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3 )
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3 )
d
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| see the answer see the solution |
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13)
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
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1 )
150 N
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2 )
3 N
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3 )
30 N
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3 )
300 N
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| see the answer see the solution |
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14)
In a common base made 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 (β) will be
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1 )
49
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2 )
50
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3 )
51
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3 )
48
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| see the answer see the solution |
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15)
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
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1 )
flow from Antimony to Bismuth at the hot junction
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2 )
flow from Bismuth to Antimony at the cold junction
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3 )
not flow through the thermocouple
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3 )
flow from Antimony to Bismuth at the cold junction
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| see the answer see the solution |
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16)
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
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1 )
ultra-violet region
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2 )
infra-red region
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3 )
visible region
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3 )
X-ray region
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| see the answer see the solution |
1 back to 1
v = α.
=> dx/dt = α
=> = α/2. t
So, x ∝ t2
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2 back to 2
F×BC = Mg×AC
Flsin45° = Mg(l-lcos45°)
=> Mg(-1)
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3 back to 3
Suppose partice moved up to a distance = 5
using, v2 = u2 + 2as
=> 0 = 25 - 2gy
=> 2gy = 25
Work done by the gravity = W = Gravitational force)×(displacement)×cos180°
(direction between gravitational force and displacement is 180°)
Work done by the gravity = W = -0.1gy = -1.25 J
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4 back to 4
Vmax = Aω
Period of oscillation = T = 2π/ω
=> T = 2πA/Vmax = 0.01 second
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5 back to 5
Total initial momentun = 0
Total final momentum = m1v1 + m2v2 = 0
=> 12 × 4 + 4 × v2 = 0
=> v2 = -12
Kinetic energy of 4kg mass = 1/2 × 4 × (-12)2
Kinetic energy of 4kg mass = 288 J
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6 back to 6
where,
v=velocity of wave
vs=velocity of source. It is positive if source of wave is moving away from observer.
It is negative if source of wave is moving towards observer.
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10000 = 9500(
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300
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300 + vs
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300 + vs = 285
vs = -15ms-1 [negative means that wave-source is approching the observer]
So approching velocity of wave-source = 15ms-1
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7 back to 7
Simple harmonic is defined as, x = Asinωt
velocity = dx/dt = v = Aωcosωt
Maximum velocity = Aω
Kinetic energy = 1/2 m v2 = 1/2 m A2ω2cos2ωt
Maximun Kinetic energy = 1/2 m A2ω2
=> (3/4)(1/2 m A2ω2) = 1/2 m A2ω2cos2ωt
=> (3/4) = cos2ωt
=> ωt = &pi/6
=> (2π/T)t = &pi/6
=> (2π/T)t = &pi/6
t = 1/6 s
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8 back to 8
Resonance of a open tube of air(approximate)
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Approximate frequency = f =
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nv
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2L
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where,
L: length of the cylinder
n = 1, 2, 3...
v = speed of sound
315 = nv/2L .......... (i)
400 = ((n+1)v)/(2L) ................(ii)
Solving (i) and (ii)
n = 3
Lowest frequency of string stretched between fixed points = v/2L = 315/3 = 105 Hz
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9 back to 9
Stefan-Boltzmann Law
The energy radiated by a blackbody radiator per second = P
P = AσT4
So energy released by sun = P = 4πR2σT4
Radiant power, incident on Earth, at a distance r from the Sun = Pe
Pe = (πr02/4πr2)(4πR2σT4)
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10 back to 10
Refer
Terms and definition
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11 back to 11
Fs = mgh
F × 0.2 = 0.2 × 10 × 2.2
F = 22 N
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12 back to 12
m1 d + m2 x = 0
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13 back to 13
Force = change in momentum per unit time
F = (mv-0)/0.1 = 0.15 × 20/0.1 = 30 N
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14 back to 14
Ib = Ie - Ic = 5.60 - 5.488 = 0.112
β = Ic/Ib = 5.488/0.11 = 49
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15 back to 15
Antimony appears first in Seeback’s thermal electric series.
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16 back to 16
&lambda = (1242 eVnm)/(11.2) = 1100 A
Therefore, ultra-violet region
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