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Multiple Choice Questions
Question 1
Force exerted on a body changes it's
(a) Direction of motion
(b) Momentum
(c) Kinetic energy
(d) All the above
Solution
A body acted upon by a certain force produces acceleration i.e. it undergoes change in it's velocity and momentum and kinetic energy of a moving body depends on the velocity of the body. Hence choice (d) is correct
Question 2
There are two statements
Statement A :Rate of change of momentum corresponds to force
Statement B :Rate of change of momentum corresponds to Kinetic Energy
Which one of the following is correct?
(a) A only
(b) B only
(c) Both A and B are correct
(d) Both A and B are wrong
Solution
We know that F=dp/dt which is the rate of change of momentum with time and KE=p^{2}/2m and hence rate of change of momentum does not corresponds to kinetic energy. Hence choice (a) is correct
Question 3
A truck and car are moving on a plane road with same kinetic Energy. They are brought to rest by application of brakes which provide equal retarding forces. Which one of the following statements is true?
(a) Distance traveled by truck is shorter then car before coming to rest
(b) Distance traveled by car is shorter then truck before coming to rest
(c) Distance traveled depends on individual velocity of both the vehicles
(d) Both will travel same distance before coming to rest
Solution
By work energy theorem we know
KE_{f}-KE_{i}=Work done
So in this case, Initial KE=retarding force * distance traveled
Since kinetic energy is same and retarding force is equal distance traveled will be same
Hence choice (d) is correct
Question 4
A block A of mass m
_{1} is released from top of smooth inclined plane and it slides down the plane. Another block of mass m
_{2} such that m
_{2} > m
_{1}is dropped from the same point and falls vertically downwards. Which one of the following statements will be true if the friction offered by air is negligible?
(a) Both blocks will reach ground at same time
(b) Both blocks will reach ground with the same speed
(c) speed of both the blocks when they reach ground will depend on their masses
(d) Block A reaches ground before block B
Solution
Since velocity of block when it reaches the ground is given by v=(2gh)^{1/2} and it is independent of the mass the correct choice will be (b).
Hence choice (b) is correct
Question 5
Which of the following observer is/are non inertial
(a) A child revolving in the merry go round
(b) A driver in a train moving with constant velocity v
(c) A passenger in a bus which is slowing down to a stop
(d) None of these
Solution
Since centripetal acceleration is present in merry round it is non inertial frame of reference
Since bus is slowing down to a stop, it means it is experiencing a retardation, so it is non inertial frame of reference
Hence choice (a) and (c) are correct
Question 6
There are two statements
Statement A Newton’s first law in valid from the pilot in an aircraft which is taking off
Statement B Newton’s first law in valid from the observer in a train moving with constant velocity
Which of the following is correct
(a) A only
(b) B only
(c) Both A and B are correct
(d) Both A and B are wrong
Solution
We know that Newton’s law is invalid in Non inertial frame of reference, now since acceleration is present in Ist case. So answer is B
Question 7
Choose the correct alternative
(a) The acceleration of a moving particle is always in the direction of its velocity
(b) Velocity and acceleration vectors of a moving particle may have any angle between 0 and 360 between them
(c) Velocity and acceleration vectors of a moving particle may have any angle between 0 and 180 between them
(d) If the acceleration vector is always perpendicular to the velocity vector of a moving particle, the velocity vector does not change
Solution
in circular motion, acceleration vector is perpendicular to velocity and velocity is also changing. Now acceleration and velocity can have any angle between them.
Hence (c) is the correct answer
Question 8
Consider the figure given below in which a particle of weight W resting a smooth (frictional less) inclined plain AB with the help of force F acting on the particle at angle θ with the line AB.
find the force F and normal reaction
(a) (Wcosθ)/sinα, [Wcos(α - θ)]/cosθ
(b) (Wsinα)/cosθ, [Wcos(α + θ)]/cosθ
(c) (Wsinα)/cosθ, [Wsin(α + θ)]/cosθ
(d) none of the above
Solution
Figure given below shows all the forces acting on the block
Using law of Sine
F/sin(180 - α) = N/sin(90 + α + θ) = W/sin(90 - θ)
Solving this
F = (Wsinα)/cosθ, N=[Wcos(α + θ)]/cosθ
Hence (b) is the correct answer
Question 9
For the figure given below find the maximum acceleration of the cart that is require to prevent block B from falling coefficient of
Static friction between block and cart is μs
a) g
(b) μsg
(c) g/μs
(d) none of the above
Solution
Free body diagram for forces acting on the block B is
friction force must balance with block weight
f = mg
f - mg = 0
f = mg
now N = ma
So f/N = g/a
a = g/f/N
since maximum value of f/N is μ
we must have
a ≥ g/μs if the block is not falling
Hence (c) is the correct choice
Question 10
A particle of mass M, originally at rest is subjected to a force whose direction is constant but whose magnitude varies with the time according to the relation
where F
_{0} and T are constant. The force acts only for the time interval 2T. Find the velocity v of the particle after time 2T
(a) F
_{0}/3M
_{0}
(b) 4F
_{0}/3M
_{0}
(c) F
_{0}/2M
_{0}
(d) none of the above
Solution
Hence (b) is the correct choice
Question 11
Consider the figure given below
if Coefficient of friction is μ then what is the least value of μ so the system remains at rest
a) 0.23
(b) 0.2
(c) 0.203
(d) 0.13
Solution
The tendency of motion will be towards left so friction will oppose that motion now for least μ, frictional force should be maximum (i.e. at the verge of slipping)
mgsin53 - μmgcos53 - T = 0
T - mg sin30 - μmgcos30 = 0
solving both these equations for μ we get
μ = 0.203
Hence (c) is the correct choice
Question 12
A body of mass M is kept in a elevator. The elevator experience free fall, which of these is true
(a) the body has no acceleration with reference of frame attached to elevator
(b) the body has acceleration g with reference of frame attached to earth surface
(c) resultant force on the body is zero with reference of frame attach to earth surface
(d) resultant force on the body is zero with reference of frame attach to car moving on the earth surface moving with constant velocity
Solution
(a),(b) and (d) are correct choice
Question 13
A block of mass M is kept on a friction floor of static friction μ. The body is acted by a horizontal force F horizontally but the body does not move. Now an another f act on the body vertically downwards, what will happen
(a) frictional force on the block is change
(b) normal reaction on the block change
(c) limiting frictional force changes
(d) body starts moving
Solution
(b) and (c) are correct choice
Paragraph type question
(A) A body of mass m moves along the positive x-axis, it starts at velocity v
_{0} at t = o and it is the origin initially. It is acted by the force such that
F = -kv
Question 14
find the time in which it will come to rest
(a) t → ∞
(b) t = kt/m
(c) t = mt/k
(d) none of the above
Solution
F = -kv
m dv/dt = -kv
dv/v = -kdt/m
Integrating
ln v = -kt/m + c
now t = 0, v = v_{0}
So v = v_{0} exp (-kt/m)
So at t → ∞
v = 0
Hence (a) is correct choice
Question 15
Find the velocity of body as a function of time
(a) v
_{0} exp (-kt/m)
(b) v
_{0} exp (-mt/k)
(c) v
_{0} exp (-t)
(d) none of the above
Solution
Self explanatory from previous solution
Hence (a) is correct choice
Question 16
Find the value of x at which it velocity become 0
(a) mv
_{0}/k
(b) mv
_{0}/k
^{2}
(c) m/kv
_{0}
(d) none of the above
Solution
m dv/dt = -kv
m vdv/dx = -kv
dv = -kdx/m
Integrating
v = v_{0} - rx/m
at v = 0
x = mv_{0}/k
Hence (a) is correct choice
Question 17
Find the velocity as a function of distance
(a) v
_{0} + kx/m
(b) v
_{0} - kx/m
(c) v
_{0} - x/m
(d) none of the above
Solution
Self explanatory from previous solution
Hence (b) is correct choice
(B) A 6 kg object is subject to three forces
F_{1} = 20
i + 30
j N
F_{2} = 8
i - 50
j N
F_{3} = 2
i + 2
j N
Question 18
find the acceleration of object
(a) 5
i + 3
j
(b) 5
i - 3
j
(c) 3
i + 5
j
(d) 3
i - 5
j
Solution
F = F_{1} + F_{2} + F_{3} = 30i - 18j
a = 5i - 3j
Hence (b) is correct choice
Question 19
Which of the following expression is correct if at t = 0, object is at origin and velocity is
v_{0} =
i +
j
(a)
r =
i(2.5 t
^{2} + t) +
j(t - 1.5t
^{2})
(b)
r =
i(2.5 t
^{2} - t) +
j(t + 1.5t
^{2})
(c)
r =
it -
jt
^{2}
(d) none of the above
Solution
r = (i + j)t + 1/2(5i - 3j)t^{2}
r = i(2.5 t^{2} + t) + j(t - 1.5t^{2})
Hence (a) is correct choice
(C)A boy pushes a mass with a force F. Coefficient of friction between body and floor is μ
_{m} and between boy shoe and floor is μ
_{B}. There mass are M(block) and m(boy) respectively.
Question 20
What maximum force boy can apply without slipping
(a) μ
_{m}mg
(b) μ
_{m}Mg
(c) μ
_{B}mg
(d) none of the above
Solution
Self explanatory.
Hence (c) is correct choice
Question 21
What is the condition required to move the block without slipping
(a) μ
_{B}/μ
_{m} > M/m
(b) μ
_{B}/μ
_{m} > m/M
(c) μ
_{B}/μ
_{m} < M/m
(d) μ
_{B}/μ
_{m} < m/M
Solution
Block will have
F_{max} > μ_{m}Mg
μ_{B}mg > μ_{m}Mg
So μ_{B}/μ_{m} > M/m
Hence (a) is correct choice
Matrix match type
Question 22
Column A
A boy of weight W is standing in an elevator. find the force of the boy feel
(a) when the elevator stand still
(b) when the elevator moving with constant velocity (v m/s) downward
(c) when the elevator moving with constant velocity (v m/s) upward
(d) moving up with acceleration (a m/s
^{2 })
(e) moving down with acceleration (a m/s
^{2 })
Column B
(P) F = W
(Q) F>W
(R) F< W
(S) no appropriate match
Solution
(a) = (P)
(b) = (P)
(c) = (P)
(d) = (Q)
(e) = (R)
Question 23
Let
S
_{1} = frame of reference at rest
S
_{2} = frame of reference at constant velocity
S
_{3} = frame of reference at constant acceleration
S
_{4} = frame of reference at uniform circular motion
A block A is at rest as seen from frame of reference S
_{1}
Column A
(a) S1
(b) S2
(c) S3
(d) S4
Column B
(P) ΣF ≠ 0
(Q) ΣF =0
(R) a = 0
(S) a ≠ 0
where ΣF is the resultant force and a is the acceleration of the body
Solution
(a) -> (Q), (R)
(b) -> (Q), (R)
(c) -> (Q), (S)
(d) -> (Q), (S)
Multiple choice Questions
Question 24
A particle of weight W resting a smooth(frictional less) inclined plain AB with the help of force F acting on the particle at angle θ with the line AB
find the force F and normal reaction
(a) (Wcosθ)/sinα, [Wcos(α - θ)]/cosθ
(b) (Wsinα)/cosθ, [Wcos(α + θ)]/cosθ
(c) (Wsinα)/cosθ, [Wsin(α + θ)]/cosθ
(d) none of the above
Solution
Using law of Sine
F/sin(180 - α) = N/sin(90 + α + θ) = W/sin(90 - θ)
Solving this
F = (Wsin α)/cosθ, [Wcos(α + θ)]/cos θ
Question 25
The motion of a particle of mass m is given by
$x = 0$ for t < 0 s,
$x(t) = A sin 4 \pi t$ for 0 < t <(1/4) s (A > 0)
and x = 0 for t >(1/4) s. Which of the following statements is true?
(a) The force at t = (1/8) s on the particle is -$16 \pi^2 A m$.
(b) The particle is acted upon by on impulse of magnitude $4 \pi ^2 A m$ at t = 0 s and t = (1/4) s.
(c) The particle is not acted upon by any force.
(d) The particle is not acted upon by a constant force.
(e) There is no impulse acting on the particle.
Solution
for 0 < t <(1/4)
$x(t) = A sin 4 \pi t$
$\frac {dx}{dt} = 4\pi A cos 4\pi t$
$\frac {d^2x}{dt^2} = -16\pi ^2 A sin 4\pi t$
Hence Force acting
$F= -16 m \pi ^2 A sin 4\pi t$
Now Force at t=1/8 sec
$F=-16 \pi^2 A m$
Now Impulse is given by
Impulse = Force × time
$=16 \pi^2 A m \times \frac {1}{4}=4 \pi ^2 A m$
The impulse (Change in linear momentum)
at t=0 is same as,t=1/4s
Clearly, force varies with time t so, force is also not constant.
Hence correct option are (a) ,(b) and (d)
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