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Newton's Law Of Motion – MSQ - Physics MCQ


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10 Questions MCQ Test Topic wise Tests for IIT JAM Physics - Newton's Law Of Motion – MSQ

Newton's Law Of Motion – MSQ for Physics 2024 is part of Topic wise Tests for IIT JAM Physics preparation. The Newton's Law Of Motion – MSQ questions and answers have been prepared according to the Physics exam syllabus.The Newton's Law Of Motion – MSQ MCQs are made for Physics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Newton's Law Of Motion – MSQ below.
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Newton's Law Of Motion – MSQ - Question 1
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A painter is applying force himself to raise him and the box with an acceleration of 5 m/s2 by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100kg and that of box is 50kg. If g = 10 m/s2, then :

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Detailed Solution for Newton's Law Of Motion – MSQ - Question 1

For painter :
.....(1)
For the system;

..(2)
where; m = 100 kg

and
The correct answers are: tension in the rope is 1125 N, force of contact between the painter and the floor is 375 N

*Multiple options can be correct
Newton's Law Of Motion – MSQ - Question 2
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Two blocks A and B of equal masses m kg each are connected by a light thread, which passes over a massless pulley as shown in the figure. Both the blocks lies on wedge of mass m kg. Assume friction to be absent everywhere and both the blocks to be always in contact with the wedge. The wedge lying over smooth horizontal surface is pulled towards right with constant acceleration a (m/s2). 

(g is acceleration due to gravity)

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Detailed Solution for Newton's Law Of Motion – MSQ - Question 2


The FBD of A and B are
Applying Newton's second law to block A and B along normal to inclined surface

Solving  
For NA to be non zero

or 

The correct answers are:

Normal reaction on A is  

Normal reaction on B is 

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Newton's Law Of Motion – MSQ - Question 3
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A light inextensible string connects a block of mass m and top of wedge of mass M. The string is parallel to inclined surface and the inclined surface makes and angle θ with horizontal as show in the figure. All surfaces are smooth. Now a constant horizontal force of minimum magnitude F is applied to wedge towards right such that the normal reaction on block exerted by wedge just becomes zero.

The magnitude of acceleration of wedge is

Detailed Solution for Newton's Law Of Motion – MSQ - Question 3

If a is acceleration of wedge, from

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Newton's Law Of Motion – MSQ - Question 4
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Two smooth blocks are placed at a smooth corner as shown in the figure. Both the blocks are having mass m. We apply a force F on the small block m. Block A presses the block B in the normal direction, due to which pressing force on vertical wall will increase, and pressing force on the horizontal wall decrease, as we increase F (θ = 37º  with horizontal). As soon as the pressing force on the horizontal wall by block B becomes zero, it will loose the contact with the ground. If the value of F is further increased, the block B will accelerate in upward direction and simultaneously the block A will move toward right.

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Detailed Solution for Newton's Law Of Motion – MSQ - Question 4

For equilibrium of block (A)

To lift block B from ground

To keep regular contact

The correct answer is: aB = 3a/4

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Newton's Law Of Motion – MSQ - Question 5
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Two blocks of mass M and m, are used to compress two different massless springs as shown in the figure. The left spring is compressed by 3cm, while the right spring is compressed by an unknown amount. The system is at rest and all contact surfaces are smooth. Which of the following statements are true?

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Detailed Solution for Newton's Law Of Motion – MSQ - Question 5

Let F be the force exerted by mass m on mass M.
FBD of mass M
F = k1x1 = 2 × 3 = 6N

FBD of mass m

k2x2 = F = 6N to the left
Hence the force exerted on block of mass m be the right spring (k2x2) is 6 N to the left. From FBD, the normal reaction (F) between blocks is 6 N.
As system is at rest, net force on block of mass m is zero.
The correct answers are: The force exerted on block of mass m by the right spring is 6N to left, The net force on block of mass m is zero, The normal force exerted by block of mass m on block of mass M is 6 N

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Newton's Law Of Motion – MSQ - Question 6
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Two blocks A and B of equal mass m are connected through a massless string and arrangement as shown in the figure. The wedge is fixed on horizontal surface. Friction is absent everywhere. When the system is released from rest.

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Detailed Solution for Newton's Law Of Motion – MSQ - Question 6


Let a be acceleration of system and T be tension in, the string.
F.B.D. of block A

F.B.D. of block B

Adding equation (1) and (2) we get


from equation (1) ;

T = mg/4
The correct answers are: tension in string is mg/4, acceleration of A is 3g/4

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Newton's Law Of Motion – MSQ - Question 7
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Figure shows two block A and B connected to an ideal pulley string system. In this system when bodies are released then : (neglect friction and take g = 10m/s2)

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Detailed Solution for Newton's Law Of Motion – MSQ - Question 7

Applying NLM on 40kg block

For 10kg block T = 10 × 4a
Solving a = 2m/s2
T = 80 N
The correct answers are: Acceleration of block A is 2 m/s2, Tension in string connected to block B is 80 N

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Newton's Law Of Motion – MSQ - Question 8
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In the figure shown all the surfaces are smooth. All the blocks A, B and C are movable, x-axis is horizontal and y-axis is vertical as shown. Just after the system is released from the position as shown.

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Detailed Solution for Newton's Law Of Motion – MSQ - Question 8

There is no horizontal force on block A, therefore it does not move in x-direction, whereas there is net downwards force (mg – N) is acting on it, making its acceleration along negative y-direction.
Block B moves downwards as well as in negative x-direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B, A moves in positive x-direction.

Due to the component of normal exerted by C on B, it moves in negative x-direction.

The force acting vertically downwards on block B are mg and NA (normal reaction due to block A). Hence the component of net force on block B along the inclined surface of B is greater than mg sin θ. Therefore the acceleration of ‘B’ relative to ground directed along the inclined surface of ‘C’ is greater than g sin θ.

The correct answers are: Acceleration of ‘A’ relative to ground is in negative y-direction, Acceleration of ‘A’ relative to B is in positive x-direction, The horizontal acceleration of ‘B’ relative to ground is in negative x-direction., The acceleration of ‘B’ relative to ground directed along the inclined surface of ‘C’ is greater than gsinθ

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Newton's Law Of Motion – MSQ - Question 9
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A block of mass m is placed on a wedge. The wedge can be accelerated in four manners marked as (1), (2), (3) and (4) as shown. If the normal reactions in situation (1), (2), (3) and (4) are N1, N2, N3 and N4 respectively and acceleration with which the block slides on the wedge in situations are b1, b2, b3 and b4 respectively then :

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Detailed Solution for Newton's Law Of Motion – MSQ - Question 9

1.

Balancing forces perpendicular to incline


2. 

Similarly for this case get  

3.

Similarly for this case get 

4.

Similarly for this case get, 


The correct answers are: N3 > N1 > N2 > N4 , b2 > b3 > b4 > b1

*Multiple options can be correct
Newton's Law Of Motion – MSQ - Question 10
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A cylinder of mass M and radius R is resting on two corner edges A and B as shown in the figure. The normal reaction at the edges A and B are : (Neglect friction)

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Detailed Solution for Newton's Law Of Motion – MSQ - Question 10


The correct answers are: 

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