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Newton's Law Of Motion NAT Level - 2 - Physics MCQ


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10 Questions MCQ Test Topic wise Tests for IIT JAM Physics - Newton's Law Of Motion NAT Level - 2

Newton's Law Of Motion NAT Level - 2 for Physics 2024 is part of Topic wise Tests for IIT JAM Physics preparation. The Newton's Law Of Motion NAT Level - 2 questions and answers have been prepared according to the Physics exam syllabus.The Newton's Law Of Motion NAT Level - 2 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 NAT Level - 2 below.
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Newton's Law Of Motion NAT Level - 2 - Question 1
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In the figure shown blocks A and B are kept on a wedge C. A, B and C each have mass m. All surfaces are smooth. Find the acceleration of C.


Detailed Solution for Newton's Law Of Motion NAT Level - 2 - Question 1


Let c be acceleration of wedge C.
a be acceleration of block A w.r.t. wedge C.
b be acceleration of block B w.r.t. wedge C.
Applying Newton law in horizontal direction to system of A + B + C.

Applying Newton's law of block A and B along the incline gives.

Solving (1), (2) and (3)
we get c = 0
Acceleration of wedge C is 0 
The correct answer is: 0

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Newton's Law Of Motion NAT Level - 2 - Question 2
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System is shown in the figure. Assume that cylinder remains in contact with the two wedges. The velocity of cylinder  (in m/s) is :


Detailed Solution for Newton's Law Of Motion NAT Level - 2 - Question 2

As cylinder will remain in contact with wedge A vx = 2u

As it also remain in contact with wedge B

The correct answer is: 2.64

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Newton's Law Of Motion NAT Level - 2 - Question 3
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In the figure shown the velocity of lift is 2m/s while string is winding on the motor shaft with velocity 2m/s block A is moving downwards with a velocity of 2m/s, then find out the velocity of block B (in m/s).


Detailed Solution for Newton's Law Of Motion NAT Level - 2 - Question 3

v = velocity of B w.r.t. ground


v = 8 m/s    (velocity of B w.r.t. ground)
v = 8 m/s
v' = 6 m/s    (velocity of B w.r.t. lift)
v' = 6 m/s
The correct answer is: 8

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Newton's Law Of Motion NAT Level - 2 - Question 4
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In the arrangement shown, by what acceleration (in m/s2) the boy must go up so that 100kg block remains stationary on the wedge. The wedge is fixed and friction is absent everywhere. (Take g = 10 m/s2)

Detailed Solution for Newton's Law Of Motion NAT Level - 2 - Question 4

For block to be stationary T = 800 N
If man moves up by acceleration ‘a’


The correct answer is: 6

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Newton's Law Of Motion NAT Level - 2 - Question 5
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In the figures shown P1 and P2 are massless pulleys. P1 is fixed and P2 can move. Masses of A, B and C are 9m/64, 2m and m respectively. All contacts are smooth and the string is massless  Find the acceleration of block C in m/s2.

Detailed Solution for Newton's Law Of Motion NAT Level - 2 - Question 5

Let the acceleration of B downwards be aB = a
From constraint; acceleration of A and C are 
 towards left.
aC = a/2
free body diagram of A, B and C are


solving we get

The correct answer is: 3

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Newton's Law Of Motion NAT Level - 2 - Question 6
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In the arrangement shown in the Figure, a block of mass m = 2 kg  lies on a wedge on mass M = 8 kg. The initial acceleration of the wedge (if the surfaces are smooth) given by  then x is.

Detailed Solution for Newton's Law Of Motion NAT Level - 2 - Question 6




from wedge constraint


Applying Newton’s law (wedge + block)

along horizontal direction

Applying Newton’s law on block along the incline plane

Solving equation (1); (2); (3) and (4)

The correct answer is: 23

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Newton's Law Of Motion NAT Level - 2 - Question 7
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Two block A and B each of mass m are placed on a smooth horizontal surface. Two horizontal force F and 2F are applied on both the blocks A and B respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is nF : find the value of n.

Detailed Solution for Newton's Law Of Motion NAT Level - 2 - Question 7

Acceleration of two mass system is a = F/2m leftward
FBD of block A



solving N = 3F
The correct answer is: 3

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Newton's Law Of Motion NAT Level - 2 - Question 8
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In the figure shown, the pulleys and strings are massless. The acceleration (in m/s2) of the block of mass 4m just after the system is released from rest is ...........  

Detailed Solution for Newton's Law Of Motion NAT Level - 2 - Question 8

The FBD of block is as shown
From Newton’s second law



cos θ = 4/5 and from constraint we get a = A cos θ ...(3)
solving equation (1), (2) and (3)
we get acceleration of block of mass 4m, a = 5g/11 downwards.
The correct answer is: 4.545

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Newton's Law Of Motion NAT Level - 2 - Question 9
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Inside a horizontally moving box, an experimenter finds that the when an object is placed on a smooth horizontal table and is released, it moves with an acceleration of 10 m/s2. In this box if 1 kg body is suspended with a light string, the tension (in N) in the string in equilibrium position. (w.r.t.) experimenter will be (Take g = 10 m/s2)

Detailed Solution for Newton's Law Of Motion NAT Level - 2 - Question 9

Acceleration of box = 10 m/s2
Inside the box force acting on bob are shown in the figure



The correct answer is: 14.14

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Newton's Law Of Motion NAT Level - 2 - Question 10
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In the figure shown all the contacts are smooth. Strings and spring and light. Initially A is held by someone and  B and C are at rest and in equilibrium also. Find out the acceleration of block C in m/s2 just after the block A is released. Masses of A, B and C are M, M and 2M respectively.

Detailed Solution for Newton's Law Of Motion NAT Level - 2 - Question 10

Before block A was released, the system was at rest, and all blocks were in equilibrium hence, tension in both the strings is equal to 2 Mg.
When block A is released, it will have an unbalanced force on it and hence the tension in string (2) will change to say T2. Now the arrangement will be
Since, tension in spring does not change instantaneously, hence, tension in string perpendicular will remain same i.e. 2Mg, Thus, Block C will remain at rest aC = 0.
Newton's law along the string (2),  ⇒ a = g/2
Hence acceleration of A = g/2 upward, B = g/2 toward Right, and C = 0


The correct answer is: 0

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