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Newton's Law Of Motion NAT Level - 2 - IIT JAM MCQ

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10 Questions MCQ Test - Newton's Law Of Motion NAT Level - 2

Newton's Law Of Motion NAT Level - 2 for IIT JAM 2024 is part of IIT JAM preparation. The Newton's Law Of Motion NAT Level - 2 questions and answers have been prepared according to the IIT JAM exam syllabus.The Newton's Law Of Motion NAT Level - 2 MCQs are made for IIT JAM 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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*Answer can only contain numeric values

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

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

*Answer can only contain numeric values

Newton's Law Of Motion NAT Level - 2 - Question 2

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 v_x= 2u

As it also remain in contact with wedge B

The correct answer is: 2.64

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*Answer can only contain numeric values

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

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

*Answer can only contain numeric values

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

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

*Answer can only contain numeric values

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

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 a_B = a
From constraint; acceleration of A and C are
towards left.
a_C = a/2
free body diagram of A, B and C are

solving we get

The correct answer is: 3

*Answer can only contain numeric values

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

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

*Answer can only contain numeric values

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

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

*Answer can only contain numeric values

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

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

*Answer can only contain numeric values

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

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

*Answer can only contain numeric values

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

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 a_C = 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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