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This mock test of Newton's Law Of Motion NAT Level - 2 for IIT JAM helps you for every IIT JAM entrance exam.
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*Answer can only contain numeric values

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

Solution:

Let * c* be acceleration of wedge

Applying Newton law in horizontal direction to system of

Applying Newton's law of block

Solving (1), (2) and (3)

we get

Acceleration of wedge

The correct answer is: 0

*Answer can only contain numeric values

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 :

Solution:

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

*Answer can only contain numeric values

QUESTION: 3

In the figure shown the velocity of lift is 2*m*/*s* while string is winding on the motor shaft with velocity 2*m*/*s* block *A* is moving downwards with a velocity of 2*m*/*s*, then find out the velocity of block *B* (in *m*/*s*).

Solution:

*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

The correct answer is: 8

*Answer can only contain numeric values

QUESTION: 4

In the arrangement shown, by what acceleration (in *m/s*^{2}) the boy must go up so that 100*kg* block remains stationary on the wedge. The wedge is fixed and friction is absent everywhere. (Take *g* = 10 *m/s*^{2})

Solution:

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

QUESTION: 5

In the figures shown *P*_{1} and *P*_{2} are massless pulleys. *P*_{1} is fixed and *P*_{2} 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*/*s*^{2}.

Solution:

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

QUESTION: 6

In the arrangement shown in the Figure, a block of mass * m* = 2

Solution:

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

QUESTION: 7

Two block *A* and *B* each of mass * m* are placed on a smooth horizontal surface. Two horizontal force

Solution:

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

*FBD* of block *A*

solving* N =* 3*F*

The correct answer is: 3

*Answer can only contain numeric values

QUESTION: 8

In the figure shown, the pulleys and strings are massless. The acceleration (in *m*/*s*^{2}) of the block of mass 4*m* just after the system is released from rest is ...........

Solution:

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 4*m*, a = 5g/11 downwards.

The correct answer is: 4.545

*Answer can only contain numeric values

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/s*^{2}. 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*/*s*^{2})

Solution:

Acceleration of box = 10 *m*/*s*^{2}

Inside the box force acting on bob are shown in the figure

The correct answer is: 14.14

*Answer can only contain numeric values

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*/*s*^{2} just after the block *A* is released. Masses of *A*, *B* and *C* are * M*,

Solution:

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 *T*_{2}. Now the arrangement will be

Since, tension in spring does not change instantaneously, hence, tension in string perpendicular will remain same i.e. 2*Mg*, 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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