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QUESTION: 1

Concept of pseudo force is valid only in

Solution:

Pseude force is applied only for non-inertial frame.

QUESTION: 2

A block of mass m is kept on a smooth moving wedge. If the acceleration of the wedge is equal to a, acceleration of m relative to the wedge is

Solution:

F. B. D. of block with respect to wedge.

QUESTION: 3

A boy of mass 40 kg is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is 0.8 and g = 10 m/s^{2} the horizontal force that he is applying on the pole is

Solution:

Here, μ= 0. 8

Let F be horizontal force that the boy is applying on the pole.

The various forces are acting on the boy as shown in the below figure.

Frictional force.

QUESTION: 4

In the adjoining figure, the tension in the string connecting A and B is

Solution:

QUESTION: 5

In the system shown in the adjoining figure, the acceleration of the 1 kg mass is

Solution:

Suppose a be the downward acceleration of the 4 kg mass, therefore, 2 a is the upward acceleration of the I kg mass. Hence, equations of motion are:

Adding, after multiplying the equation (i) by 2.

Thus, the acceleration of the mass

a 1 kg is g/2 upwards.

QUESTION: 6

Two masses M_{1} and M_{2} are attached to the ends of a string which passes over a pulley attached to the top of a double inclined plane of angles of inclination α and β If M_{2 }> M_{1}, the acceleration ‘a’ of the system is given by

Solution:

Suppose M_{2} moves downwards with an acceleration a and the tension in the string is T.

QUESTION: 7

Three masses of 1 kg, 6 kg and 3 kg are connected to each other with threads and are placed on a table a as shown in the figure. What is the acceleration with which the system is moving (Take g = 10 m/s^{2})

Solution:

QUESTION: 8

The pulley arrangements shown in figure are identical, the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with a constant downward force, F = 2 mg, where g is acceleration due to gravity. The acceleration of mass m in case I is

Solution:

QUESTION: 9

In the pulley-block arrangement shown in figure. Find the relation between acceleration of block A and B.

Solution:

x_{1} + x_{2} = l_{1}

Differentiating with respect to time, we get v_{1} + v_{3} = 0

Again differentiating w.r.t. time,

QUESTION: 10

Two masses of 1 kg and 5 kg are attached to the ends of a massless string passing over a pulley of negligible weight. The pulley itself is attached to a light spring balance as shown in the figure. The masses start moving during this interval, the reading of spring balance will be

Solution:

Applying the 2nd law of motion,

Equating all forces on the string carrying the 2 blocks we get T = 5g/3

The tension in spring balance = 2T

So, the reading is the mass = T/g = 10/3 Kg < 6Kg

QUESTION: 11

Two bodies of mass 4 kg and 6 kg are attached to the ends of a string passing over a pulley. A 4 kg mass is attached to the table top by another string. The tension in this string T_{1} is equal to (g = 10 m/s^{2})

Solution:

QUESTION: 12

Two bodies of mass 6 kg and 4 kg are tied to a string as shown in the adjoining figure. If the table is smooth and pulley frictionless, then acceleration of mass 6 kg will be (g = 10 m/s^{2})

Solution:

QUESTION: 13

Two masses of 8 kg and 4 kg are connected by a string as shown in the figure over a frictionless pulley. The acceleration of the system is

Solution:

QUESTION: 14

A trolley T of mass 5 kg on a horizontal smooth surface is pulled by a load of 2 kg through a uniform rope ABC of length 2 m and mass 1 kg. As the load falls from BC = 0 to BC = 2 m, its acceleration (in m/s^{2}) changes from

Solution:

Initial force = load x g = 2 x 10 = 20 N

∴ initial acceleration = Force/Mass =

Final force = (load + mass of the thread) x g = ( 2 + 1 ) x 10 = 30N

∴ Final acceleration = (30/8) m/s^{2}

QUESTION: 15

Two wooden blocks are moving on a smooth horizontal surface such that the mass m remains stationary with respect to block of mass M as shown in figure. The magnitude of force P is

Solution:

The difterent torces acting on mass m are shown in the adjoining figure. Acceleration of the system

QUESTION: 16

A block sliding along inclined plane as shown in figure. If the acceleration of chamber is ‘a’ as shown in the figure. The time required to cover a distance L along inclined plane is

Solution:

Applying pseudo force ma along the left direction of the block.

The net acceleration along the incline will be a_{effective}=acosθ+gsinθ

The time required will be L=0×t+½ a_{effective}t^{2}

⇒ √ 2L/ a_{effective}

⇒√ 2L /acosθ+gsinθ

QUESTION: 17

A block of mass 4 kg is suspended through two light spring balances A and B. Then A and B will read respectively

Solution:

Tension is uniformly transmitted if the springs are massless.

QUESTION: 18

A light string fixed at one end to a clamp on ground passes over a fixed pulley and hangs at the other side. It makes an angle of 30^{o} with the ground. A monkey of mass 5 kg climbs up the rope. The clamp can tolerate a vertical force of 40 N only. The maximum acceleration in upward direction with which the monkey can climb safely is (neglect friction and take g = 10 m/s^{2})

Solution:

Let T be the tension in the string.

The upward force exerted on the clamp = T sin 30^{o} = T/2

Given: (T/2) = 40 N or T = 80 N

If ‘a’ is the acceleration of the monkey in upward direction,

QUESTION: 19

If the normal reactional force is doubled, the coefficient of friction is

Solution:

Coefficient of static friction does not depend upon normal reaction.

QUESTION: 20

The force acting on the block pushes it, then pushing of the block will be possible along the surface, if

Solution:

In this case, vertical component of the force increases the normal reaction, i.e.,

R = mg + mg cos θ = mg(1 + cosθ)

Hence, block can be pushed along the horizontal surface when Horizontal component of force > frictional force,

QUESTION: 21

An object is placed on the surface of a smooth inclined plane of inclination θ It takes time ‘t’ to reach he bottom. If the same object is allowed to slide down a rough inclined plane of inclination θ it takes time nt to reach the bottom where n is a number greater than 1. The coefficient of friction μ is given by

Solution:

On smooth inclined plane:

Acceleration of the body = g sin θ If s be the distance travelled, then

QUESTION: 22

The frictional force between two surfaces is independent of

Solution:

The frictional force between two surfaces

where μ_{s} depends on the nature of surface and their materials. Thus, frictional force is independent of area of contact of the surfaces.

QUESTION: 23

A body of mass 2 kg rests on a rough inclined plane making an angle 30^{o} with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

Solution:

The body is at rest on the inclined plane, as μ > tan θ force of friction = mg sin 30^{o} = 2 x 9.8 x (1/2) = 9.8 N

QUESTION: 24

A man walks over a rough surface, the angle between the force of friction and the instantaneous velocity of the person is

Solution:

When a man walks on a rough surface, it is the frictional force which is responsible for motion, i.e, required angle between frictional force and instantaneous velocity is zero.

QUESTION: 25

A heavy block of mass M is slowly placed on a conveyor belt moving with a speed v. The coefficient of friction between the block and the belt is μ. Through what distance will the block slide on the belt?

Solution:

Frictional force on the block = μMg

It will be causing acceleration a =μMg/M = μg

The block slide on the ball till its velocity becomes ‘v’ given by the equation

QUESTION: 26

A uniform rope of length l lies on a table. If the coefficient of friction is m, then the maximum length l_{1}of the part of this rope which can overhang from the edge of the table without sliding down is

Solution:

QUESTION: 27

A body is moving with uniform velocity of 2 m/s on a rough level surface. The frictional force on it is 10 N. If the body moves with velocity 4 m/s, the force of friction will be

Solution:

Kinetic friction is constant, hence frictional force will remain same (= 10 N).

QUESTION: 28

The coefficient of static friction μ between block A of mass 2kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless (g = 10 m/s^{2})

Solution:

For the equilibrium, m_{B}g = μm_{A}g

QUESTION: 29

At any instant if the velocity of point A is 10 m/s then calculate the velocity of point B?

Solution:

QUESTION: 30

Frictional force acting on the block is

Solution:

Net force = 50 - 20 = 30 N

Limiting friction on the body = u (static) × m × g = 0.6 × 10 × 10 = 60 N.

F = 30 N is less than the limiting friction so the body is static. So, a = 0.

Force of friction acting on the body is static friction, f = driving force = 30 N.

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