Multiple Choice Questions (MCQs): Friction - NEET MCQ

# Multiple Choice Questions (MCQs): Friction - NEET MCQ

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## 10 Questions MCQ Test Physics Class 11 - Multiple Choice Questions (MCQs): Friction

Multiple Choice Questions (MCQs): Friction for NEET 2024 is part of Physics Class 11 preparation. The Multiple Choice Questions (MCQs): Friction questions and answers have been prepared according to the NEET exam syllabus.The Multiple Choice Questions (MCQs): Friction MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Multiple Choice Questions (MCQs): Friction below.
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Multiple Choice Questions (MCQs): Friction - Question 1

### Complete the sentence. Friction always ____________

Detailed Solution for Multiple Choice Questions (MCQs): Friction - Question 1

Frictional forces act over the common surfaces of the two bodies to avoid and restrict relative moment between those two surfaces.

Multiple Choice Questions (MCQs): Friction - Question 2

### After the body starts moving, the friction involved with motion is

Detailed Solution for Multiple Choice Questions (MCQs): Friction - Question 2

When the body is in rest it is under static friction but when it starts moving (neither rolling nor sliding), the static friction slowly chngs to kinetic friction as the coefficient of static friction start decreasing and that of kinetic friction starts increasing. In case it starts rolling motion then the friction is rolling friction & if it slides then sliding fiction.

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Multiple Choice Questions (MCQs): Friction - Question 3

### Impending motion of a body is opposed by

Detailed Solution for Multiple Choice Questions (MCQs): Friction - Question 3

The impending motion refers to the state of a body when it is on the verge of slipping. In such cases, the static friction has reached its upper limit and is given by the equation, F=μsN

Multiple Choice Questions (MCQs): Friction - Question 4

A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5 ms-2, the frictional force acting on the block is

Detailed Solution for Multiple Choice Questions (MCQs): Friction - Question 4

Limiting friction, f = μmg = 0.6 × 1 × 9.8 = 5.88 N

Applied force = F = ma = 1 × 5 = 5 N

As F < f, so force of friction = 5 N

Multiple Choice Questions (MCQs): Friction - Question 5

A particle moves in a circle with a constant speed of 6m /s. Its centripetal acceleration is 120 m/s2, The radius of the circle is

Detailed Solution for Multiple Choice Questions (MCQs): Friction - Question 5

We know that centripetal acceleration, c = v2/r
Thus we get r = v2/c
= 36 / 120
= .3 /m

Multiple Choice Questions (MCQs): Friction - Question 6

A body rests on an inclined plane and the angle of inclination is varied till the body just begins to slide down. The coefficient of friction is . What is the angle of inclination?

Detailed Solution for Multiple Choice Questions (MCQs): Friction - Question 6

At the time when the block just starts to move, we get that net force acting upon it is 0, thus we get, f - mg.sin a = 0
Where f is friction force and a is angle of incline.
We also know that f = 1√3 x N
= 1√3 x mg.cos a
Thus we get 1√3 x mg.cos a = mg.sin a
Thus we get tan a = 1√3
And a = 30

Multiple Choice Questions (MCQs): Friction - Question 7

A block of mass 2kg rests on a plane inclined at an angle of 30with the horizontal. The coefficient of friction between the block and the surface is 0.7. The frictional force acting on the block is

Detailed Solution for Multiple Choice Questions (MCQs): Friction - Question 7

Mass of the block = 2 kg

Weight of the block = mg = 2 × 9.8 = 19.6 N

The component of the weight along normal = 19.6 Cos 30° = 16.97 N

Hence, the normal force (N) = 16. 97 N

Now, the component of the weight along the inclined plane = 19.6 Sin 30°

Along the inclined plane = 9.8 N

Now, the friction = μ N = 0.7 × 16.97 = 11.87 N

Since, 9.8 N is less than the maximum value of the friction (11.87 N)

Hence, the frictional force = 9.8 N

Multiple Choice Questions (MCQs): Friction - Question 8

A particle moves in a circle of radius 0.30m with a constant speed of 6m/s. Its centripetal acceleration is

Detailed Solution for Multiple Choice Questions (MCQs): Friction - Question 8

We know that centripetal acceleration, a = v x v /r
Here v = 6m/s and r = 0.3m
Thus a = 36 / 0.3 m/s2
= 120 m/s2

Multiple Choice Questions (MCQs): Friction - Question 9

When a wheel rolls on a level road, the direction of frictional force at the point of contact of wheel and ground is:

Detailed Solution for Multiple Choice Questions (MCQs): Friction - Question 9

Frictional force is the opposing force which plays between two surfaces and it destroys the relative motion between them. Frictional force is a non-conservative force. The force produced by two surfaces that contact and slide against each other, that force is called the frictional force. These forces are affected by the nature of the surface and amount of force acting on them.

In case of a bicycle, the front wheel of the bicycle is connected to a rod passing through its centre. The force acting on the wheel about its central axis by the force coming from the rest of the bicycle is zero. Front wheel obtains linear velocity by pedalling but it cannot rotate it.

Wheel or ball can also be rolled by pushing on it. The frictional force prevents the wheel from sliding forward at the point of contact. Here, the frictional force prevents the wheel from sliding forward and it is in the opposite direction.

So, in the case of the wheel, the point P which is in contact with the ground tries to go backward due to rotation. Frictional force will oppose this motion. Hence it will move forward.
Hence the direction of frictional force at the point P of the wheel is in forward direction.
Note: Frictional force opposes the motion. Here static friction holds a wheel or a ball on the surface. Frictional force is equal and opposite in direction to the applied force parallel to the contacting surfaces. The resistance due to the rolling body on a surface is called rolling friction. Torque is a force that acts on a body that is undergoing rotation.

Multiple Choice Questions (MCQs): Friction - Question 10

A force of 20 N is applied on a body of mass 4 kg kept on a rough surface having coefficient of friction 0.1. Find acceleration of body. Take g = 10 m/s2

Detailed Solution for Multiple Choice Questions (MCQs): Friction - Question 10

## Physics Class 11

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## Physics Class 11

102 videos|411 docs|121 tests