Description

This mock test of Friction Force MCQ Level - 2 for Physics helps you for every Physics entrance exam.
This contains 10 Multiple Choice Questions for Physics Friction Force MCQ Level - 2 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Friction Force MCQ Level - 2 quiz give you a good mix of easy questions and tough questions. Physics
students definitely take this Friction Force MCQ Level - 2 exercise for a better result in the exam. You can find other Friction Force MCQ Level - 2 extra questions,
long questions & short questions for Physics on EduRev as well by searching above.

QUESTION: 1

A box of mass 8kg is placed on a rough inclined plane of inclination *θ.* Its downward motion can be prevented by applying an upward pull *F* and it can be made to slide upwards by applying a force 2*F.* The coefficient of friction between the box and the inclined plane is

Solution:

During downward motion :

*F* = *mg* Sin*θ* - *mg* Cos*θ*

During upward motion :

2*F* = *mg *Sin*θ* + μ*mg *Cos*θ*

Solving above two equations, we get *m =* 1/3 tan*θ*

The correct answer is:

QUESTION: 2

The coefficient of friction between 4*kg* and 5*kg* blocks is 0.2 and between 5*kg* blocks and ground is 0.1 respectively. Choose the correct statement :

Solution:

So block '*Q*' is moving due to force while block '*P*' due to friction.

Friction direction on both *P* + *Q* blocks as shown.

First block '*Q*' will move and P will move with '*Q*' so by *FBD* taking '*P*' and '*Q*' as system

When applied force is 4 *N* then *FBD
*

i.e., a

4

So acceleration

Slipping will start at when

The correct answer is: Maximum acceleration of 4

QUESTION: 3

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 *m*/*s*^{2}, the friction force acting on the block is :

Solution:

Maximum friction

Pseudo force = *ma* = 5*N*

acceleration due to the applied Pseudo force

So* *required friction force is only 5N although maximum friction force available is 6N.

The correct answer is: 5 N

QUESTION: 4

In the following figure, find the direction of friction on the blocks and ground respectively.

Solution:

Direction of friction is such that it opposes the relative velocity.

The correct answer is: Block *A* (left), block *B* (right due to block *A*, right due to ground)

QUESTION: 5

A chain of length *L* is placed on a horizontal surface as shown in figure. At any instant x is the length of chain on rough surface and the remaining portion lies on smooth surface. Initially *x* = 0. A horizontal force *P* is applied to the chain (as shown in the figure). In the duration *x* changes from *x* = 0 to *x* = *L*, for chain to move with constant speed :

Solution:

For chain to move with constant speed *P* needs to be equal to frictional forces on the chain. As the length of chain on the rough surface increases.

Hence the friction force *f _{k}* = µ

The correct answer is: the magnitude of

QUESTION: 6

In the arrangement shown in the figure mass of the block *B* and *A* are 2*m*, 8m respectively. Surface between *B* and floor is smooth. The block *B* is connected to block *C* by means of a pulley. If the whole system is released then the minimum value of mass of the block *C* so that the block *A* remains stationary with respect to *B* is :

(Co-efficient of friction between *A* and *B* is μ).

Solution:

*FBD* of *A* and *C*

If acceleration of '*C*' is a

For block '*A*' N = 8 ma ...(1)

for block *A* to remain stationary with respect to block *B*,

8mg = μN (Limiting condition) ...(2)

a = g/μ

and acceleration a can be written by the equation of system (*A* + *B* + *C*)

The correct answer is:

QUESTION: 7

The two blocks, *m* = 10*kg* and *M* = 50*kg* are free to move as shown. The coefficient of static friction between the blocks is 0.5 and there is no friction between *M* and the ground. A minimum horizontal force *F* is applied to hold m against *M* that is equal to :

Solution:

Limiting condition for *m* to not slip in vertical downward direction,

Same normal force would acceleration *M*, thus

Taking *m* + *M* as system

The correct answer is: 240 *N*

QUESTION: 8

A given object takes *n* times more time to slide down 45° rough inclined plane as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the objects and the incline is.

Solution:

For smooth plane *a* = *g* sin*θ*

For rough plane, *a' = g*(sin*θ - μ*cos*θ*)

∴ t' = nt

when θ = 45° , sin*θ* = cos*θ = 1/√2*

On solving we get,

The correct answer is:

QUESTION: 9

A force *F* = *t* is applied to a block *A* as shown in figure, where *t* is time in *seconds*. The force is applied at *t* = 0 *seconds* when the system was at rest. Which of the following graph correctly give the frictional force between *A* and horizontal surface as a function of time *t*. [Assume that at *t* = 0, tension in the string connecting the two blocks is zero].

Solution:

Let *m _{A}* and

As the force

As

As

The correct answer is:

QUESTION: 10

Given *m _{A}* = 20

Solution:

Limiting friction between *A* and *B* = 60 *N*

Limiting friction between *B* and *C =* 90 *N*

Limiting friction between *C* and ground = *50 N*

Since limiting friction is least between *C* and ground, slipping will occur at first between *C* and ground. This will occur when *F = 50 N.*

The correct answer is: 60 *N*

### Force of friction

Video | 35:50 min

### Notes : Newton's Force & Friction

Doc | 24 Pages

### L6 : Contact Force: Friction - Force & Pressure, Science, Class 8

Video | 08:07 min

### Force of friction keeping velocity constant

Video | 09:10 min

- Friction Force MCQ Level - 2
Test | 10 questions | 45 min

- Friction Force NAT Level - 2
Test | 10 questions | 45 min

- Friction Force MCQ Level - 1
Test | 10 questions | 30 min

- Friction Force NAT Level - 1
Test | 10 questions | 30 min

- Friction Force â€“ MSQ
Test | 10 questions | 45 min