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Friction Obj-2 HC Verma Solutions Vol-1 Ch-6

Friction Obj-2 HC Verma Solutions Vol-1 Chapter-6 Vol-1 Concept of Physics for Class-11. Step by Step Solutions of Objective -2 (MCQ-2) Questions of Chapter-6 Friction  (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Friction Obj-2 (MCQ-2) HC Verma Solutions Ch-6 Vol-1 Concept of Physics for Class-11

Board ISC and other board
Publications Bharti Bhawan Publishers
Chapter-6 Friction 
Class 11
Vol  1st
writer HC Verma
Book Name Concept of Physics
Topics Solution of Objective-2 (MCQ-2) Questions
Page-Number 97

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Objective-I

Objective-II

Exercise


Friction  Obj-2 (MCQ-2) HC Verma

HC Verma Solutions of Ch-6  Vol-1 Concept of Physics for Class-11

(Page-97)

Question-1 :-

Let F, FN and f denote the magnitudes of the contact force, normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero.
(a) F > FN
(b) F > f
(c) FN > f
(d) FN − f < F < FN + f.

Answer-1 :-

The options (a), (b) and (d) are correct

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Explanation:

Let F, FN and f denote the magnitudes of the contact force, normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero.

 

The system is in equilibrium condition when F = f.
Hence, the net horizontal force is zero.
f = μFN
F > FN
f = FN and 0 ≤ μ ≤ 1
Therefore, we can say that F > f. So the net horizontal force is nonzero.
F > f, and so the net horizontal force is zero.
FN > f ⇒ FN > μFN ⇒ μ < 1
Here, the given relation between F and f i.e
F > f and f = μFN  will not be satisfied So it cannot be said that the net horizontal force is zero or nonzero.
FN − f < F < FN + f
∵ f = μFN

Let F, FN and f denote the magnitudes of the contact force, normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero. 2

For the above relation, we can say that F ≠ f and so the net horizontal force is nonzero.

Question-2 :-

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The contact force exerted by a body A on another body B is equal to the normal force between the bodies We conclude that
(a) the surface must be frictionless
(b) the force of friction between the bodies is zero
(c) the magnitude of normal force equal that of friction
(d) the bodies may be rough but they don’t slip on each other.

Answer-2 :-

The options  (b) and (d) are correct

Explanation:

The contact force exerted by a body A on another body B is equal to the normal force between the bodies. Therefore, we can conclude that the force of friction between the bodies is zero or the bodies may be rough but they don’t slip on each other.

Question-3 :-

Mark the correct statements about the friction between two bodies.
(a) Static friction is always greater than the kinetic friction.
(b) Coefficient of static friction is always greater than the coefficient of kinetic friction.
(c) Limiting friction is always greater than the kinetic friction.
(d) Limiting friction is never less than static friction.

Answer-3 :-

The options (b) , (c) and (d) are correct

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Explanation:

All the above statements are correct. The static friction is sometimes less than the kinetic friction.

Question-5 :-

Consider a vehicle going on a horizontal road towards east. Neglect any force by the air. The frictional force on the vehicle by the road
(a) is towards east if the vehicle is accelerating
(b) is zero if the vehicle is moving with a uniform velocity
(c) must be towards east
(d) must be towards west.

Answer-5 :-

The options (a) and (b) are correct
Explanation:

When the vehicle is accelerating, the force is applied (by the tyre on the road) in west direction .That causes a net resultant  frictional force acting in east direction. Due to this  force of friction only ,the  car is moving in east direction.
When the vehicle is moving with a uniform velocity, the force of friction on the wheels of the vehicle by the road is zero.

—: End of Friction Obj-2 (MCQ-2) HC Verma Solutions Ch-6 Vol-I :–


Return to — HC Verma Solutions Vol-1 Concept of Physics

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