Question

CASE BASED MCQs-1
When bodies are in contact, there are mutual contact forces satisfying the third law of motion. The component of contact force normal to the surfaces in contact is called normal reaction. The component parallel to the surfaces in contact is called friction.


In the above figure, 8 kg and 6 kg are hanging stationary from a rough pulley and are about to move. They are stationary due to roughness of the pulley.
1. Which force is acting between pulley and rope?
(a) Gravitational force
(b) Tension force
(c) Frictional force
(d) Buoyant force
2. The normal reaction acting on the system is

(a) 8 g (c) 2 g
(b) 6 g (d) 14 g
3. The tension is more on side having mass of
(a) 8 kg (c) 6 kg
(b) Same on both (d) Nothing can be said
4. The force of friction acting on the rope is
(a) 20 N (c) 40 N
(b) 30 N (d) 50 N
5. Coefficient of friction of the pulley is
(a) 1/6 (c)1/5
(b) 1/5 (d) 1/4​

Answer 1

Answer:

1. The answer must be (c) Frictional force.

2.The answer must be (d) 14g.

3.The answer must be (a) 8g.

4.The answer must be (b) 30 N.

5.The answer must be (c) 1/5.

Answer 2

Answer:

5 .coefficient of friction of the pulley is 1/7