CGPET 2022 Solved Paper

Let (

1

n

) th part of chain hangs in equilibrium. Then, (1−

1

n

) th part lies on table.
In equilibrium, downward gravitational pull = Force of friction.
⇒(

1

n

)⋅L⋅x⋅g=(1−

1

n

)⋅L⋅x⋅g⋅µ
Here, x= mass per unit length, L= length of complete chain.
So, n−1=

1

µ

⇒n=1+

1

µ

=1+

1

0.25

=5
∴

1

5

th part of chain =20% of total length can over hang.