A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum friction of the length of the chain that can hang over one edge of the table is

(a) 20%

(b) 25%

(c) 35%

(d) 15%

As per the given question

we have : coefficient of friction = o.25

Let total length of chain be “L” and l be the length which is hang over

remaining part which will be on table will be [L-l] .

Now as it is not mentioned in question that the chain is at rest or sliding, Lets consider it is at rest.

If chain is at rest than , Tension = mg (of hanging part) USING FREE BODY DIAGRAM

T = mg than m = T/g

F = uR

As Linear mass desity = m/l

for hanging part , m = lamda l , and remaining one as m = lamda [L-l]

now substitute in equation of equilibrium as friction force on [L-l] length will be equal to tension.

and thus l is calculated

aftr solving we will get : l/L = 1/5

ie l/L x 100 = 20%

Hence ans will be 20% of total length can hang over the edge [for sytem to be in euilibrium