A uniform chain of length $$l\ $$ has one of its end attached to the wall a point A, while $$\frac{3l}{4}$$of the length of the chain is lying on table as shown in figure. Find the minimum co-efficient of friction between table and chain so that chain remains in equilibrium is.

Let the mass per unit length of the chain be $$\lambda$$.

The length of the inclined part is  $$l_i=\frac{l}{4}$$

Hence, its weight is  $$W_i=\lambda\frac{l}{4}g$$

Let the tension at the junction of the inclined and horizontal parts be $$T$$.

For the inclined part, the tension at the junction is horizontal. The tension at A makes an angle $$37^\circ$$ with the vertical.

For equilibrium of the inclined part,  $$T_A\cos37^\circ=W_i$$

Therefore,  $$T_A=\frac{W_i}{\cos37^\circ}$$

The horizontal component of this tension is balanced by $$T$$:  $$T=T_A\sin37^\circ$$

Hence,  $$T=W_i\tan37^\circ$$

Using  $$\tan37^\circ=\frac{3}{4}$$

we get  $$T=\lambda\frac{l}{4}g\times\frac{3}{4}$$

$$T=\frac{3\lambda lg}{16}$$

Now, the length of the chain lying on the table is  $$l_t=\frac{3l}{4}$$

Therefore, the normal reaction is  $$N=\lambda\frac{3l}{4}g$$

For limiting equilibrium,  $$f=\mu N$$

The friction balances the tension $$T$$:  $$\mu N=T$$

Therefore,  $$\mu=\frac{T}{N}$$

$$\mu=\frac{\frac{3\lambda lg}{16}}{\frac{3\lambda lg}{4}}$$

$$\mu=\frac{1}{4}$$

Therefore,  $${\mu_{\min}=\frac14}$$

Hence, the correct option is B.