A boy and a man carry a uniform rod of length ‘L’ horizontally in such a way that the boy get 1/4 of the load. If the boy is at one end of the rod, the distance of the man from the other end is

Therefore, for rotational equilibrium, the couple about the C.M. must be zero. It implies that.

$$\begin{gathered} \frac{\mathrm{W}}{4}\times \frac{\mathrm{L}}{2}=\frac{3\mathrm{W}}{4}\times \mathrm{x} \\ \mathrm{x}=\frac{\mathrm{L}}{6} \end{gathered}$$

And the distance between man and the other end is

$\frac{\mathrm{L}}{2}-\mathrm{x}=\frac{\mathrm{L}}{2}-\frac{\mathrm{L}}{6}=\frac{\mathrm{L}}{3}$

Hence the correct answer is $\frac{\mathrm{L}}{3}.$