Question

A boy and a man carry a uniform rod of length $L$ horizontally in such a way that the boy gets $\left(\frac{1}{4}\right)^{t h}$ of the load. If the boy is at one end of the rod, the distance of the man from the other end is

• A. $\frac{L}{3}$
• B. $\frac{L}{4}$
• C. $\frac{2 L}{3}$
• D. $\frac{3 L}{4}$

Answer 1

Answer: (A)

Let normal reaction on the rod due to boy is $N_{1}$ and due to man is $N_{2}$
Given, $N_{1}=\frac{W}{4}$
$\therefore $ $N_{2}=W-\frac{W}{4}=\frac{3 W}{4}$

Let $x$ be the distance of the man from the other end.
For rotational equilibrium net torque must be zero.
$\Rightarrow N_{1}\times \frac{L}{2}=N_{2}\times \left(\frac{L}{2} - x\right)$
$\Rightarrow $ $\frac{W}{4}\times \frac{L}{2}=\frac{3 W}{4}\times \left(\frac{L}{2} - x\right)$
$\Rightarrow $ $x=\frac{L}{3}$