Q. A boy and a man carry a uniform rod of length $L$, horizontally in such a way that boy gets $\frac{1}{4}$ th 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) $\frac{2L}{3}$
D) $\frac{3L}{4}$
Explanation
(a) Weight of the rod=$w$
Reaction of boy $R_{B} = \frac{w}{4}$
Reaction of man $R_{M} = \frac{3w}{4}$
[image:7_image134.emf]
As the rod is in rotational equilibrium
$\therefore$ $\sum_{}^{}{\tau = 0}$
$R_{B} \times \frac{L}{2} - R_{M} \times x = 0$
$\Rightarrow$ $\frac{w}{4} \times \frac{L}{2} - \frac{3w}{4} \times x = 0$
$\Rightarrow$ $x = \frac{L}{6}$
$\therefore$ Distance from other end, $y = \frac{L}{2} - x$
$\Rightarrow$ $y = \frac{L}{2} - \frac{L}{6} = \frac{2L}{6} = \frac{L}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.