A boy and a man carry a uniform rod of length $L$, horizontally in such a way that the boy gets ${\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{th}$ of the load. If the boy is at one end of the rod, the distance of the man from the other end is

• Correct answer
  A

  $\frac{L}{3}$
• B

  $\frac{L}{4}$
• C

  $\frac{2L}{3}$
• D

  $\frac{3L}{4}$

Solution

Weight of the rod=$w$

Reaction of boy $R_B=\frac{w}{4}$

Reaction of man $R_M=\frac{3w}{4}$




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, $y=\frac{L}{2}-x$

$\Rightarrow$ $y=\frac{L}{2}-\frac{L}{6}=\frac{2L}{6}=\frac{L}{3}$

Hence, the correct answer is option: 'A'.