question_answer

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)  \[\frac{3L}{4}\]

Correct Answer: A

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 \[\therefore \]  \[\Sigma \overset{\to }{\mathop{\tau }}\,=0\]                 \[{{R}_{B}}\times \frac{L}{2}-{{R}_{M}}\times x=0\] \[\Rightarrow \]               \[\frac{w}{4}\times \frac{L}{2}-\frac{3w}{4}\times x=0\] \[\Rightarrow \]                                               \[x=\frac{L}{6}\] \[\therefore \]Distance from other end,\[y=\frac{L}{2}-x\] \[\Rightarrow \]               \[y=\frac{L}{2}-\frac{L}{6}=\frac{2L}{6}=\frac{L}{3}\]