question_answer

A boy and a man carry a uniform rod of length  \[L,\]horizontally in such a way that boy gets\[\frac{\text{1}}{\text{4}}\text{th}\] 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 \[\therefore \] \[\sum{{\vec{\tau }}}=0\] \[{{R}_{B}}\times \frac{L}{2}-{{R}_{M}}\times x=0\] \[\Rightarrow \] \[\frac{w}{4}\times \frac{L}{2}-\frac{3w}{4}\times x=0\] \[\Rightarrow \] \[x=\frac{L}{6}\] \[\therefore \]Distance from other end,\[y=\frac{L}{2}-x\] \[\Rightarrow \] \[y=\frac{L}{2}-\frac{L}{6}=\frac{2L}{6}=\frac{L}{3}\]