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,\[y=\frac{L}{2}-x\] \[\Rightarrow \] \[y=\frac{L}{2}-\frac{L}{6}=\frac{2L}{6}=\frac{L}{3}\]You need to login to perform this action.
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