A) \[\frac{3g}{2L}\]
B) \[\frac{g}{L}\]
C) \[\frac{2g}{L}\]
D) \[\frac{2g}{3L}\]
Correct Answer: A
Solution :
Taking torque about P \[Mg=\frac{L}{2}=\left( \frac{M{{L}^{2}}}{3} \right)\alpha \] Hence \[\alpha =\frac{3g}{2L}\]You need to login to perform this action.
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