A) \[\frac{3g}{2L}\]
B) \[\frac{g}{L}\]
C) \[\frac{2g}{L}\]
D) \[\frac{2g}{3L}\]
Correct Answer: A
Solution :
Torque on the rod=Moment of weight of the rod about P \[\tau =mg\frac{L}{2}\] ?(i) \[\because \]Moment of inertia of rod about \[P=\frac{M{{L}^{2}}}{3}\] ?(ii) As \[\tau =la\] From Eqs. (i) and (ii), we get \[Mg\frac{L}{2}=\frac{M{{L}^{2}}}{3}\alpha \] \[\therefore \] \[\alpha =\frac{3g}{2L}\]You need to login to perform this action.
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