A) \[\frac{2g}{3l}\]
B) \[mg\frac{l}{2}\]
C) \[\frac{3}{2}gl\]
D) \[\frac{3g}{2l}\]
Correct Answer: D
Solution :
The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is \[I=\frac{m{{l}^{2}}}{3}\] where m is mass of rod and I is length. Torque \[(\tau =I\alpha )\] acting on centre of gravity of rod is given by \[\tau =mg\frac{l}{2}\] or \[I\alpha =mg\frac{l}{2}\] or \[\frac{m{{l}^{2}}}{3}\alpha =mg\frac{l}{2}\] or \[\alpha =\frac{3g}{2l}\]
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