A) \[\frac{3g}{2l}\]
B) \[\frac{2l}{3g}\]
C) \[\frac{3g}{2{{l}^{2}}}\]
D) \[mg\frac{l}{2}\]
Correct Answer: A
Solution :
The moment of inertia of the uniform rod about an axis through one end and perpendicular to length is \[I=\frac{m{{l}^{2}}}{3}\] where m is mass of rod and \[l\] its 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 \[\frac{m{{l}^{2}}}{3}\alpha =mg\frac{l}{2}\] \[\therefore \] \[\alpha =\frac{3g}{2l}\]You need to login to perform this action.
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