A) \[\frac{2g}{3\sqrt{2}l}\]
B) \[\frac{3g}{4\sqrt{2}l}\]
C) \[\frac{\sqrt{3}g}{2l}\]
D) \[\frac{3g}{2\sqrt{2\,}l}\]
Correct Answer: D
Solution :
\[Mg\cdot \frac{l}{2}\sin \theta =I\alpha \] \[I=\frac{M{{l}^{2}}}{3}\] \[\Rightarrow \] \[\alpha =\frac{mgl\,\sin \theta }{2}\times \frac{3}{m{{l}^{2}}}\] \[=\frac{3g}{2{{l}^{2}}}\sin \theta \] At \[\theta =45{}^\circ ,\]\[\alpha =\frac{3g}{2\sqrt{2}l}\]You need to login to perform this action.
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