A) \[\frac{m{{l}^{2}}\omega }{12t}\]
B) \[\frac{m{{l}^{2}}\omega }{3t}\]
C) \[\frac{m{{l}^{2}}\omega }{t}\]
D) \[\frac{4m{{l}^{2}}\omega }{3t}\]
Correct Answer: B
Solution :
Since, \[\tau =I\alpha \] \[\Rightarrow \] \[\tau =\left[ \frac{m\,{{(2l)}^{2}}}{12} \right]\left( \frac{\omega }{t} \right)\] \[\Rightarrow \] \[\tau =\frac{m\times 4{{l}^{2}}\times \omega }{12\times t}\] \[\Rightarrow \] \[\tau =\frac{4m{{l}^{2}}\omega }{12t}=\left( \frac{m{{l}^{2}}\omega }{3t} \right)\]
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