A) 25
B) 5/4
C) 5
D) 1/4
Correct Answer: C
Solution :
From the formula, rotational K.E. \[{{E}_{rotational}}=\frac{1}{2}I{{\omega }^{2}}=\frac{{{L}^{2}}}{2I}\] \[(\because L=I\omega )\] Therefore, \[{{L}^{2}}=2EI\] Hence \[L_{A}^{2}=2{{E}_{A}}{{I}_{A}}\] ?(i) and \[L_{B}^{2}=2{{E}_{B}}{{I}_{B}}\] ?(ii) From equation (i) and (ii) \[\frac{{{L}_{A}}}{{{L}_{B}}}=\sqrt{\frac{2{{E}_{A}}{{I}_{B}}}{2{{E}_{B}}{{I}_{A}}}}=\sqrt{\frac{100}{4}}=5\] \[\left( \begin{align} & \because \,\,\,\,\,\,\,\,\frac{{{E}_{B}}}{{{E}_{A}}}=4 \\ & \text{and}\,\,\,\frac{{{I}_{A}}}{{{I}_{B}}}=100 \\ \end{align} \right)\]You need to login to perform this action.
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