A) \[L=R\]
B) \[L=\sqrt{3}R\]
C) \[L=\frac{R}{\sqrt{3}}\]
D) \[L=\sqrt{\frac{3}{2}}R\]
Correct Answer: B
Solution :
\[MI\]of a cylinder about its centre and parallel to its length \[=\frac{M{{R}^{2}}}{2}\] \[MI\]about its centre and perpendicular to its length \[=M\left( \frac{{{l}^{2}}}{12}+\frac{{{R}^{2}}}{4} \right)\] According to problem, \[\frac{M{{L}^{2}}}{12}+\frac{M{{R}^{2}}}{4}=\frac{M{{R}^{2}}}{2}\] Solving these \[L=\sqrt{3}R\]You need to login to perform this action.
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