The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is $I_{1}$ . The same rod is bent into a ring and its moment of inertia about a diameter is $I_{2}$ . If $\frac{I_{1}}{I_{2}}$ is $\frac{x π^{2}}{3}$ , then the value of x will be _________.

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answer is 8.

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Detailed Solution

$I_{1} = \frac{{mL}^{2}}{3}$ ;

$L = 2 πr \Rightarrow \frac{L}{r} = 2 π$ ;

$I_{2} = \frac{{mr}^{2}}{2}$ ;

$\begin{array}{l} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{2}{3} {(\frac{L}{r})}^{2} \\ = \frac{2}{3} \times 4 π^{2} = \frac{8 π^{2}}{3} \\ \Rightarrow x = 8 \end{array}$

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