The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is n times higher than the moment of inertia of the given ring. Here, n =_______.
Consider all the bodies have equal masses.

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answer is 4.

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Detailed Solution

$I_{1} = \frac{{MR}_{1}^{2}}{4}, I_{2} = \frac{{MR}_{2}^{2}}{2}, I_{3} = \frac{2 {MR}_{1}^{2}}{5}$

According to problem
$\frac{I_{1}}{I_{2}} = 2.5 \Rightarrow \frac{\frac{{MR}_{1}^{2}}{4}}{\frac{{MR}_{2}^{2}}{2}} = \frac{5}{2} \Rightarrow \frac{R_{1}^{2}}{R_{2}^{2}} = 5 \dots$ (1)
Now we are provided with information that
$\begin{aligned} \frac{I_{3}}{I_{2}} = n \\ \Rightarrow \frac{\frac{2 {MR}_{1}^{2}}{5}}{\frac{{MR}_{2}^{2}}{2}} = n \Rightarrow \frac{4 R_{1}^{2}}{5 R_{2}^{2}} = n \end{aligned}$

From Eq’, (1) and (2)

$\Rightarrow n = 4$

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