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, $\mathrm{n}=$ _______
(Consider all the bodies have equal masses)

Question

Aicharya · Accepted Answer

$\mathrm{I}_1=\frac{\mathrm{MR}_1^2}{4}, \mathrm{I}_2=\frac{\mathrm{MR}_2^2}{2}, \mathrm{I}_3=\frac{2 \mathrm{MR}_1^2}{5}$
According to problem
$\frac{\mathrm{I}_1}{\mathrm{I}_2}=2.5 \Rightarrow \frac{\frac{\mathrm{MR}_1^2}{4}}{\frac{\mathrm{MR}_2^2}{2}}=\frac{5}{2} \Rightarrow \frac{\mathrm{R}_1^2}{\mathrm{R}_2^2}=5 \ldots$
Now we are provided with information that
$\begin{aligned}
& \frac{\mathrm{I}_3}{\mathrm{I}_2}=\mathrm{n} \\ & \Rightarrow \frac{\frac{2 \mathrm{MR}_1^2}{5}}{\frac{\mathrm{MR}_2^2}{2}}=\mathrm{n} \Rightarrow \frac{4 \mathrm{R}_1^2}{5 \mathrm{R}_2^2}=\mathrm{n}
\end{aligned}$
From Eq', (1) and (2)
$\Rightarrow n=4$

The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of…

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