MHT CET · Physics · Rotational Motion
A solid sphere of mass \(M\), radius \(R\) has moment of inertia ' \(I\) ' about its diameter. It is recast into a disc of thickness ' \(t\) ' whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains ' \(I\) '. Radius of the disc will be
- A \(\frac{4 \mathrm{R}}{\sqrt{11}}\)
- B \(\frac{3 R}{4}\)
- C \(\frac{2 \mathrm{R}}{\sqrt{15}}\)
- D \(\frac{2 R}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \mathrm{R}}{\sqrt{15}}\)
Step-by-step Solution
Detailed explanation
For solid sphere, \(\mathrm{I}=\frac{2}{5} \mathrm{MR}^2\)
For disc, \(\mathrm{I}^{\prime}=\frac{\mathrm{MR}^{\prime 2}}{2}+\mathrm{MR}^{\prime 2}=\frac{3}{2} \mathrm{MR}^{\prime 2}\)
\(\begin{aligned}& \because I^{\prime}=I \\& \therefore \frac{3}{2} M R^{\prime 2}=\frac{2}{5} M^2 \\& R^{\prime}=\frac{2}{\sqrt{15}} R\end{aligned}\)
For disc, \(\mathrm{I}^{\prime}=\frac{\mathrm{MR}^{\prime 2}}{2}+\mathrm{MR}^{\prime 2}=\frac{3}{2} \mathrm{MR}^{\prime 2}\)
\(\begin{aligned}& \because I^{\prime}=I \\& \therefore \frac{3}{2} M R^{\prime 2}=\frac{2}{5} M^2 \\& R^{\prime}=\frac{2}{\sqrt{15}} R\end{aligned}\)
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