MHT CET · Physics · Rotational Motion
A solid sphere of mass ' \(m\) ', radius ' \(R\) ', having moment of inertia about an axis passing through center of mass as ' \(I\) ' is recast into a disc of thickness ' \(t\) ' whose moment of inertia about an axis passing through the rim (edge) \& perpendicular to plane remains ' \(I\) '. Then the radius of disc is
- A \(\frac{2 \mathrm{R}}{\sqrt{15}}\)
- B \(\left(\sqrt{\frac{2}{15}}\right) \mathrm{R}\)
- C \(\frac{4 \mathrm{R}}{\sqrt{15}}\)
- D \(\frac{\mathrm{R}}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{2 \mathrm{R}}{\sqrt{15}}\)
Step-by-step Solution
Detailed explanation
The moment of inertia of a solid sphere
\(=\frac{2}{5} \mathrm{MR}^2\)
Moment of inertia of a disc through its rim,
\(=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2\)
Since both the moment of inertias are equal,
\(\therefore \quad \frac{2}{5} \mathrm{MR}^2=\frac{3}{2} \mathrm{Mr}^2\), where r is the radius of the disc
\(\therefore \quad r=\frac{2 R}{\sqrt{15}}\)
\(=\frac{2}{5} \mathrm{MR}^2\)
Moment of inertia of a disc through its rim,
\(=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2\)
Since both the moment of inertias are equal,
\(\therefore \quad \frac{2}{5} \mathrm{MR}^2=\frac{3}{2} \mathrm{Mr}^2\), where r is the radius of the disc
\(\therefore \quad r=\frac{2 R}{\sqrt{15}}\)
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