JEE Advanced · Physics · 8. Rotational Motion
A solid sphere of radius \(R\) has moment of inertia \(J\) about its qeometrical axis. If it is melted into a disc of radius \(r\) and thickness \(t\). If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to \(I\), then the value of \(r\) is equal to
- A \(\frac{2}{\sqrt{15}} R\)
- B \(\frac{2}{\sqrt{5}} R\)
- C \(\frac{3}{\sqrt{15}} R\)
- D \(\frac{\sqrt{3}}{\sqrt{15}} R\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{\sqrt{15}} R\)
Step-by-step Solution
Detailed explanation
\(\frac{2}{3} M R^2=\frac{1}{2} M r^2+M r^2\) or \(\frac{2}{3} M R^2=\frac{3}{2} M r^2 \Rightarrow r=\frac{2}{\sqrt{15}} R\)
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