Question

A solid sphere is rotating about a diameter at an angular velocity \(\omega\). If it cools so that its radius reduces to \(\left(\frac{1}{n}\right)^{\text{th}}\) of its original value, its angular velocity becomes

• A. \(\frac{\omega }{n}\)
• B. \(\frac{\omega }{n^{2}}\)
• C. \(n\omega\)
• D. \(n^{2}\omega\)

Hint:

No external torque acts, so angular momentum is conserved; the moment of inertia of a solid sphere is proportional to the square of its radius.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
With no external torque, angular momentum \(L=I\omega\) stays constant.

Step 2: Moment of Inertia
For a solid sphere \(I=\dfrac{2}{5}MR^{2}\), so \(I\propto R^{2}\).

Step 3: Calculation
When the radius becomes \(\dfrac{R}{n}\), \(I'=\dfrac{I}{n^{2}}\).
From \(I\omega=I'\omega'\) we get \(\omega'=\dfrac{I}{I'}\omega=n^{2}\omega\).

Final Answer:
The angular velocity becomes n^2 times the original value. \[ \boxed{\text{(D) }n^{2}\omega} \]