Question

A rigid spherical body is spinning around an axis without any external torque. Due to rise of temperature its volume increases by 3% . Then percentage change in its angular speed is

• A. -2
• B. -1
• C. -3
• D. 1

Answer 1

The Correct Option is A

Volume increases by 3% means, its radius increases by 1% Therefore I increases and $\omega$ decreases $\omega \,\propto\,\frac{1}{I}\Rightarrow\,\propto\,\frac{1}{R^2}$ $\therefore \,change\,in\,\omega\,is\,-2%$

Answer 2

Volume Increase and Radius Relationship:
The volume of the sphere increases by 3%. 
Volume \(V \propto R^3\).
If volume increases by 3%, \(V' = 1.03 V\).
Radius \(R \propto V^{1/3}\)
Therefore, a new radius \(R' = (1.03)^{1/3} R \approx 1.01 R\).

Moment of Inertia and Radius Relationship:
Moment of inertia \(I \propto R^2\).
If the radius increases by 1%, \(R' = 1.01 R\).
A new moment of inertia \(I' \propto (1.01 R)^2 = 1.01^2 I \approx 1.02 I\).

Angular Speed and Moment of Inertia Relationship:
Angular momentum L is conserved:
\(L = I \omega = I' \omega'\)
\(I \omega = 1.02 I \omega'\)
\(\omega' = \frac{\omega}{1.02}\)

Percentage Change in Angular Speed:
Calculate the percentage change:
\(\text{Percentage change} = \left( \frac{\omega' - \omega}{\omega} \right) \times 100\%\)

\(\text{Percentage change} = \left( \frac{\frac{\omega}{1.02} - \omega}{\omega} \right) \times 100\%\)

\(\text{Percentage change} = \left( \frac{\omega - 1.02 \omega}{\omega \cdot 1.02} \right) \times 100\%\)

\(\text{Percentage change} = \left( \frac{-0.02 \omega}{\omega \cdot 1.02} \right) \times 100\%\)

\(\text{Percentage change} = -\frac{0.02}{1.02} \times 100\% \approx -1.96\%\)
So, the angular speed decreases by approximately 2%.

So, the correct option is (A): 2%.