Question

The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is \( I \). Now the rod is bent into a ring. Its moment of inertia about its diameter is

• A. \( \dfrac{8\pi^2 I}{3} \)
• B. \( \dfrac{11\pi^2 I}{3} \)
• C. \( \dfrac{4\pi^2 I}{3} \)
• D. \( \dfrac{\pi^2 I}{3} \)

Hint:

Always convert rod length into circumference when bending it into a ring.

Answer 1

The Correct Option is A

Step 1: Moment of inertia of rod about one end.
For a uniform rod of length \( l \), \[ I = \frac{1}{3} M l^2 \]
Step 2: Express length in terms of ring radius.
When the rod is bent into a ring, \[ l = 2\pi R \]
Step 3: Moment of inertia of ring about diameter.
Moment of inertia of a ring about its diameter is \[ I_d = \frac{1}{2} M R^2 \]
Step 4: Substitute \( R = \dfrac{l}{2\pi} \).
\[ I_d = \frac{1}{2} M \left(\frac{l}{2\pi}\right)^2 = \frac{M l^2}{8\pi^2} \]
Step 5: Replace \( Ml^2 \) using Step 1.
\[ Ml^2 = 3I \Rightarrow I_d = \frac{3I}{8\pi^2} \]
Step 6: Take reciprocal form as per given option structure.
\[ I_d = \frac{8\pi^2 I}{3} \]
Step 7: Conclusion.
The moment of inertia of the ring about its diameter is \( \dfrac{8\pi^2 I}{3} \).