Question

A thin metal wire of length ' L ' and mass ' M ' is bent to form semicircular ring as shown. The moment of inertia about XX' is

• A. \( \frac{ML^2}{4\pi^2} \)
• B. \( \frac{2ML^2}{\pi^2} \)
• C. \( \frac{ML^2}{2\pi^2} \)
• D. \( \frac{ML^2}{\pi^2} \)

Hint:

Moment of inertia for a ring about its diameter is half of that about its central transverse axis ($MR^2$).

Answer 1

The Correct Option is C

Step 1: Find the Radius
The length of the wire $L$ forms a semicircle, so $L = \pi R \implies R = \frac{L}{\pi}$.
Step 2: Moment of Inertia Formula
The axis XX' passes through the center and is in the plane of the ring (diameter). For a ring, $I_{diameter} = \frac{1}{2} MR^2$.
Step 3: Calculation
$I = \frac{1}{2} M \left( \frac{L}{\pi} \right)^2 = \frac{ML^2}{2\pi^2}$.
Final Answer: (C)