Question

Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is '$I_1$'. The same rod is bent into a ring and its moment of inertia about the diameter is '$I_2$'. Then $I_1/I_2$ is ______.

• A. $\frac{3\pi^2}{2}$
• B. $\frac{2\pi^2}{3}$
• C. $\frac{\pi^2}{3}$
• D. $\frac{\pi^2}{9}$

Hint:

Read carefully! It asks for the MOI of the ring about its diameter ($\frac{1}{2}MR^2$), NOT its central axis ($MR^2$). This factor of $1/2$ is the most common trap in this classic problem.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
$I_1 = \frac{ML^2}{12}$ for a rod. For a ring, the length $L$ becomes the circumference $2\pi R$.

Step 2: Formula Application:
$L = 2\pi R \implies R = \frac{L}{2\pi}$.
$I_2$ (diameter) $= \frac{1}{2} MR^2 = \frac{1}{2} M \left( \frac{L}{2\pi} \right)^2 = \frac{ML^2}{8\pi^2}$.

Step 3: Explanation:
$\frac{I_1}{I_2} = \frac{ML^2/12}{ML^2/8\pi^2} = \frac{8\pi^2}{12} = \frac{2\pi^2}{3}$.

Step 4: Final Answer:
The ratio is $\frac{2\pi^2}{3}$.