Question

Moment of inertia of a rod about an axis passing through its 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 the ratio \( \dfrac{I_2}{I_1} \) is

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

Hint:

When a rod is bent into a ring, always equate its length to the circumference of the ring.

Answer 1

The Correct Option is D

Step 1: Moment of inertia of the rod.
For a uniform rod of length \(L\) and mass \(m\), about an axis through its centre and perpendicular to its length,
\[ I_1 = \frac{1}{12} mL^2 \]
Step 2: Converting the rod into a ring.
When the rod is bent into a ring, its circumference equals the length of the rod:
\[ 2\pi R = L \Rightarrow R = \frac{L}{2\pi} \]
Step 3: Moment of inertia of the ring.
Moment of inertia of a ring about its diameter is:
\[ I_2 = \frac{1}{2} mR^2 \]
Step 4: Substituting value of \(R\).
\[ I_2 = \frac{1}{2} m \left(\frac{L}{2\pi}\right)^2 = \frac{mL^2}{8\pi^2} \]
Step 5: Taking the ratio.
\[ \frac{I_2}{I_1} = \frac{\frac{mL^2}{8\pi^2}}{\frac{1}{12}mL^2} = \frac{12}{8\pi^2} = \frac{3}{2\pi^2} \]
Step 6: Conclusion.
\[ \boxed{\dfrac{I_2}{I_1} = \dfrac{3}{2\pi^2}} \]