Question

If \(I_1\) is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and \(I_2\) is the moment of inertia of the ring about an axis perpendicular to the plane of the ring and passing through its centre, formed by bending the rod, then:

• A. \(\dfrac{I_1}{I_2} = \dfrac{3}{\pi^2}\)
• B. \(\dfrac{I_1}{I_2} = \dfrac{2}{\pi^2}\)
• C. \(\dfrac{I_1}{I_2} = \dfrac{\pi^2}{2}\)
• D. \(\dfrac{I_1}{I_2} = \dfrac{\pi^2}{3}\)

Hint:

Key formulas to remember:
Rod (about centre, perpendicular): \(I = \dfrac{1}{12}ML^2\)
Ring (about centre, perpendicular): \(I = MR^2\)
When a rod is bent into a ring: \(L = 2\pi R\) Always convert geometry correctly before comparing moments of inertia.

Answer 1

The Correct Option is D

Step 1: Moment of inertia of the thin rod. Let the mass of the rod be \(M\) and its length be \(L\). The moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass is: \[ I_1 = \frac{1}{12} M L^2 \]
Step 2: Geometry of the ring formed by bending the rod. When the rod is bent into a ring: \[ \text{Circumference of ring} = L = 2\pi R \] \[ \Rightarrow R = \frac{L}{2\pi} \]
Step 3: Moment of inertia of the ring. The moment of inertia of a ring about an axis perpendicular to its plane and passing through its centre is: \[ I_2 = M R^2 \] Substitute \(R = \dfrac{L}{2\pi}\): \[ I_2 = M \left(\frac{L}{2\pi}\right)^2 = \frac{M L^2}{4\pi^2} \]
Step 4: Find the ratio \( \dfrac{I_1}{I_2} \). \[ \frac{I_1}{I_2} = \frac{\dfrac{1}{12} M L^2}{\dfrac{M L^2}{4\pi^2}} \] Cancel \(M L^2\): \[ \frac{I_1}{I_2} = \frac{1}{12} \times 4\pi^2 = \frac{\pi^2}{3} \] Final Answer: \[ \boxed{\dfrac{I_1}{I_2} = \dfrac{\pi^2}{3}} \]