The moment of inertia of a uniform rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I2. Then I_1/I_2 is

Q: The moment of inertia of a uniform rod about a perpendicular axis passing through one end is I1 . The same rod is bent into a ring and its moment of inertia about a diameter is I2 . Then I1/I2 is

(a) $\frac{\pi^2}{3}$

(b) $\frac{2 \pi^2}{3}$

(c) $\frac{4 \pi^2}{3}$

(d) $\frac{8 \pi^2}{3}$

Ans: (d)

Sol: Let mass of the rod is M & Length of rod is L .

Moment of Inertia of rod about a perpendicular axis passing through one end will be

$I_1 = \frac{ML^2}{3}$ …(i)

Suppose radius of ring = R

Then , L = 2 π R

R = L/2 π

Moment of Inertia of ring about a perpendicular axis passing through its center will be $ M R^2 $

Moment of inertia about its diameter $ I_2 = \frac{MR^2}{2}$ (By using Perpendicular axis Theorem )

$ I_2 = \frac{M}{2} (\frac{L}{2\pi})^2 = \frac{ML^2}{8 \pi^2} $ …(ii)

$\frac{I_1}{I_2} = \frac{8 \pi^2}{3}$