Question

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I 1. The same rod is bent into a ring and its moment of inertia about a diameter is I 2. If
$\frac{I_1}{I_2}$ is $\frac{xπ²}{3}$
then the value of x will be ______.

Answer 1

The correct answer is 8
$I_1 = \frac{ML^2}{3} ...... (i)$
For ring :
$I_2 = \frac{MR^2}{2}$and 2π R = L
$⇒ I_2 = \frac{M}{2} (\frac{L²}{4π²}) ....... (ii)$
From equation (i) and (ii) , we get
$⇒ \frac{I_1}{I_2} = \frac{8π²}{3}$
⇒ x = 8