Practice portal › Rotational Motion › Moment of Inertia

The moment of inertia of a uniform thin 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. If (I₁)/(I₂)=(xπ²)/3, then the value of x will be ______.

Asked in JEE Main 29th June 2nd Shift 2022 · Standard bodies and axis theorems

Answer: 8

Step-by-step solution

Let the rod have mass M and length L.

Rod about a perpendicular axis through one end: I₁=(ML²)/3.

Bend it into a ring. The length becomes the circumference: L=2π r, so r=L/(2π).

Ring about a diameter: I₂=1/2Mr²=1/2M(L²)/(4π²)=(ML²)/(8π²).

(I₁)/(I₂)=(ML²/3)/(ML²/(8π²))=(8π²)/3.

x=8.

More Moment of Inertia questionsAll Moment of Inertia questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer