Question

Two rings of same mass \( M \) and radius \( R \) are so placed that their centre is common and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to any one ring is

• A. \( \frac{3MR^2}{2} \)
• B. \( \frac{MR^2}{2} \)
• C. \( \frac{2MR^2}{3} \)
• D. \( MR^2 \)

Hint:

For a system of two perpendicular rings, their total moment of inertia is the sum of their individual moments of inertia.

Answer 1

The Correct Option is A

Step 1: Understanding the concept of moment of inertia.
For two rings with their planes perpendicular to each other, the total moment of inertia about an axis passing through their center is the sum of the individual moments of inertia of each ring. For each ring, the moment of inertia about the center is: \[ I = MR^2 \] Step 2: Using the perpendicular axis theorem.
By the perpendicular axis theorem, the total moment of inertia for two perpendicular rings is: \[ I_{\text{total}} = I_1 + I_2 = \frac{MR^2}{2} + \frac{MR^2}{2} = \frac{3MR^2}{2} \] Thus, the correct answer is (A) \( \frac{3MR^2}{2} \).