Question

From a disc of mass \( M \) and radius \( R \), a circular hole of diameter \( R \) is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

• A. \( \frac{11MR^2}{32} \)
• B. \( \frac{7MR^2}{32} \)
• C. \( \frac{9MR^2}{32} \)
• D. \( \frac{13MR^2}{32} \)

Hint:

When removing a part of a symmetrical object like a disc, subtract the moment of inertia of the removed part from the original moment of inertia.

Answer 1

The Correct Option is B

Step 1: Understanding the moment of inertia.
The moment of inertia of a disc is given by \( I_{\text{disc}} = \frac{1}{2} MR^2 \). When a hole is cut from the disc, the moment of inertia of the remaining part can be calculated by subtracting the moment of inertia of the hole from the moment of inertia of the full disc.
Step 2: Moment of inertia of the hole.
The moment of inertia of a circular hole of radius \( R/2 \) (since the diameter of the hole is \( R \)) is given by \( I_{\text{hole}} = \frac{1}{2} M_{\text{hole}} R^2 \). Subtracting this from the original disc's moment of inertia gives the result: \[ I = \frac{7MR^2}{32} \] Step 3: Conclusion.
The correct answer is (B), \( \frac{7MR^2}{32} \).