Question

From a uniform circular disc of mass M and radius R, part AOC is removed as shown in the figure. The M.I. of the remaining portion of the disc about an axis passing through the centre and perpendicular to the plane of the disc is

• A. \(\frac{MR^{2}}{3}\)
• B. \(\frac{MR^{2}}{9}\)
• C. \(\frac{MR^{2}}{6}\)
• D. \(\frac{MR^{2}}{12}\)

Hint:

The removed sector is one-third of the disc, so the rest has two-thirds of the mass at the same distances from the axis.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The removed sector subtends \(120^{\circ}\), which is one-third of the full disc. The remaining part subtends \(240^{\circ}\).

Step 2: Mass of the remaining part
The remaining mass is \(\dfrac{2M}{3}\). Every element of a sector is at the same distances from the centre as in a full disc.

Step 3: Calculation
A sector of mass \(m\) has \(I=\dfrac{1}{2}mR^{2}\) about the central axis.
\(I=\dfrac{1}{2}\left(\dfrac{2M}{3}\right)R^{2}=\dfrac{MR^{2}}{3}\)

Final Answer:
The moment of inertia of the remaining portion is \(\frac{MR^{2}}{3}\). \[ \boxed{\text{(A) }\dfrac{MR^{2}}{3}} \]