Question

Moment of inertia of a disc of mass \(M\) and radius 'R' about any of its diameter is \(MR^2/4\). The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, \((x/2)MR^2\). The value of \(x\) is

• A. \(1\)
• B. \(3\)
• C. \(5\)
• D. \(7\)

Hint:

Use the perpendicular axis theorem, then the parallel axis theorem.

Answer 1

The Correct Option is B

Step 1: Plan:
Apply the parallel axis theorem directly with the known centre value.

Step 2: Steps:
The standard result for a disc about its central normal axis is $\frac12MR^2$. Moving the axis a distance $R$ adds $MR^2$. Total $= \frac32MR^2 = \frac{3}{2}MR^2$, so $x = 3$.

Final Answer:
The value of $x$ is $3$, option (B). \[ \boxed{3} \]