Question

A Solid sphere has mass M and radius R. Its moment of inertia about a parallel axis passing through a point at a distance R/3 from its centre is

• A. \(8\text{MR}^2/11\)
• B. \(11\text{MR}^2/15\)
• C. \(23\text{MR}^2/45\)
• D. \(13\text{MR}^2/20\)

Hint:

Use the parallel axis theorem with I_cm = 2MR^2/5.

Answer 1

The Correct Option is C

Step 1: Write in decimals:
$I_{cm} = 0.4MR^2$ and $Md^2 = \tfrac19MR^2 = 0.111MR^2$.

Step 2: Add:
$0.4 + 0.111 = 0.511$, and $\tfrac{23}{45} = 0.511$.

Step 3: Match:
So the moment of inertia is $\tfrac{23}{45}MR^2$, option (C).

Final Answer:
The moment of inertia is 23 MR^2 / 45. \[ \boxed{\text{(C) }\dfrac{23}{45}MR^2} \]