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: Understanding the Concept:
The moment of inertia of a solid sphere about an axis through its centre is \(\dfrac25MR^2\). For a parallel axis at distance \(d\) from the centre, \(I = I_{cm} + Md^2\).

Step 2: Apply the theorem:
With \(d = \dfrac R3\):
\[ I = \frac25MR^2 + M\left(\frac R3\right)^2 = \frac25MR^2 + \frac19MR^2 \]

Step 3: Add the fractions:
\[ \frac25 + \frac19 = \frac{18 + 5}{45} = \frac{23}{45} \]
\[ I = \frac{23}{45}MR^2 \]

Step 4: Check:
Option (C), \(\dfrac{23}{45}MR^2\), is the value that results.

Final Answer:
I = 2/5 MR^2 + M(R/3)^2 = 23 MR^2 / 45. \[ \boxed{\text{(C) }\dfrac{23}{45}MR^2} \]