Question

Moment of inertia of a solid sphere about its diameter is 'I'. It is then casted into 27 small spheres of same diameter. The moment of inertia of each small sphere about its diameter is ______.

• A. $I/44$
• B. $I/188$
• C. $I/204$
• D. $I/243$

Hint:

Scaling rules for spheres: If cut into $N$ equal spheres, radius scales by $N^{-1/3}$, and mass scales by $N^{-1}$. Since $I \propto M \cdot R^2$, the new MOI is $I' = I \cdot (N^{-1}) \cdot (N^{-1/3})^2 = I \cdot N^{-5/3}$. Here $27^{-5/3} = (3^3)^{-5/3} = 3^{-5} = 1/243$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When a large sphere is cast into 27 small ones, the total volume remains constant. $V_{big} = 27 \times V_{small}$. Also, $M_{big} = 27 \times m_{small}$.

Step 2: Formula Application:
Volume $\propto R^3 \implies R^3 = 27r^3 \implies R = 3r$. Moment of Inertia $I = \frac{2}{5}MR^2$.

Step 3: Explanation:
$I_{small} = \frac{2}{5}m r^2 = \frac{2}{5} (\frac{M}{27}) (\frac{R}{3})^2$ $I_{small} = \frac{2}{5} \frac{M}{27} \frac{R^2}{9} = \frac{1}{27 \times 9} (\frac{2}{5}MR^2)$ $I_{small} = \frac{1}{243} I$.

Step 4: Final Answer:
The moment of inertia of each small sphere is $I/243$.