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 Question:
A large solid sphere is melted and reformed ("casted") into 27 smaller identical spheres. We must find the Moment of Inertia (MOI) of one small sphere in terms of the original sphere's MOI.

Step 2: Detailed Explanation:
Let the large sphere have mass $M$, radius $R$, and MOI $I$.
For a solid sphere, the MOI about its diameter is:
$I = \frac{2}{5} M R^2$
1. Determine the radius of a small sphere ($r$):
Since the total volume is conserved during casting:
$\text{Volume}_{\text{large}} = 27 \times \text{Volume}_{\text{small}}$
$\frac{4}{3} \pi R^3 = 27 \times \left( \frac{4}{3} \pi r^3 \right)$
$R^3 = 27 r^3$
Taking the cube root of both sides:
$R = 3r \implies r = \frac{R}{3}$
2. Determine the mass of a small sphere ($m$):
Since the total mass is conserved and split equally among 27 spheres:
$m = \frac{M}{27}$
3. Calculate the MOI of one small sphere ($i$):
$i = \frac{2}{5} m r^2$
Substitute the expressions for $m$ and $r$ in terms of $M$ and $R$:
$i = \frac{2}{5} \left( \frac{M}{27} \right) \left( \frac{R}{3} \right)^2$
$i = \frac{2}{5} \left( \frac{M}{27} \right) \left( \frac{R^2}{9} \right)$
Pull the fraction denominators out to the front:
$i = \left( \frac{1}{27 \times 9} \right) \times \left( \frac{2}{5} M R^2 \right)$
We recognize that $\left( \frac{2}{5} M R^2 \right)$ is precisely the original MOI $I$:
$i = \left( \frac{1}{243} \right) \times I$
$i = \frac{I}{243}$

Step 3: Final Answer:
The moment of inertia of each small sphere is $I/243$, matching option (d).