A solid sphere is suspended from a wire and is made to execute torsional oscillations with a period of 2 second. The torque required to twist the wire through unit radian is `6 xx10^(-2)` N/m. Calculate the moment of inertia of the sphere about the axis of rotation.

Explanation

To solve the problem, we need to calculate the moment of inertia (I) of a solid sphere executing torsional oscillations. We are given the period (T) of the oscillation and the torsional constant (C) of the wire.

### Step-by-Step Solution:

1. **Identify the given values:**
- Period of oscillation, \( T = 2 \) seconds
- Torsional constant, \( C = 6 \times 10^{-2} \) N/m

2. **Use the formula for the period of torsional oscillation:**
The formula relating the period \( T \) to the moment of inertia \( I \) and the torsional constant \( C \) is:
\[ T = 2\pi \sqrt{\frac{I}{C}} \]

3. **Rearrange the formula to solve for \( I \):**
To isolate \( I \), we can square both sides of the equation:
\[ T^2 = (2\pi)^2 \frac{I}{C} \]
This simplifies to:
\[ I = \frac{C T^2}{(2\pi)^2} \]

4. **Substitute the known values into the equation:**
Substitute \( T = 2 \) seconds and \( C = 6 \times 10^{-2} \) N/m into the equation:
\[ I = \frac{(6 \times 10^{-2}) \times (2)^2}{(2\pi)^2} \]

5. **Calculate \( (2\pi)^2 \):**
\[ (2\pi)^2 = 4\pi^2 \approx 39.478 \]

6. **Calculate \( T^2 \):**
\[ T^2 = (2)^2 = 4 \]

7. **Now substitute these values back into the equation for \( I \):**
\[ I = \frac{(6 \times 10^{-2}) \times 4}{39.478} \]

8. **Perform the multiplication:**
\[ I = \frac{24 \times 10^{-2}}{39.478} = \frac{0.24}{39.478} \]

9. **Calculate the final value of \( I \):**
\[ I \approx 0.00608 \text{ kg m}^2 \]

10. **Convert to scientific notation:**
\[ I \approx 6.08 \times 10^{-3} \text{ kg m}^2 \]

### Final Answer:
The moment of inertia of the sphere about the axis of rotation is approximately \( 6.08 \times 10^{-3} \) kg m².

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