A uniform circular disc of moment of inertia `7.35 xx 10^(-3)kg m^(2)` is suspended by a wire of length 1 m and executes torsional oscillations. The couple per unit twist of the wire is `2 xx 10^(-2)` N/m. Calculate the period of oscillation ?

Explanation

To calculate the period of oscillation for the given uniform circular disc executing torsional oscillations, we can use the formula for the period \( T \) of torsional oscillations:

\[ T = 2\pi \sqrt{\frac{I}{C}} \]

where:
- \( I \) is the moment of inertia of the disc,
- \( C \) is the couple per unit twist of the wire.

### Step-by-step Solution:

1. **Identify the given values:**
- Moment of inertia \( I = 7.35 \times 10^{-3} \, \text{kg m}^2 \)
- Couple per unit twist \( C = 2 \times 10^{-2} \, \text{N/m} \)

2. **Substitute the values into the formula:**
\[ T = 2\pi \sqrt{\frac{7.35 \times 10^{-3}}{2 \times 10^{-2}}} \]

3. **Calculate the fraction inside the square root:**
\[ \frac{7.35 \times 10^{-3}}{2 \times 10^{-2}} = \frac{7.35}{2} \times 10^{-1} = 3.675 \times 10^{-1} \]

4. **Take the square root:**
\[ \sqrt{3.675 \times 10^{-1}} \approx 0.607 \, \text{(approximately)} \]

5. **Multiply by \( 2\pi \):**
\[ T \approx 2\pi \times 0.607 \approx 3.81 \, \text{seconds (approximately)} \]

6. **Final Calculation:**
\[ T \approx 3.81 \, \text{seconds} \]

### Final Answer:
The period of oscillation \( T \) is approximately **3.81 seconds**.