A 94.2 cm long sitar string of 1mm diameter is tuned by stretching it 1.0 cm. Calculate (i) the tension (ii) the work done in stretching the wire (Young's modulus of the material of the wire is `6 xx 10^(10) N//m^(2)` ).

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Area of the Wire
The diameter of the wire is given as 1 mm. First, we need to convert this into meters and then calculate the radius.
- Diameter (d) = 1 mm = 1 × 10^(-3) m
- Radius (r) = d/2 = (1 × 10^(-3) m) / 2 = 0.5 × 10^(-3) m

Now, we can calculate the cross-sectional area (A) of the wire using the formula for the area of a circle:
\[ A = \pi r^2 \]
\[ A = \pi (0.5 \times 10^{-3})^2 \]
\[ A = \pi (0.25 \times 10^{-6}) \]
\[ A \approx 7.85 \times 10^{-7} \, \text{m}^2 \]

### Step 2: Calculate the Tension in the Wire
We can use Young's modulus (Y) to find the tension (T) in the wire. The formula for Young's modulus is:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress = \( \frac{F}{A} \)
- Strain = \( \frac{\Delta L}{L_0} \)

Rearranging the formula to find the force (tension):
\[ F = Y \cdot A \cdot \frac{\Delta L}{L_0} \]

Given:
- Young's modulus (Y) = \( 6 \times 10^{10} \, \text{N/m}^2 \)
- Initial length (L_0) = 94.2 cm = 0.942 m
- Elongation (\(\Delta L\)) = 1 cm = 0.01 m

Now substituting the values:
\[ F = 6 \times 10^{10} \cdot 7.85 \times 10^{-7} \cdot \frac{0.01}{0.942} \]
\[ F \approx 6 \times 10^{10} \cdot 7.85 \times 10^{-7} \cdot 0.01061 \]
\[ F \approx 500 \, \text{N} \]

### Step 3: Calculate the Work Done in Stretching the Wire
The work done (W) in stretching the wire can be calculated using the formula for elastic potential energy:
\[ W = \frac{1}{2} F \Delta L \]

Substituting the values:
\[ W = \frac{1}{2} \cdot 500 \cdot 0.01 \]
\[ W = \frac{1}{2} \cdot 5000 \]
\[ W = 2500 \, \text{J} = 2.5 \, \text{J} \]

### Final Answers:
(i) The tension in the wire is **500 N**.
(ii) The work done in stretching the wire is **2.5 J**.