Speed of a transverse wave on a straight wire ( mass 6.0 g, length 60 cm and area of cross - section 1.0 ` mm^(2 ) ` ) is ` 90 ms ^( - 1 ) `. If the Young's modulus of wire is ` 16 xx 10 ^( 11) Nm ^( - 2 ) ` , the extension of wire over its natural length is :

Options

A. `0.04 ` mm
B. ` 0.03 ` mm
C. ` 0.01 ` mm
D. `0.02 `mm

Answer

To find the extension of the wire over its natural length, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Given Values to SI Units**:
- Mass of the wire, \( m = 6 \, \text{g} = 6 \times 10^{-3} \, \text{kg} \)
- Length of the wire, \( L = 60 \, \text{cm} = 60 \times 10^{-2} \, \text{m} = 0.6 \, \text{m} \)
- Area of cross-section, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Speed of the wave, \( v = 90 \, \text{m/s} \)
- Young's modulus, \( Y = 16 \times 10^{11} \, \text{N/m}^2 \)

2. **Calculate the Linear Mass Density (\( \mu \))**:
\[ \mu = \frac{m}{L} = \frac{6 \times 10^{-3} \, \text{kg}}{0.6 \, \text{m}} = 1 \times 10^{-2} \, \text{kg/m} \]

3. **Relate Wave Speed to Tension and Linear Mass Density**:
The speed of a transverse wave on a string is given by:
\[ v = \sqrt{\frac{F}{\mu}} \]
Rearranging this gives:
\[ F = v^2 \cdot \mu \]
Substituting the known values:
\[ F = (90 \, \text{m/s})^2 \cdot (1 \times 10^{-2} \, \text{kg/m}) = 8100 \, \text{N} \]

4. **Calculate the Extension (\( \Delta L \)) Using Young's Modulus**:
Young's modulus is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Rearranging this gives:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
Substituting the values:
\[ \Delta L = \frac{8100 \, \text{N} \cdot 0.6 \, \text{m}}{1 \times 10^{-6} \, \text{m}^2 \cdot 16 \times 10^{11} \, \text{N/m}^2} \]

5. **Calculate \( \Delta L \)**:
\[ \Delta L = \frac{4860}{16 \times 10^5} = \frac{4860}{1600000} = 0.0030375 \, \text{m} = 30.375 \times 10^{-6} \, \text{m} = 0.030375 \, \text{mm} \]

6. **Final Answer**:
The extension of the wire over its natural length is approximately:
\[ \Delta L \approx 0.03 \, \text{mm} \]