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 1: Identify the given values
- Mass of the wire (m) = 6.0 g = 6.0 × 10^(-3) kg
- Length of the wire (L) = 60 cm = 0.60 m
- Area of cross-section (A) = 1.0 mm² = 1.0 × 10^(-6) m²
- Speed of the transverse wave (v) = 90 m/s
- Young's modulus (Y) = 16 × 10^(11) N/m²

### Step 2: Calculate the mass per unit length (μ)
The mass per unit length (μ) is given by:
\[ \mu = \frac{m}{L} \]
Substituting the values:
\[ \mu = \frac{6.0 \times 10^{-3} \text{ kg}}{0.60 \text{ m}} = 0.01 \text{ kg/m} \]

### Step 3: Calculate the tension (T) in the wire
Using the formula for wave speed in a string:
\[ v = \sqrt{\frac{T}{\mu}} \]
We can rearrange it to find T:
\[ T = \mu v^2 \]
Substituting the values:
\[ T = 0.01 \text{ kg/m} \times (90 \text{ m/s})^2 = 0.01 \times 8100 = 81 \text{ N} \]

### Step 4: Use Young's modulus to find the extension (ΔL)
The formula relating Young's modulus, stress, and strain is:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\Delta L / L} \]
Rearranging gives us:
\[ \Delta L = \frac{T \cdot L}{A \cdot Y} \]
Substituting the values:
\[ \Delta L = \frac{81 \text{ N} \cdot 0.60 \text{ m}}{1.0 \times 10^{-6} \text{ m}^2 \cdot 16 \times 10^{11} \text{ N/m}^2} \]

### Step 5: Calculate ΔL
Calculating the denominator:
\[ A \cdot Y = 1.0 \times 10^{-6} \cdot 16 \times 10^{11} = 16 \times 10^{5} = 1.6 \times 10^{6} \]
Now substituting back into the equation for ΔL:
\[ \Delta L = \frac{81 \cdot 0.60}{1.6 \times 10^{6}} = \frac{48.6}{1.6 \times 10^{6}} \approx 3.0375 \times 10^{-5} \text{ m} \]

### Step 6: Convert ΔL to mm
To convert from meters to millimeters:
\[ \Delta L \approx 3.0375 \times 10^{-5} \text{ m} \times 1000 \approx 0.030375 \text{ mm} \]

### Final Answer
The extension of the wire over its natural length is approximately:
\[ \Delta L \approx 0.030375 \text{ mm} \approx 0.03 \text{ mm} \]