Speed of a transverse wave on a straight wire (mass 6 g, length 120 cm and area of cross section `1.2 mm^(2)` is 100 m/s) . If the Young's modulus of wire is `10^(12) Nm^(-2)` the extension of wire over its natural length is :

Options

A. 0.05 mm
B. 0.04mm
C. 0.03mm
D. 0.02mm

Explanation

To find the extension of the wire over its natural length, we can use the relationship between tension, Young's modulus, and the extension of the wire. Let's break down the solution step by step.

### Step 1: Understand the given parameters
- Mass of the wire (m) = 6 g = \(6 \times 10^{-3}\) kg
- Length of the wire (L) = 120 cm = 1.2 m
- Area of cross-section (A) = 1.2 mm² = \(1.2 \times 10^{-6}\) m²
- Speed of the wave (v) = 100 m/s
- Young's modulus (Y) = \(10^{12}\) N/m²

### Step 2: Calculate the mass per unit length (μ)
The mass per unit length (μ) can be calculated using the formula:
\[ \mu = \frac{m}{L} \]
Substituting the values:
\[ \mu = \frac{6 \times 10^{-3} \text{ kg}}{1.2 \text{ m}} = 5 \times 10^{-3} \text{ kg/m} \]

### Step 3: Relate wave speed to tension (T)
The speed of a transverse wave on a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Squaring both sides gives:
\[ v^2 = \frac{T}{\mu} \implies T = \mu v^2 \]
Substituting the values:
\[ T = (5 \times 10^{-3} \text{ kg/m}) \times (100 \text{ m/s})^2 = (5 \times 10^{-3}) \times 10^4 = 50 \text{ N} \]

### Step 4: Calculate the extension (ΔL)
Using the formula for extension in terms of tension, area, length, and Young's modulus:
\[ \Delta L = \frac{T \cdot L}{A \cdot Y} \]
Substituting the values:
\[ \Delta L = \frac{50 \text{ N} \cdot 1.2 \text{ m}}{1.2 \times 10^{-6} \text{ m}^2 \cdot 10^{12} \text{ N/m}^2} \]
Calculating the denominator:
\[ 1.2 \times 10^{-6} \cdot 10^{12} = 1.2 \times 10^6 \]
Now substituting back:
\[ \Delta L = \frac{60}{1.2 \times 10^6} = 5 \times 10^{-5} \text{ m} \]

### Step 5: Convert extension to mm
To convert meters to millimeters:
\[ \Delta L = 5 \times 10^{-5} \text{ m} = 0.05 \text{ mm} \]

### Final Answer
The extension of the wire over its natural length is **0.05 mm**.

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