The area of cross-section of a wire of length 1.1 meter is `1mm^(2)`. It is loaded with 1 kg. if young's modulus of copper is `1.1xx10^(11)N//m^(2)` then the increase in length will be (if `g=10m//s^(2)`)-

Options

A. 0.01mm
B. 0.075mm
C. 0.1mm
D. 0.15mm

Answer

To solve the problem, we will use the formula for the increase in length (ΔL) of a wire under tension, which is derived from Young's modulus (Y). The formula is given by:

\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

Where:
- \( \Delta L \) = increase in length
- \( F \) = force applied on the wire
- \( L \) = original length of the wire
- \( A \) = area of cross-section of the wire
- \( Y \) = Young's modulus of the material

### Step 1: Calculate the Force (F)
The force applied on the wire can be calculated using the weight of the load, which is given by:

\[ F = m \cdot g \]

Where:
- \( m = 1 \, \text{kg} \) (mass)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:

\[ F = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10 \, \text{N} \]

### Step 2: Convert the Area of Cross-Section (A)
The area of cross-section is given as \( 1 \, \text{mm}^2 \). We need to convert this to square meters:

\[ A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Substitute the Values into the Formula
Now we have all the values needed to calculate the increase in length (ΔL). The original length \( L \) is given as \( 1.1 \, \text{m} \), and Young's modulus \( Y \) for copper is given as \( 1.1 \times 10^{11} \, \text{N/m}^2 \).

Substituting these values into the formula:

\[ \Delta L = \frac{F \cdot L}{A \cdot Y} = \frac{10 \, \text{N} \cdot 1.1 \, \text{m}}{1 \times 10^{-6} \, \text{m}^2 \cdot 1.1 \times 10^{11} \, \text{N/m}^2} \]

### Step 4: Simplify the Equation
Now, simplifying the equation:

\[ \Delta L = \frac{11 \, \text{N} \cdot \text{m}}{1.1 \times 10^{5} \, \text{N}} = \frac{11}{1.1 \times 10^{5}} \, \text{m} \]

Calculating this gives:

\[ \Delta L = 1 \times 10^{-4} \, \text{m} = 0.1 \, \text{mm} \]

### Final Answer
The increase in length of the wire is \( 0.1 \, \text{mm} \).

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