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 Young's modulus, which relates stress and strain in a material. The formula is given by:

\[ Y = \frac{\text{Stress}}{\text{Strain}} \]

Where:
- Young's modulus \( Y \) is given as \( 1.1 \times 10^{11} \, \text{N/m}^2 \)
- Stress \( \sigma \) is defined as the force \( F \) applied per unit area \( A \)
- Strain \( \epsilon \) is defined as the change in length \( \Delta L \) per original length \( L_0 \)

### Step 1: Calculate the force applied on the wire
The force \( F \) due to the weight of the load can be calculated using the formula:

\[ F = m \cdot g \]

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

Calculating the force:

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

### Step 2: Convert the area of cross-section from mm² to m²
The area \( A \) 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: Calculate the stress on the wire
Stress \( \sigma \) is calculated using the formula:

\[ \sigma = \frac{F}{A} \]

Substituting the values:

\[ \sigma = \frac{10 \, \text{N}}{1 \times 10^{-6} \, \text{m}^2} = 10^{7} \, \text{N/m}^2 \]

### Step 4: Relate stress to strain using Young's modulus
Using the formula for Young's modulus:

\[ Y = \frac{\sigma}{\epsilon} \]

We can rearrange this to find strain \( \epsilon \):

\[ \epsilon = \frac{\sigma}{Y} \]

Substituting the values we have:

\[ \epsilon = \frac{10^{7} \, \text{N/m}^2}{1.1 \times 10^{11} \, \text{N/m}^2} \]

Calculating \( \epsilon \):

\[ \epsilon = \frac{10^{7}}{1.1 \times 10^{11}} = \frac{10^{7}}{1.1} \times 10^{-11} = 9.09 \times 10^{-5} \]

### Step 5: Calculate the increase in length \( \Delta L \)
The increase in length \( \Delta L \) can be calculated using the formula:

\[ \Delta L = \epsilon \cdot L_0 \]

Where \( L_0 = 1.1 \, \text{m} \):

\[ \Delta L = 9.09 \times 10^{-5} \times 1.1 \]

Calculating \( \Delta L \):

\[ \Delta L = 1.000 \times 10^{-4} \, \text{m} = 0.0001 \, \text{m} \]

### Step 6: Convert \( \Delta L \) from meters to millimeters
To convert meters to millimeters:

\[ \Delta L = 0.0001 \, \text{m} \times 1000 \, \text{mm/m} = 0.1 \, \text{mm} \]

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

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