A force of 200 N is applied at one end of a wire of length 2m and having area of cross-section `10^(-2) cm^(2)`. The other end of the wire is rigidly fixed. If coefficient of linear expansion of the wire `alpha=8xx10^(-6)//.^(@)C` and Young's modulus `Y=2.2xx10^(11)N//m^(2)` and its temperature is increased by `5^(@)C`, then the increase in the tension of the wire will be

Options

A. 4.2 N
B. 4.4 N
C. 2.4 N
D. 8.8 N

Explanation

To solve the problem, we need to calculate the increase in tension in the wire due to the applied force and the increase in temperature. We will use the relationship between Young's modulus, stress, strain, and thermal expansion.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Force (F) = 200 N
- Length of wire (L₀) = 2 m
- Area of cross-section (A) = 10^(-2) cm² = 10^(-2) × 10^(-4) m² = 10^(-6) m²
- Coefficient of linear expansion (α) = 8 × 10^(-6) /°C
- Young's modulus (Y) = 2.2 × 10^(11) N/m²
- Change in temperature (ΔT) = 5°C

2. **Calculate the Change in Length (ΔL) due to Thermal Expansion:**
The change in length due to thermal expansion can be calculated using the formula:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]
Substituting the values:
\[ \Delta L = 2 \, \text{m} \cdot (8 \times 10^{-6} \, /°C) \cdot (5 \, °C) = 2 \cdot 8 \times 10^{-6} \cdot 5 = 8 \times 10^{-5} \, \text{m} \]

3. **Calculate the Stress (σ) in the Wire:**
Stress is defined as force per unit area:
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{200 \, \text{N}}{10^{-6} \, \text{m}^2} = 2 \times 10^{8} \, \text{N/m}^2 \]

4. **Calculate the Strain (ε) in the Wire:**
Strain is defined as the change in length divided by the original length:
\[ \epsilon = \frac{\Delta L}{L_0} \]
Substituting the values:
\[ \epsilon = \frac{8 \times 10^{-5} \, \text{m}}{2 \, \text{m}} = 4 \times 10^{-5} \]

5. **Relate Stress and Strain using Young's Modulus:**
Young's modulus (Y) is defined as the ratio of stress to strain:
\[ Y = \frac{\sigma}{\epsilon} \]
Rearranging gives:
\[ \sigma = Y \cdot \epsilon \]

6. **Calculate the Increase in Tension (ΔTension):**
The increase in tension due to thermal expansion can be expressed as:
\[ \Delta Tension = Y \cdot A \cdot \Delta L / L_0 \]
Substituting the values:
\[ \Delta Tension = 2.2 \times 10^{11} \, \text{N/m}^2 \cdot 10^{-6} \, \text{m}^2 \cdot \frac{8 \times 10^{-5} \, \text{m}}{2 \, \text{m}} \]
\[ = 2.2 \times 10^{11} \cdot 10^{-6} \cdot 4 \times 10^{-5} = 8.8 \, \text{N} \]

7. **Final Result:**
The increase in tension of the wire is **8.8 N**.