A steel wire , of uniform area `2mm^2` , is heated up to `50^@C` and is stretched by tying its ends rigidly . The change in tension , when the temperature falls from `50^@C " to " 30^@C` is (Take `Y=2xx10^(11)Nm^(-2) , alpha=1.1xx10^(-5^@C-1))`

Options

A. (a)`1.5xx10^(10)N`
B. (b)`5 N`
C. (c)`88 N`
D. (d)`2.510^(10)N`

Answer

To solve the problem, we need to calculate the change in tension in a steel wire when its temperature decreases from \(50^\circ C\) to \(30^\circ C\). We will use the formula that relates change in tension to Young's modulus, the area of the wire, the coefficient of linear expansion, and the change in temperature.

### Step-by-Step Solution:

1. **Identify the given values:**
- Area \(A = 2 \, \text{mm}^2 = 2 \times 10^{-6} \, \text{m}^2\)
- Young's modulus \(Y = 2 \times 10^{11} \, \text{N/m}^2\)
- Coefficient of linear expansion \(\alpha = 1.1 \times 10^{-5} \, \text{°C}^{-1}\)
- Change in temperature \(\Delta \theta = 50^\circ C - 30^\circ C = 20^\circ C\)

2. **Use the formula for change in tension:**
The change in tension \(\Delta T\) in the wire can be calculated using the formula:
\[ \Delta T = Y \cdot A \cdot \alpha \cdot \Delta \theta \]

3. **Substitute the values into the formula:**
\[ \Delta T = (2 \times 10^{11} \, \text{N/m}^2) \cdot (2 \times 10^{-6} \, \text{m}^2) \cdot (1.1 \times 10^{-5} \, \text{°C}^{-1}) \cdot (20 \, \text{°C}) \]

4. **Calculate step-by-step:**
- First, calculate \(A \cdot \alpha\):
\[ A \cdot \alpha = (2 \times 10^{-6}) \cdot (1.1 \times 10^{-5}) = 2.2 \times 10^{-11} \, \text{m}^2/\text{°C} \]
- Now multiply by \(Y\) and \(\Delta \theta\):
\[ \Delta T = (2 \times 10^{11}) \cdot (2.2 \times 10^{-11}) \cdot (20) \]
- Calculate \(2 \times 10^{11} \cdot 2.2 \times 10^{-11}\):
\[ 2 \times 2.2 = 4.4 \quad \text{and} \quad 10^{11} \cdot 10^{-11} = 10^{0} = 1 \]
Thus, \(2 \times 10^{11} \cdot 2.2 \times 10^{-11} = 4.4\).
- Now multiply by \(20\):
\[ \Delta T = 4.4 \cdot 20 = 88 \, \text{N} \]

5. **Final Result:**
The change in tension when the temperature falls from \(50^\circ C\) to \(30^\circ C\) is \(88 \, \text{N}\).