On increasing the length by 0.5 mm in a steel wire of length 2 m and area of cross section 2`mm^2`, the force required is [Y for steel `=2.2xx10^11(N)/(m^2)`]

Options

A. `1.1xx10^5N`
B. `1.1xx10^4N`
C. `1.1xx10^3N`
D. `1.1xx10^2N`

Explanation

To find the force required to increase the length of a steel wire by 0.5 mm, we can use the formula for Young's modulus (Y):

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

Where:
- Stress = \(\frac{F}{A}\) (Force per unit area)
- Strain = \(\frac{\Delta L}{L_0}\) (Change in length per original length)

Given:
- Original length, \(L_0 = 2 \, \text{m}\)
- Change in length, \(\Delta L = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m}\)
- Area of cross-section, \(A = 2 \, \text{mm}^2 = 2 \times 10^{-6} \, \text{m}^2\)
- Young's modulus for steel, \(Y = 2.2 \times 10^{11} \, \text{N/m}^2\)

### Step 1: Calculate Strain
Strain is calculated as:
\[ \text{Strain} = \frac{\Delta L}{L_0} = \frac{0.5 \times 10^{-3} \, \text{m}}{2 \, \text{m}} = 0.25 \times 10^{-3} \]

### Step 2: Rearranging Young's Modulus Formula
From the Young's modulus formula, we can rearrange to find the force \(F\):
\[ Y = \frac{F/A}{\Delta L/L_0} \implies F = Y \cdot \text{Strain} \cdot A \]

### Step 3: Substitute Values
Now substitute the values into the equation:
\[ F = 2.2 \times 10^{11} \, \text{N/m}^2 \cdot (0.25 \times 10^{-3}) \cdot (2 \times 10^{-6} \, \text{m}^2) \]

### Step 4: Calculate the Force
Calculating the force:
\[ F = 2.2 \times 10^{11} \cdot 0.25 \times 10^{-3} \cdot 2 \times 10^{-6} \]
\[ F = 2.2 \times 0.25 \times 2 \times 10^{11 - 3 - 6} \]
\[ F = 1.1 \times 10^{2} \, \text{N} \]
\[ F = 110 \, \text{N} \]

### Final Answer
The force required to increase the length of the steel wire by 0.5 mm is **110 N**.

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