A steel wire of diameter 1 mm, and length 2 m is stretched by applying a force of 2.2 kg wt. The stress is `(g=10m//s^(2), Y=2xx10^(11)N//m^(2))`

Options

A. `7xx10^(7)N//m^(2)`
B. `3xx10^(6)N//m^(2)`
C. `21xx10^(8)N//m^(2)`
D. `28xx10^(6)N//m^(2)`

Explanation

To solve the problem of finding the stress generated in a steel wire when a force is applied, we can follow these steps:

### Step 1: Identify the Given Data
- Diameter of the wire, \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Length of the wire, \( L = 2 \, \text{m} \)
- Force applied, \( F = 2.2 \, \text{kg} \, \text{wt} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)

### Step 2: Calculate the Force in Newtons
The force in Newtons can be calculated using the formula:
\[ F = m \cdot g \]
where \( m \) is the mass in kg and \( g \) is the acceleration due to gravity.

Substituting the values:
\[ F = 2.2 \, \text{kg} \times 10 \, \text{m/s}^2 = 22 \, \text{N} \]

### Step 3: Calculate the Radius of the Wire
The radius \( r \) is half of the diameter:
\[ r = \frac{d}{2} = \frac{1 \times 10^{-3} \, \text{m}}{2} = 0.5 \times 10^{-3} \, \text{m} = 5 \times 10^{-4} \, \text{m} \]

### Step 4: Calculate the Cross-Sectional Area of the Wire
The area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (5 \times 10^{-4} \, \text{m})^2 = \pi (25 \times 10^{-8} \, \text{m}^2) = 25\pi \times 10^{-8} \, \text{m}^2 \]

### Step 5: Calculate the Stress
Stress \( \sigma \) is defined as the force per unit area:
\[ \sigma = \frac{F}{A} \]
Substituting the values of \( F \) and \( A \):
\[ \sigma = \frac{22 \, \text{N}}{25\pi \times 10^{-8} \, \text{m}^2} \]

### Step 6: Simplify the Expression
Calculating the numerical value:
\[ \sigma = \frac{22}{25\pi} \times 10^{8} \, \text{N/m}^2 \]
Using \( \pi \approx 3.14 \):
\[ \sigma \approx \frac{22}{25 \times 3.14} \times 10^{8} \approx \frac{22}{78.5} \times 10^{8} \approx 0.28 \times 10^{8} \, \text{N/m}^2 \]
\[ \sigma \approx 28 \times 10^{6} \, \text{N/m}^2 \]

### Final Answer
The stress generated in the steel wire is approximately:
\[ \sigma \approx 28 \times 10^{6} \, \text{N/m}^2 \]