A load of 4 kg is suspended from a ceiling through a steel wire of length 2 m and radius 2 mm. It is found that the length of the wire increase by 0.032 mm as equilibrium is achieved. What would be the Young's modulus of steel ? (Take , `g=3.1 pi m s^(-2)`)

Options

A. `2.0 xx 10^(11) Nm^(-2)`
B. `2.82 xx 10^(11) Nm^(-2)`
C. `0.20 xx 10^(11) Nm^(-2)`
D. `0.028 xx 10^(11) Nm^(-2)`

Explanation

To find the Young's modulus of steel, we will use the formula:

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

### Step 1: Calculate the Force (F) acting on the wire
The force acting on the wire due to the suspended load can be calculated using the formula:

\[ F = m \cdot g \]

Where:
- \( m = 4 \, \text{kg} \) (mass of the load)
- \( g = 3.1 \pi \, \text{m/s}^2 \)

Calculating the force:

\[ F = 4 \, \text{kg} \cdot (3.1 \pi) \approx 4 \cdot 9.73 \approx 38.92 \, \text{N} \]

### Step 2: Calculate the Cross-sectional Area (A) of the wire
The cross-sectional area of the wire can be calculated using the formula for the area of a circle:

\[ A = \pi r^2 \]

Where:
- \( r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

Calculating the area:

\[ A = \pi (2 \times 10^{-3})^2 = \pi (4 \times 10^{-6}) = 4\pi \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the Strain
Strain is defined as the change in length divided by the original length:

\[ \text{Strain} = \frac{\Delta L}{L_0} \]

Where:
- \( \Delta L = 0.032 \, \text{mm} = 0.032 \times 10^{-3} \, \text{m} \)
- \( L_0 = 2 \, \text{m} \)

Calculating the strain:

\[ \text{Strain} = \frac{0.032 \times 10^{-3}}{2} = 0.016 \times 10^{-3} \]

### Step 4: Calculate the Stress
Stress is defined as the force divided by the area:

\[ \text{Stress} = \frac{F}{A} \]

Substituting the values we calculated:

\[ \text{Stress} = \frac{38.92}{4\pi \times 10^{-6}} \]

Calculating the stress:

\[ \text{Stress} \approx \frac{38.92}{4 \times 3.14 \times 10^{-6}} \approx \frac{38.92}{1.256 \times 10^{-5}} \approx 3.1 \times 10^6 \, \text{N/m}^2 \]

### Step 5: Calculate Young's Modulus (Y)
Now we can calculate Young's modulus using the values of stress and strain:

\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{3.1 \times 10^6}{0.016 \times 10^{-3}} \]

Calculating Young's modulus:

\[ Y \approx \frac{3.1 \times 10^6}{0.016 \times 10^{-3}} \approx 1.9375 \times 10^{11} \, \text{N/m}^2 \]

Rounding off, we get:

\[ Y \approx 2 \times 10^{11} \, \text{N/m}^2 \]

### Final Answer:
The Young's modulus of steel is approximately:

\[ Y \approx 2 \times 10^{11} \, \text{N/m}^2 \]

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