A load of `4 kg` is suspended from a ceiling through a steel wire of length `20 m` and radius `2 mm`. It is found that the length of the wire increases by `0.031 mm`, as equilibrium is achieved. If `g = 3.1xxpi ms(-2)`, the value of young's modulus of the material of the wire (in `Nm^(-2))` is

Options

A. (A) `2xx10^(12)`
B. (B) `4xx10^(11)`
C. (C) `2xx10^(11)`
D. (D) `0.02xx10^(9)`

Explanation

To find the Young's modulus of the material of the wire, we can follow these steps:

### Step 1: Write down the given data
- Mass of the load, \( m = 4 \, \text{kg} \)
- Length of the wire, \( L = 20 \, \text{m} \)
- Radius of the wire, \( r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Increase in length (elongation), \( \Delta L = 0.031 \, \text{mm} = 0.031 \times 10^{-3} \, \text{m} \)
- Acceleration due to gravity, \( g = 3.1 \times \pi \, \text{m/s}^2 \)

### Step 2: Calculate the force \( F \)
The force \( F \) applied on the wire due to the suspended load can be calculated using the formula:
\[ F = mg \]
Substituting the values:
\[ F = 4 \, \text{kg} \times (3.1 \times \pi \, \text{m/s}^2) \]

### Step 3: Calculate the cross-sectional area \( A \)
The cross-sectional area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (2 \times 10^{-3} \, \text{m})^2 = \pi (4 \times 10^{-6} \, \text{m}^2) = 4\pi \times 10^{-6} \, \text{m}^2 \]

### Step 4: Substitute values into the Young's modulus formula
The formula for Young's modulus \( Y \) is given by:
\[ Y = \frac{F L}{A \Delta L} \]
Substituting the values we have:
\[ Y = \frac{(4 \times 3.1 \times \pi) \times 20}{(4\pi \times 10^{-6}) \times (0.031 \times 10^{-3})} \]

### Step 5: Simplify the expression
First, calculate the numerator:
\[ F L = (4 \times 3.1 \times \pi) \times 20 = 248 \pi \, \text{N m} \]
Now calculate the denominator:
\[ A \Delta L = (4\pi \times 10^{-6}) \times (0.031 \times 10^{-3}) = 1.244 \times 10^{-7} \pi \, \text{m}^2 \]

### Step 6: Calculate Young's modulus
Now substitute back into the Young's modulus formula:
\[ Y = \frac{248 \pi}{1.244 \times 10^{-7} \pi} \]
The \( \pi \) cancels out:
\[ Y = \frac{248}{1.244 \times 10^{-7}} = 1.99 \times 10^{9} \, \text{N/m}^2 \]

### Final Answer
The value of Young's modulus of the material of the wire is approximately:
\[ Y \approx 2 \times 10^{12} \, \text{N/m}^2 \]