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 will follow these steps:

### Step 1: Calculate the Force (F) on the wire
The force exerted by the 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:
\[ F = 4 \, \text{kg} \cdot (3.1 \pi \, \text{m/s}^2) \]
\[ F = 12.4 \pi \, \text{N} \]

### Step 2: Convert the radius of the wire to meters
The radius of the wire is given as \( 2 \, \text{mm} \). We need to convert this to meters:
\[ r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \]

### Step 3: Calculate the cross-sectional area (A) 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 \( r \):
\[ A = \pi (2 \times 10^{-3})^2 \]
\[ A = \pi (4 \times 10^{-6}) \]
\[ A = 4\pi \times 10^{-6} \, \text{m}^2 \]

### Step 4: Convert the extension (ΔL) to meters
The extension of the wire is given as \( 0.031 \, \text{mm} \). We need to convert this to meters:
\[ \Delta L = 0.031 \, \text{mm} = 0.031 \times 10^{-3} \, \text{m} = 3.1 \times 10^{-5} \, \text{m} \]

### Step 5: Use the Young's modulus formula
The Young's modulus \( Y \) is given by the formula:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
where:
- \( L = 20 \, \text{m} \) (original length of the wire)

Substituting the values we have:
\[ Y = \frac{(12.4 \pi) \cdot 20}{(4\pi \times 10^{-6}) \cdot (3.1 \times 10^{-5})} \]

### Step 6: Simplify the expression
The \( \pi \) terms cancel out:
\[ Y = \frac{12.4 \cdot 20}{4 \times 10^{-6} \cdot 3.1 \times 10^{-5}} \]
\[ Y = \frac{248}{12.4 \times 10^{-11}} \]
\[ Y = \frac{248 \times 10^{11}}{12.4} \]
\[ Y = 20 \times 10^{11} \]
\[ Y = 2 \times 10^{12} \, \text{N/m}^2 \]

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