A wire of length 1 m and area of cross section `2xx10^(-6)m^(2)` is suspended from the top of a roof at one end and a load of 20 N is applied at the other end. If the length of the wire is increased by `0.5xx10^(-4)m`, calculate its Young’s modulus (in `10^(11)N//m^(2))`.

Explanation

To calculate Young's modulus for the given wire, we will use the formula:

\[ Y = \frac{F \cdot L}{\Delta L \cdot A} \]

Where:
- \( Y \) = Young's modulus
- \( F \) = applied force (load)
- \( L \) = original length of the wire
- \( \Delta L \) = change in length of the wire
- \( A \) = area of cross-section of the wire

### Step 1: Identify the given values
- Length of the wire, \( L = 1 \, \text{m} \)
- Area of cross-section, \( A = 2 \times 10^{-6} \, \text{m}^2 \)
- Load applied, \( F = 20 \, \text{N} \)
- Increase in length, \( \Delta L = 0.5 \times 10^{-4} \, \text{m} \)

### Step 2: Substitute the values into the Young's modulus formula
Now, substituting the values into the formula:

\[ Y = \frac{20 \, \text{N} \cdot 1 \, \text{m}}{0.5 \times 10^{-4} \, \text{m} \cdot 2 \times 10^{-6} \, \text{m}^2} \]

### Step 3: Calculate the denominator
First, calculate the denominator:

\[ \Delta L \cdot A = 0.5 \times 10^{-4} \, \text{m} \cdot 2 \times 10^{-6} \, \text{m}^2 = 1 \times 10^{-10} \, \text{m}^3 \]

### Step 4: Substitute and simplify
Now substitute this back into the equation for \( Y \):

\[ Y = \frac{20 \, \text{N}}{1 \times 10^{-10} \, \text{m}^3} \]

\[ Y = 20 \times 10^{10} \, \text{N/m}^2 \]

### Step 5: Convert to the required units
To express \( Y \) in terms of \( 10^{11} \, \text{N/m}^2 \):

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

### Final Answer
Thus, the Young's modulus of the wire is:

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