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 solve the problem, we need to calculate the Young's modulus of the wire using the given parameters. Let's break down the solution step by step.

### Step 1: Write down the known values
- Length of the wire (L) = 1 m
- Area of cross-section (A) = 2 × 10^(-6) m²
- Force applied (F) = 20 N
- Change in length (ΔL) = 0.5 × 10^(-4) m

### Step 2: Calculate the stress
Stress (σ) is defined as the force applied per unit area. The formula for stress is:
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{20 \, \text{N}}{2 \times 10^{-6} \, \text{m}^2} \]
Calculating this gives:
\[ \sigma = \frac{20}{2 \times 10^{-6}} = 10 \times 10^{6} \, \text{N/m}^2 = 10^7 \, \text{N/m}^2 \]

### Step 3: Calculate the strain
Strain (ε) is defined as the change in length per unit original length. The formula for strain is:
\[ \epsilon = \frac{\Delta L}{L} \]
Substituting the values:
\[ \epsilon = \frac{0.5 \times 10^{-4} \, \text{m}}{1 \, \text{m}} \]
Calculating this gives:
\[ \epsilon = 0.5 \times 10^{-4} = 5 \times 10^{-5} \]

### Step 4: Calculate Young's modulus
Young's modulus (Y) is defined as the ratio of stress to strain. The formula for Young's modulus is:
\[ Y = \frac{\sigma}{\epsilon} \]
Substituting the values we calculated:
\[ Y = \frac{10^7 \, \text{N/m}^2}{5 \times 10^{-5}} \]
Calculating this gives:
\[ Y = \frac{10^7}{5 \times 10^{-5}} = 2 \times 10^{12} \, \text{N/m}^2 \]

### Step 5: Convert Young's modulus to the required form
The problem asks for Young's modulus in terms of \(10^{11} \, \text{N/m}^2\). We can express \(2 \times 10^{12} \, \text{N/m}^2\) as:
\[ Y = 20 \times 10^{11} \, \text{N/m}^2 \]
Thus, in terms of \(10^{11} \, \text{N/m}^2\), the answer is:
\[ Y = 20 \]

### Final Answer
The Young's modulus of the wire is \(20 \times 10^{11} \, \text{N/m}^2\).

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