A stress of 1 kg `mm^(-2)` is applied to a wire of which Young's modulus is `10^(11) Nm^(-2)` and `1.1xx10^(11) Nm^(-2)`. Find the percentage increase in length.

Answer

To solve the problem of finding the percentage increase in length of a wire when a stress of 1 kg/mm² is applied, and given that the Young's modulus is \(10^{11} \, \text{N/m}^2\), we can follow these steps:

### Step 1: Convert Stress to SI Units
The given stress is \(1 \, \text{kg/mm}^2\). We need to convert this to \( \text{N/m}^2\) (Pascals) for consistency with the Young's modulus units.

1 kg = 9.8 N (approximately, due to gravity), and \(1 \, \text{mm}^2 = 10^{-6} \, \text{m}^2\).

Thus,
\[ \text{Stress} = 1 \, \text{kg/mm}^2 = 1 \times 9.8 \, \text{N} / (1 \times 10^{-6} \, \text{m}^2) = 9.8 \times 10^{6} \, \text{N/m}^2. \]

### Step 2: Use Young's Modulus to Find Strain
Young's modulus \(Y\) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}}. \]
Rearranging this gives:
\[ \text{Strain} = \frac{\text{Stress}}{Y}. \]

Substituting the values:
\[ \text{Strain} = \frac{9.8 \times 10^{6} \, \text{N/m}^2}{10^{11} \, \text{N/m}^2} = 9.8 \times 10^{-5}. \]

### Step 3: Relate Strain to Change in Length
Strain is also defined as the change in length (\(\Delta L\)) divided by the original length (\(L\)):
\[ \text{Strain} = \frac{\Delta L}{L}. \]

### Step 4: Calculate Percentage Increase in Length
To find the percentage increase in length, we multiply the strain by 100:
\[ \text{Percentage Increase} = \text{Strain} \times 100 = 9.8 \times 10^{-5} \times 100 = 9.8 \times 10^{-3} \, \text{or} \, 0.0098 \, \text{(in percentage)}. \]

### Final Answer
The percentage increase in length of the wire is approximately \(0.0098\%\).

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