What percent of length of a wire will increses by applying a stress fo `1kg. Wt//mm^(2)` on it.
`[Y = 1xx10^(11) Nm^(-2) "and" 1 kg wt = 9.8N]`

Options

A. `0.0078%`
B. `0.0088%`
C. `0.0098%`
D. `0.0067%`

Explanation

To solve the problem of finding the percentage increase in the length of a wire when a stress of \(1 \, \text{kg wt/mm}^2\) is applied, we will follow these steps:

### Step 1: Understand the relationship between stress, strain, and Young's modulus
The relationship is given by the formula:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
where \(Y\) is Young's modulus, stress is the force per unit area, and strain is the change in length per original length (\(\frac{\Delta L}{L}\)).

### Step 2: Convert the given stress from kg wt/mm² to N/m²
Given:
- \(1 \, \text{kg wt} = 9.8 \, \text{N}\)
- \(1 \, \text{mm}^2 = 10^{-6} \, \text{m}^2\)

Thus, the stress in SI units is:
\[ \text{Stress} = 1 \, \text{kg wt/mm}^2 = \frac{9.8 \, \text{N}}{1 \times 10^{-6} \, \text{m}^2} = 9.8 \times 10^6 \, \text{N/m}^2 \]

### Step 3: Calculate the strain using Young's modulus
We are given:
- \(Y = 1 \times 10^{11} \, \text{N/m}^2\)

Using the formula for strain:
\[ \text{Strain} = \frac{\text{Stress}}{Y} = \frac{9.8 \times 10^6 \, \text{N/m}^2}{1 \times 10^{11} \, \text{N/m}^2} \]
Calculating this gives:
\[ \text{Strain} = 9.8 \times 10^{-5} \]

### Step 4: Calculate the percentage increase in length
The percentage increase in length is given by:
\[ \text{Percentage Increase} = \left(\frac{\Delta L}{L}\right) \times 100 = 9.8 \times 10^{-5} \times 100 \]
Calculating this gives:
\[ \text{Percentage Increase} = 9.8 \times 10^{-3} = 0.0098 \% \]

### Final Answer
The percentage increase in the length of the wire when a stress of \(1 \, \text{kg wt/mm}^2\) is applied is approximately \(0.0098\%\).

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