A stress of 1 kg `wt//mm^(2)` is applied to a wire whose Young's modulus is `10^(12)"dyne"//cm^(2)`. The percentage increase in its length is

Options

A. 0.98
B. `98xx10^(-4)`
C. `9.8xx10^(-6)`
D. `9.8xx10^(-5)`

Explanation

To solve the problem, we need to calculate the percentage increase in the length of a wire when a stress of 1 kg wt/mm² is applied, given that the Young's modulus of the wire is \(10^{12} \text{ dyne/cm}^2\).

### Step-by-Step Solution:

1. **Convert Stress from kg wt/mm² to Newton/m²**:
- Given stress = 1 kg wt/mm².
- We know that 1 kg wt = 9.8 N (approximately).
- Also, 1 mm² = \(10^{-6}\) m².
- Therefore, the stress in SI units:
\[ \text{Stress} = \frac{9.8 \text{ N}}{10^{-6} \text{ m}^2} = 9.8 \times 10^6 \text{ N/m}^2 \]

2. **Convert Young's Modulus from dyne/cm² to N/m²**:
- Given Young's modulus = \(10^{12} \text{ dyne/cm}^2\).
- We know that 1 dyne = \(10^{-5}\) N and 1 cm² = \(10^{-4}\) m².
- Therefore, the Young's modulus in SI units:
\[ \text{Young's Modulus} = 10^{12} \times 10^{-5} \text{ N} / 10^{-4} \text{ m}^2 = 10^{12} \times 10^{1} \text{ N/m}^2 = 10^{13} \text{ N/m}^2 \]

3. **Calculate Strain using Young's Modulus**:
- Young's modulus (E) is defined as:
\[ E = \frac{\text{Stress}}{\text{Strain}} \]
- Rearranging gives us:
\[ \text{Strain} = \frac{\text{Stress}}{E} \]
- Substituting the values:
\[ \text{Strain} = \frac{9.8 \times 10^6 \text{ N/m}^2}{10^{13} \text{ N/m}^2} = 9.8 \times 10^{-7} \]

4. **Calculate Percentage Increase in Length**:
- Percentage increase in length is given by:
\[ \text{Percentage Increase} = \text{Strain} \times 100 \]
- Therefore:
\[ \text{Percentage Increase} = 9.8 \times 10^{-7} \times 100 = 9.8 \times 10^{-5} \% \]

### Final Result:
The percentage increase in the length of the wire is \(9.8 \times 10^{-5} \%\).