What is the percentage increase in length of a wire of diameter 2.5 mm, stretched by a force of 100 kg wt ? Young's modulus of elasticity of wire `=12.5xx10^(11) dyn e//cm^(2)`.

Options

A. `0.16%`
B. `0.32%`
C. `0.08%`
D. `0.12%`

Answer

To solve the problem of finding the percentage increase in length of a wire when a force is applied, we can follow these steps:

### Given Data:
- Diameter of the wire, \( d = 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m} \)
- Force applied, \( F = 100 \, \text{kg wt} = 1000 \, \text{N} \) (using \( g \approx 10 \, \text{m/s}^2 \))
- Young's modulus, \( Y = 12.5 \times 10^{11} \, \text{dyn/cm}^2 \)

### Step 1: Convert Young's Modulus to SI Units
Young's modulus in SI units is expressed in \( \text{N/m}^2 \). We convert \( \text{dyn/cm}^2 \) to \( \text{N/m}^2 \):
\[ 1 \, \text{dyn} = 10^{-5} \, \text{N}, \quad 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \]
Thus,
\[ Y = 12.5 \times 10^{11} \, \text{dyn/cm}^2 = 12.5 \times 10^{11} \times 10^{-5} \, \text{N} / 10^{-4} \, \text{m}^2 = 12.5 \times 10^{10} \, \text{N/m}^2 \]

### Step 2: Calculate the Area of Cross-Section of the Wire
The area \( A \) of the cross-section of the wire can be calculated using the formula for the area of a circle:
\[ A = \frac{\pi d^2}{4} \]
Substituting the diameter:
\[ A = \frac{\pi (2.5 \times 10^{-3})^2}{4} = \frac{\pi (6.25 \times 10^{-6})}{4} \approx \frac{3.14 \times 6.25 \times 10^{-6}}{4} \approx 4.91 \times 10^{-7} \, \text{m}^2 \]

### Step 3: Calculate the Strain
Using the formula for Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \quad \Rightarrow \quad \text{Strain} = \frac{\text{Stress}}{Y} \]
Where stress \( \sigma \) is given by:
\[ \sigma = \frac{F}{A} \]
Thus,
\[ \text{Strain} = \frac{F/A}{Y} \]

### Step 4: Calculate the Percentage Increase in Length
The percentage increase in length can be calculated using:
\[ \text{Percentage Increase} = \text{Strain} \times 100 = \left( \frac{F}{A \cdot Y} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Increase} = \left( \frac{1000 \, \text{N}}{(4.91 \times 10^{-7} \, \text{m}^2) \cdot (12.5 \times 10^{10} \, \text{N/m}^2)} \right) \times 100 \]

### Step 5: Calculate the Final Result
Now, we compute the above expression:
\[ \text{Percentage Increase} = \left( \frac{1000}{(4.91 \times 10^{-7}) \cdot (12.5 \times 10^{10})} \right) \times 100 \]
Calculating the denominator:
\[ (4.91 \times 10^{-7}) \cdot (12.5 \times 10^{10}) \approx 6.1375 \times 10^{3} \]
Thus,
\[ \text{Percentage Increase} \approx \left( \frac{1000}{6.1375 \times 10^{3}} \right) \times 100 \approx 0.163 \% \]

### Final Answer
The percentage increase in length of the wire is approximately **0.16%**.