In the Searle's method to determine the Young's modulus of a wire, a steel wire of length `156cm`
and diameter `0.054cm` is taken as experinmental wire. The average increases in length for `1.5kg wt` is found to be `0.50cm`. Then the Young's modulus of the wire is

Options

A. `3.002xx10^(11) N//m^(2)`
B. `1.002xx10^(11) N//m^(2)`
C. `2.002xx10^(11) N//m^(2)`
D. `2.5xx10^(11) N//m^(2)`

Explanation

To determine the Young's modulus of the steel wire using Searle's method, we can follow these steps:

### Step 1: Convert Measurements to SI Units
- **Length of the wire (L)**: Given as 156 cm, we convert this to meters:
\[ L = 156 \, \text{cm} = 156 \times 10^{-2} \, \text{m} = 1.56 \, \text{m} \]

- **Diameter of the wire (d)**: Given as 0.054 cm, we convert this to meters:
\[ d = 0.054 \, \text{cm} = 0.054 \times 10^{-2} \, \text{m} = 0.00054 \, \text{m} \]

### Step 2: Calculate the Radius of the Wire
- The radius (r) is half of the diameter:
\[ r = \frac{d}{2} = \frac{0.00054}{2} = 0.00027 \, \text{m} \]

### Step 3: Calculate the Cross-Sectional Area of the Wire
- The cross-sectional area (A) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (0.00027)^2 \]
\[ A \approx 3.14 \times (0.00027)^2 \approx 3.14 \times 0.0000000729 \approx 2.27 \times 10^{-7} \, \text{m}^2 \]

### Step 4: Calculate the Stress on the Wire
- The force (F) applied is given by the weight (W) of the load. The weight can be calculated as:
\[ W = mg = 1.5 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 14.715 \, \text{N} \]
- Stress (σ) is defined as force per unit area:
\[ \sigma = \frac{F}{A} = \frac{14.715}{2.27 \times 10^{-7}} \approx 6.48 \times 10^{7} \, \text{N/m}^2 \]

### Step 5: Calculate the Strain on the Wire
- The strain (ε) is defined as the change in length (ΔL) divided by the original length (L):
\[ \Delta L = 0.50 \, \text{cm} = 0.005 \, \text{m} \]
\[ \epsilon = \frac{\Delta L}{L} = \frac{0.005}{1.56} \approx 0.003205 \]

### Step 6: Calculate Young's Modulus
- Young's modulus (Y) is defined as the ratio of stress to strain:
\[ Y = \frac{\sigma}{\epsilon} = \frac{6.48 \times 10^{7}}{0.003205} \approx 2.02 \times 10^{10} \, \text{N/m}^2 \]

### Final Result
Thus, the Young's modulus of the wire is approximately:
\[ Y \approx 2.02 \times 10^{10} \, \text{N/m}^2 \]