In searle's experiment, the diameter of the wire, as measured by a screw gauge of least count 0.001 cm is 0.500 cm. The length, measured by a scale of least count 0.1cm is 110.0cm. When a weight of 40N is suspended from the wire, its extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the Young's modulus of the meterial of the wire from this data.

Explanation

To solve the problem of finding the Young's modulus of the material of the wire in Searle's experiment, we will follow these steps:

### Step 1: Gather the given data
- Diameter of the wire, \( d = 0.500 \, \text{cm} \) (with least count \( \delta d = 0.001 \, \text{cm} \))
- Length of the wire, \( L = 110.0 \, \text{cm} \) (with least count \( \delta L = 0.1 \, \text{cm} \))
- Weight suspended from the wire, \( F = 40 \, \text{N} \)
- Extension of the wire, \( \Delta L = 0.125 \, \text{cm} \) (with least count \( \delta \Delta L = 0.001 \, \text{cm} \))

### Step 2: Convert all measurements to meters
- Diameter: \( d = 0.500 \, \text{cm} = 0.005 \, \text{m} \)
- Length: \( L = 110.0 \, \text{cm} = 1.10 \, \text{m} \)
- Extension: \( \Delta L = 0.125 \, \text{cm} = 0.00125 \, \text{m} \)

### Step 3: Calculate the cross-sectional area of the wire
The area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \frac{\pi d^2}{4} \]
Substituting the diameter:
\[ A = \frac{\pi (0.005)^2}{4} = \frac{\pi (0.000025)}{4} \approx 1.9635 \times 10^{-5} \, \text{m}^2 \]

### Step 4: Calculate the Young's modulus
The formula for Young's modulus \( Y \) is given by:
\[ Y = \frac{F/A}{\Delta L/L} \]
This can be rearranged to:
\[ Y = \frac{4F L}{\pi d^2 \Delta L} \]
Substituting the values:
\[ Y = \frac{4 \times 40 \times 1.10}{\pi (0.005)^2 \times 0.00125} \]
Calculating the numerator:
\[ 4 \times 40 \times 1.10 = 176 \]
Calculating the denominator:
\[ \pi (0.005)^2 \times 0.00125 \approx 3.1416 \times 0.000025 \times 0.00125 \approx 9.817 \times 10^{-8} \]
Thus,
\[ Y = \frac{176}{9.817 \times 10^{-8}} \approx 1.792 \times 10^9 \, \text{N/m}^2 \]

### Step 5: Calculate the error in Young's modulus
To find the relative error in \( Y \):
\[ \frac{\delta Y}{Y} = \frac{\delta L}{L} + \frac{2 \delta d}{d} + \frac{\delta \Delta L}{\Delta L} \]
Substituting the values:
- \( \delta L = 0.1 \, \text{cm} = 0.001 \, \text{m} \)
- \( \delta d = 0.001 \, \text{cm} = 0.00001 \, \text{m} \)
- \( \delta \Delta L = 0.001 \, \text{cm} = 0.00001 \, \text{m} \)

Calculating:
\[ \frac{\delta L}{L} = \frac{0.001}{1.10} \approx 0.000909 \]
\[ \frac{2 \delta d}{d} = \frac{2 \times 0.00001}{0.005} = 0.004 \]
\[ \frac{\delta \Delta L}{\Delta L} = \frac{0.00001}{0.00125} = 0.008 \]
Adding these:
\[ \frac{\delta Y}{Y} \approx 0.000909 + 0.004 + 0.008 \approx 0.012909 \]
Calculating \( \delta Y \):
\[ \delta Y = 0.012909 \times 1.792 \times 10^9 \approx 2.31 \times 10^7 \, \text{N/m}^2 \]

### Final Result
Thus, the Young's modulus \( Y \) is given by:
\[ Y = 1.792 \times 10^9 \pm 2.31 \times 10^7 \, \text{N/m}^2 \]