In Searl's experiment, which is used to find Young's Modulus of elasticity, the diameter of experimental wire is `D = 0.05cm` (measured by a scale of least count `0.001cm`) and length is `L = 110cm` (measured by a scale of least count `0.1cm`). A weight of `50N` causes an extension of `X = 0.125 cm` (measured by a micrometer of least count `0.001cm`). find the maximum possible error in the values of Young's modulus. Screw gauge and meter scale are free error.

Explanation

To find the maximum possible error in the values of Young's modulus using Searl's experiment, we will follow these steps:

### Step 1: Understand the formula for Young's Modulus
The formula for Young's Modulus (Y) is given by:

\[ Y = \frac{F \cdot L}{A \cdot X} \]

Where:
- \( F \) = Force applied (in Newtons)
- \( L \) = Original length of the wire (in meters)
- \( A \) = Cross-sectional area of the wire (in square meters)
- \( X \) = Extension of the wire (in meters)

### Step 2: Calculate the Cross-sectional Area
The cross-sectional area \( A \) of the wire can be calculated using the diameter \( D \):

\[ A = \frac{\pi D^2}{4} \]

### Step 3: Convert all measurements to consistent units
Given:
- Diameter \( D = 0.05 \, \text{cm} = 0.0005 \, \text{m} \)
- Length \( L = 110 \, \text{cm} = 1.1 \, \text{m} \)
- Extension \( X = 0.125 \, \text{cm} = 0.00125 \, \text{m} \)
- Force \( F = 50 \, \text{N} \)

### Step 4: Calculate Young's Modulus
Substituting the values into the Young's Modulus formula:

\[ Y = \frac{50 \cdot 1.1}{\left(\frac{\pi (0.0005)^2}{4}\right) \cdot 0.00125} \]

Calculating the area:

\[ A = \frac{\pi (0.0005)^2}{4} = \frac{\pi \cdot 0.00000025}{4} \approx 1.9635 \times 10^{-7} \, \text{m}^2 \]

Now substituting back into the Young's Modulus formula:

\[ Y = \frac{50 \cdot 1.1}{(1.9635 \times 10^{-7}) \cdot 0.00125} \approx 2.24 \times 10^{11} \, \text{N/m}^2 \]

### Step 5: Calculate the maximum possible error
The relative error in Young's modulus is given by:

\[ \frac{\Delta Y}{Y} = 2 \frac{\Delta D}{D} + \frac{\Delta X}{X} + \frac{\Delta L}{L} \]

Where:
- \( \Delta D \) = Least count error in diameter = \( 0.001 \, \text{cm} = 0.00001 \, \text{m} \)
- \( D = 0.0005 \, \text{m} \)
- \( \Delta X \) = Least count error in extension = \( 0.001 \, \text{cm} = 0.00001 \, \text{m} \)
- \( X = 0.00125 \, \text{m} \)
- \( \Delta L \) = Least count error in length = \( 0.1 \, \text{cm} = 0.001 \, \text{m} \)
- \( L = 1.1 \, \text{m} \)

Calculating each term:

1. \( \frac{\Delta D}{D} = \frac{0.00001}{0.0005} = 0.02 \)
2. \( \frac{\Delta X}{X} = \frac{0.00001}{0.00125} = 0.008 \)
3. \( \frac{\Delta L}{L} = \frac{0.001}{1.1} \approx 0.000909 \)

Now substituting these values into the relative error formula:

\[ \frac{\Delta Y}{Y} = 2(0.02) + 0.008 + 0.000909 \approx 0.048909 \]

### Step 6: Calculate the maximum possible error
Now, to find \( \Delta Y \):

\[ \Delta Y = Y \cdot \frac{\Delta Y}{Y} = 2.24 \times 10^{11} \cdot 0.048909 \approx 1.09 \times 10^{10} \, \text{N/m}^2 \]

### Final Answer
The maximum possible error in the values of Young's modulus is approximately:

\[ \Delta Y \approx 1.09 \times 10^{10} \, \text{N/m}^2 \]