In an experiment on the determination of young's Modulus of a wire by Searle's method , following data is available:
Normal length of the wire `(L) = 110 cm`
Diameter of the wire `(d) = 0.01 cm`
Elongation in the wire `(l) = 0.125 cm`
This elongation is for a tension of `50 N`. The least counts for corresponding quantities are `0.01 cm , 0.00005 cm, and 0.001 cm` , respectively. Calculate the maximum error in calculating the value of Young's modulus (Y)`.

Explanation

To calculate the maximum error in determining Young's modulus (Y) using the given data, we can follow these steps:

### Step 1: Understand the formula for Young's Modulus
Young's modulus (Y) is defined as:
\[ Y = \frac{F \cdot L}{A \cdot l} \]
where:
- \( F \) = Force applied (in Newtons)
- \( L \) = Original length of the wire (in cm)
- \( A \) = Cross-sectional area of the wire (in cm²)
- \( l \) = Elongation of the wire (in cm)

### Step 2: Calculate the Cross-sectional Area (A)
The cross-sectional area \( A \) of the wire can be calculated using the diameter \( d \):
\[ A = \frac{\pi d^2}{4} \]
Given \( d = 0.01 \, \text{cm} \):
\[ A = \frac{\pi (0.01)^2}{4} = \frac{\pi \times 0.0001}{4} \approx 7.85 \times 10^{-5} \, \text{cm}^2 \]

### Step 3: Identify the least counts and calculate relative errors
The least counts for the measurements are:
- Length \( L \): \( \Delta L = 0.01 \, \text{cm} \)
- Elongation \( l \): \( \Delta l = 0.001 \, \text{cm} \)
- Diameter \( d \): \( \Delta d = 0.00005 \, \text{cm} \)

#### Relative error in each quantity:
1. **Relative error in Length \( L \)**:
\[ \text{Relative error in } L = \frac{\Delta L}{L} = \frac{0.01}{110} \approx 9.09 \times 10^{-5} \]

2. **Relative error in Elongation \( l \)**:
\[ \text{Relative error in } l = \frac{\Delta l}{l} = \frac{0.001}{0.125} = 0.008 \]

3. **Relative error in Diameter \( d \)**:
\[ \text{Relative error in } d = \frac{\Delta d}{d} = \frac{0.00005}{0.01} = 0.005 \]

### Step 4: Calculate the maximum error in Young's Modulus
The formula for the maximum error in \( Y \) is given by:
\[ \frac{\Delta Y}{Y} = \frac{\Delta L}{L} + \frac{\Delta l}{l} + 2 \cdot \frac{\Delta d}{d} \]
Substituting the values:
\[ \frac{\Delta Y}{Y} = 9.09 \times 10^{-5} + 0.008 + 2 \cdot 0.005 \]
\[ = 9.09 \times 10^{-5} + 0.008 + 0.01 \]
\[ = 0.0180909 \]

### Step 5: Convert to percentage error
To convert the relative error to percentage:
\[ \text{Percentage error} = \frac{\Delta Y}{Y} \times 100 \approx 1.80909 \% \]

### Final Answer
The maximum error in calculating the value of Young's modulus \( Y \) is approximately **1.81%**.

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