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 provided data, we will follow these steps:

### Step 1: Understand the formula for Young's modulus
Young's modulus (Y) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress = \(\frac{F}{A}\) (Force per unit area)
- Strain = \(\frac{l}{L}\) (Elongation per unit original length)

### Step 2: Express Young's modulus in terms of the given quantities
From the definitions, we can express Young's modulus as:
\[ Y = \frac{F \cdot L}{A \cdot l} \]
Where:
- \(F\) = Tension (Force) = 50 N
- \(L\) = Normal length of the wire = 110 cm = 1.1 m
- \(A\) = Cross-sectional area of the wire = \(\frac{\pi d^2}{4}\)
- \(l\) = Elongation = 0.125 cm = 0.00125 m

### Step 3: Substitute the area in terms of diameter
The area \(A\) can be expressed as:
\[ A = \frac{\pi d^2}{4} \]
Substituting this into the formula for Young's modulus gives:
\[ Y = \frac{4F \cdot L}{\pi d^2 \cdot l} \]

### Step 4: Calculate the relative error in Young's modulus
To find the maximum error in Y, we need to consider the contributions from the uncertainties in \(L\), \(l\), and \(d\). The relative error in Y can be expressed as:
\[ \frac{\Delta Y}{Y} = \frac{\Delta L}{L} + \frac{\Delta l}{l} + 2\frac{\Delta d}{d} \]
Where:
- \(\Delta L\) = least count for length = 0.01 cm = 0.0001 m
- \(\Delta l\) = least count for elongation = 0.001 cm = 0.00001 m
- \(\Delta d\) = least count for diameter = 0.00005 cm = 0.0000005 m

### Step 5: Substitute the values into the error formula
Now we can substitute the values into the error formula:
\[ \frac{\Delta Y}{Y} = \frac{0.0001}{1.1} + \frac{0.00001}{0.00125} + 2 \cdot \frac{0.0000005}{0.0001} \]

Calculating each term:
1. \(\frac{\Delta L}{L} = \frac{0.0001}{1.1} \approx 0.0000909\)
2. \(\frac{\Delta l}{l} = \frac{0.00001}{0.00125} = 0.008\)
3. \(2 \cdot \frac{\Delta d}{d} = 2 \cdot \frac{0.0000005}{0.0001} = 0.01\)

### Step 6: Sum the relative errors
Now, we sum these contributions:
\[ \frac{\Delta Y}{Y} \approx 0.0000909 + 0.008 + 0.01 \approx 0.0180909 \]

### Step 7: Convert to percentage
To express this as a percentage, we multiply by 100:
\[ \Delta Y \approx 0.0180909 \times 100 \approx 1.80909\% \]

### Step 8: Round off the final answer
Rounding off gives us:
\[ \Delta Y \approx 1.81\% \]

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

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