In Searle's apparatus diameter of the wire was measured 0.05 cm by screw gauge of least count 0.001 cm. The length of wire was measured 110 cm by meter scale of least count 0.1 cm. An external load of 50 N was applied. The extension in length of wire was measured 0.125 cm by micrometer of least count 0.001 cm. Find the maximum possible error in measurement of Young's modulus.

Answer

To find the maximum possible error in the measurement of Young's modulus using Searle's apparatus, we can follow these steps:

### Step 1: Understand the Formula for Young's Modulus
The formula for Young's modulus (Y) is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress = \(\frac{F}{A}\) (Force per unit area)
- Strain = \(\frac{\Delta L}{L}\) (Change in length per original length)

### Step 2: Calculate Stress and Strain
- **Force (F)**: Given as \(50 \, \text{N}\).
- **Diameter (d)**: Measured as \(0.05 \, \text{cm} = 0.0005 \, \text{m}\).
- **Area (A)**: \(A = \frac{\pi}{4} d^2\).
- **Original Length (L)**: Measured as \(110 \, \text{cm} = 1.1 \, \text{m}\).
- **Change in Length (\(\Delta L\))**: Measured as \(0.125 \, \text{cm} = 0.00125 \, \text{m}\).

### Step 3: Calculate the Area
Using the diameter to find the area:
\[ A = \frac{\pi}{4} (0.0005)^2 = \frac{\pi}{4} (0.00000025) \approx 1.9635 \times 10^{-7} \, \text{m}^2 \]

### Step 4: Calculate Stress
Now, calculate the stress:
\[ \text{Stress} = \frac{F}{A} = \frac{50}{1.9635 \times 10^{-7}} \approx 2.548 \times 10^8 \, \text{N/m}^2 \]

### Step 5: Calculate Strain
Now, calculate the strain:
\[ \text{Strain} = \frac{\Delta L}{L} = \frac{0.00125}{1.1} \approx 0.00113636 \]

### Step 6: Calculate Young's Modulus
Now we can calculate Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{2.548 \times 10^8}{0.00113636} \approx 2.245 \times 10^{11} \, \text{N/m}^2 \]

### Step 7: Calculate Maximum Possible Error in Young's Modulus
The formula for the maximum possible error in Young's modulus is:
\[ \frac{\Delta Y}{Y} = \frac{2 \Delta d}{d} + \frac{\Delta L}{L} + \frac{\Delta x}{\Delta L} \]
Where:
- \(\Delta d\) is the least count of diameter = \(0.001 \, \text{cm} = 0.00001 \, \text{m}\)
- \(\Delta L\) is the least count of length = \(0.1 \, \text{cm} = 0.001 \, \text{m}\)
- \(\Delta x\) is the least count of extension = \(0.001 \, \text{cm} = 0.00001 \, \text{m}\)

Substituting the values:
\[ \frac{\Delta Y}{Y} = \frac{2 \times 0.00001}{0.0005} + \frac{0.001}{1.1} + \frac{0.00001}{0.00125} \]

Calculating each term:
1. \(\frac{2 \times 0.00001}{0.0005} = 0.04\)
2. \(\frac{0.001}{1.1} \approx 0.000909\)
3. \(\frac{0.00001}{0.00125} = 0.008\)

Combining these:
\[ \frac{\Delta Y}{Y} \approx 0.04 + 0.000909 + 0.008 \approx 0.0489 \]

### Step 8: Convert to Percentage
To convert to percentage:
\[ \Delta Y \approx 0.0489 \times 100 \approx 4.89\% \]

### Final Answer
The maximum possible error in the measurement of Young's modulus is approximately **4.89%**.

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