To determine the maximum fractional error in the measurement of Young's modulus (Y) using Searle's apparatus, we need to consider the formula for Young's modulus and the errors associated with the measurements taken.
### Step-by-Step Solution:
1. **Understanding Young's Modulus**:
Young's modulus (Y) is defined as:
\[
Y = \frac{F/A}{\Delta L/L}
\]
where:
- \( F \) is the force applied (weight of the load, \( W \)),
- \( A \) is the cross-sectional area of the wire,
- \( \Delta L \) is the extension of the wire,
- \( L \) is the original length of the wire.
2. **Cross-Sectional Area Calculation**:
The cross-sectional area \( A \) of the wire can be calculated using the diameter \( d \):
\[
A = \frac{\pi d^2}{4}
\]
3. **Error Propagation in Young's Modulus**:
The fractional error in Y can be derived from the errors in the measurements of \( W \), \( L \), and \( d \). The formula for the maximum fractional error in Y is given by:
\[
\frac{\Delta Y}{Y} = \frac{\Delta W}{W} + \frac{\Delta L}{L} + 2 \frac{\Delta d}{d}
\]
where:
- \( \Delta W \) is the error in the weight measurement,
- \( \Delta L \) is the error in the length measurement,
- \( \Delta d \) is the error in the diameter measurement.
4. **Calculating Errors**:
- **For Length \( L \)**:
- Given \( L = 30 \, \text{cm} = 300 \, \text{mm} \) (measured with a millimeter scale).
- Least count of the scale = 1 mm.
- Therefore, the fractional error in length:
\[
\frac{\Delta L}{L} = \frac{1 \, \text{mm}}{300 \, \text{mm}} = \frac{1}{300}
\]
- **For Diameter \( d \)**:
- Given \( d = 1 \, \text{mm} \) (measured with a screw gauge).
- Least count of the screw gauge = 0.01 mm.
- Therefore, the fractional error in diameter:
\[
\frac{\Delta d}{d} = \frac{0.01 \, \text{mm}}{1 \, \text{mm}} = 0.01
\]
- **For Weight \( W \)**:
- Assuming the weight is measured accurately, we can take a reasonable estimate for the error. If not specified, we can assume a least count of 1 gram (0.001 kg).
- Therefore, the fractional error in weight:
\[
\frac{\Delta W}{W} = \frac{0.001 \, \text{kg}}{W} \quad \text{(where W is the weight in kg)}
\]
5. **Combining Errors**:
Now we can combine these errors into the formula for maximum fractional error in Y:
\[
\frac{\Delta Y}{Y} = \frac{\Delta W}{W} + \frac{1}{300} + 2 \times 0.01
\]
\[
\frac{\Delta Y}{Y} = \frac{\Delta W}{W} + \frac{1}{300} + 0.02
\]
6. **Final Expression**:
The maximum fractional error in the measurement of Young's modulus \( Y \) is:
\[
\frac{\Delta Y}{Y} = \frac{\Delta W}{W} + \frac{1}{300} + 0.02
\]
