During Searle's experiment, zero of the Vernier sacle lies between `3.20 xx 10^(-2),` and `3.25 xx 10^(-2)m` of the main scale. The `20^(th)` division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of `2kg` is applied to the wire, the zero of the vernier scale still lies between `3.20 xx 10^(-2),` and `3.25 xx 10^(-2)m` of the main scale but now the `45^(th)` division of Vernier scale coincide with one of the main scale divisions. the length of the thin metallic wire is `2m` and its cross-sectional ares is `8 xx 10^(-7)m^(2)`. the least count of the Vernier scale is `1.0 xx 10^(-5)m`. the maximum percentage error in the Young's modulus of the wire is

Explanation

To solve the problem regarding Searle's experiment and the calculation of the maximum percentage error in the Young's modulus of the wire, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Conditions**:
- Initial position of the Vernier scale: The zero of the Vernier scale lies between \(3.20 \times 10^{-2} \, \text{m}\) and \(3.25 \times 10^{-2} \, \text{m}\).
- Initial coinciding division: The \(20^{th}\) division of the Vernier scale coincides with a main scale division.
- Final position after applying \(2 \, \text{kg}\) load: The zero of the Vernier scale still lies between \(3.20 \times 10^{-2} \, \text{m}\) and \(3.25 \times 10^{-2} \, \text{m}\), but now the \(45^{th}\) division of the Vernier scale coincides with a main scale division.
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