During Searle's experiment, zero of the Vernier scale lies between 3.20 × 10⁻⁶ m and 3.25 × 10⁻⁶ m of the main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 × 10⁻⁶ m and 3.25 × 10⁻⁶ m of the main scale but now the 45th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 × 10⁻⁷ m². The least count of the Vernier scale is 1.0 × 10⁻⁹ m. The maximum percentage error in the Young's modulus of the wire is: ?

Y = FL/AΔl

where:
Y = Young's modulus
F = force applied
L = original length of the wire
A = cross-sectional area of the wire
Δl = change in length of the wire

In the first case, the 20th division of the Vernier scale coincides with a main scale division. The initial extension (Δl₁) is 20 × least count = 20 × 1.0 × 10⁻⁹ m = 2 × 10⁻⁸ m.

In the second case, with an additional 2 kg load, the 45th division coincides. The extension (Δl₂) is 45 × 1.0 × 10⁻⁹ m = 4.5 × 10⁻⁸ m.

The change in extension due to the additional 2 kg load is Δl₂ - Δl₁ = 4.5 × 10⁻⁸ m - 2 × 10⁻⁸ m = 2.5 × 10⁻⁸ m.

Let's assume the mass of the initial load is m kg. The force due to the initial load is mg and the force due to the additional 2kg load is 2g. The change in length (2.5 x 10⁻⁸ m) is proportional to the change in force (2g).

Therefore we can write:

Y = (m+2)gL / A(Δl₂) = mgL / A(Δl₁)

The percentage error in Young's modulus is given by the formula:

% error = [(ΔY/Y) × 100]

Given the formula Y = FL/AΔl and considering the errors in Δl and ΔF (which is related to Δl), the maximum percentage error is given by:

% error = [(ΔΔl/Δl) + (ΔF/F)] × 100

ΔΔl is 2.5x10⁻⁸ m
Δl₁ is 2x10⁻⁸ m
Δl₂ is 4.5x10⁻⁸ m
The average Δl is (Δl₁ + Δl₂)/2 = (2x10⁻⁸ + 4.5x10⁻⁸)/2 = 3.25x10⁻⁸ m

% error ≈ [(2.5 x 10⁻⁸ m / 3.25 x 10⁻⁸ m) + (2g / mg)] × 100

Since we don't know the initial mass 'm', we can't determine the exact percentage error. However, if we assume the additional mass is small compared to the initial mass, the second term becomes negligible and the percentage error is approximately [(2.5/3.25) x 100] ≈ 77%

This calculation is an approximation and depends on the assumptions made about the initial load and error propagation. Without knowing the initial mass (m), a precise calculation of the percentage error in Young's modulus isn't possible. A more sophisticated error analysis would be required for a completely accurate answer. However, based on the given information and reasonable approximations the closet answer is 7 or 8. Therefore the most likely answer is 8 considering the approximation in calculation.