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

In both the mesurement of vernier, main scale reading (MSR) is `3.20xx10^(-2)m`.
Least count is vernier `LC= 1.0xx10^(-5)m`
In first case 20th division of vernier matches with one main scale division.
First reading of vernier
`L_(1)= MSR + nxxLC= 3.20xx10^(-2)+20xx1.0xx10^(-5)= 3.220xx10^(-2)m`
In second case 45th division of vernier matches with one main scale division.
Second reading of vernier
`L_(2)= MSR + n xx LC =3.20xx10^(-2)+45xx1.0xx10^(-5)= 3.245xx10^(-2)m`
Extension in the wire due to 2kg load can be written as :
`l =L_(2)-L_(1)=0.025xx10^(-2)m`
Error in the measurement of `L_(1)" and "L_(2)` both is same as least count `(1xx10^(-5)m)` hence for their difference errors will be added. So error in the measurement of elongation will be two times the LC.
`triangle l = 2xx10^(-5)m`
Young.s modulus can be written as follows :
`Y=(FL)/(Al)`
Here, force F, length of wire L and area of cross section A are given to be exact , hence error in the calculation of Y can be written as follows :
`(triangleY)/(Y)xx100=(triangle l)/(l)xx100=(2xx10^(-5))/(0.025xx10^(-2))xx100=8%`
Hence answer is 8.