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

Answer

`4`

Explanation

Using searle's method young modules is calculated
`y=(F//A)/((Deltal)/(l))`
`(dy)/(y)=(dF)/(F)+(dA)/(A)+(dl)/(l)+(d(Deltal))/(Deltal)`
Only `Delta l` calculations have error % error os
`y=(dy)/(y)xx100=(d Deltal)/(Deltal)xx100`
`=(1xx10^(-5))/(25xx10^(-5))xx100=4%`
`(dy)/(y)xx100=(d Deltal)/(Deltal)xx100=(1xx10^(-5))/(25xx10^(-5))xx100=4%`