During Searle's experiment, zero of the Vernier scale lies between `3.20 xx 10^(-2) m` and `3.25 xx 10^(-2) m` of the main scale. The `20^(th)` division of the Vernier scale exactly coincide 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) m` and `3.25 xx 10^(-2) m` of the main scale but now the `45^(th)` division of Vernier scale coincides with one of the main scale divisions. The lengths of the thin metallic wire is 2m and its cross-sectional area 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)/((Delta l)/l)`
`(dy)/y=(dF)/F+(dA)/A+(dl)/l+(d(Deltal))/(Deltal)`
Only `Delta l` calculations have error `%` error of
`y=(dy)/yxx100=(d Delta l)/(Delta l)xx100=(1xx10^(-5))/(25xx10^(-5))xx100=4 %`