During Searle's experiment zero of the Vernieer 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 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 `3.30xx 10^(-2)m` and `3.25 xx 10^(-2)m` of the main scale but now the `45^(th)` division of Vernier scale coincies with one of the main scale division THe length 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 %`