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

Here `F =2 kg wt =2xx9.8 N , A=8xx10^(-7) m^(2)` .
Original length `l =2 m`,
Insrease in length,
`Delta l =(3.20xx10^(2) m +45xx1.0xx10^(-5) m) - (3.20xx10^(2)m + 20xx1.0xx10^(-5)m)`
`=25xx1.0xx10^(-5) =25xx10^(-5)m`
`Y=(F)/(A) (l)/(Delta l)`
`Delta Y=(F)/(A) l[-((1)/(Delta l))^(2)]Delta(Delta l)`
`(Delta Y)/(Y) = -(Delta(Delta l))/(Delta l)`
`% error =(Delta Y)/(Y)xx100 =(Delta(Delta l))/(Delta l)xx100 = ((1.0xx10^(-5)))/(25xx10^(-5))xx100 =4%`