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 .

Explanation

Observation - `1`
Let weight use is `W_(1)`, extension `l_(1)`
`y = (W_(1)//A)/(l_(1)//L) rArr W_(1) = (yAl_(1))/(L), l_(1) = 3.2 xx 10^(-2) + 20 xx 10^(-5)`
Observation -`2`
Let weight used is `W_(2)` extension `l_(2)`
`y = (W_(2)//A)/(l_(2)//L) rArr W_(1) = (yAl_(2))/(L), l_(1) = 3.2 xx 10^(-2) + 45 xx 10^(-5)`
`W_(2) - W_(1) = (yA)/(L)(l_(2) - l_(1)) rArr y = ((W_(2) - W_(1))//L)/(yA(l_(2) - l_(1)))`
`((Deltay)/(y))_(mass) = (Deltal_(2) + Deltal_(1))/(l_(2) - l_(1)) = (2 xx 10^(-5))/(25 xx 10^(-5))`
`((Deltay)/(y))_(max) xx 100% = (2)/(25) xx 100% = 8%`