When a wire of length `10m` is subjected to a force of `100N` along its length, the lateral strain produced is `0.01xx10^(-3)`. The poisson's ratio was found to be `0.4`. If area of cross section of wire is `0.25 m^(2)`, its young's modulus is

Explanation

Poisson's ratio
`sigma = ("lateral of strain")/("longitudinal strian") = (-(Delta r)/(r))/((Delta l)/(l))`
`therefore = (Delta l)/(l) = (-(Delta r)/(r))/(sigma) = (10^(-5))/(0.4) = 25xx10^(-6)`
From
`Y = (Fl)/(A Delta l) = (F)/(A((Delta l)/(l))) = (100)/(25xx10^(-3)xx25xx10^(-6))`
`= 1.6xx10^(8) Nm^(2)`