When a wire of length 10 m is subjected to a force of 100 N along its length, the lateral strain produced is `0.01 xx 10^(-3)` . The Poisson's ratio was found to be 0.4. If the area of cross-section of wire is 0.025 `m^2`, its Young's modulus is

Options

A. `1.6xx10^8 N m^(-2)`
B. `2.5xx10^10 N m^(-2)`
C. `1.25 xx 10^11 N m^(-2)`
D. `16xx10^9 N m^(-2)`

Answer

A

Explanation

Poisson.s ratio = `"Lateral strain"/"Longitudinal strain"`
Longitudinal strain = `"Lateral strain"/"Poisson.s ratio"`
`= (0.01xx10^(-3))/0.4` …(i)
Young.s modulus , `Y="Normal stress"/"Longitudinal strain"` (Using (i))
`Y=F/(Axx((0.01xx10^(-3))/0.4))`
`=(100xx0.4)/(0.025xx0.01xx10^(-3))N m^(-2) = 1.6xx10^8 N m^(-2)`