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 m`. 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. `1.25xx10^11 N m^-2`
D. `16xx10^9 N m^-2`

Answer

A

Explanation

Poisson,s ratio,`mu=("Lateral strain")/("Logitudinal strain")`
`0.4=(0.01xx10^(-3))/(Deltal//l)`
`therefore (Deltal)/l =4xx10^(-5)`
Young's modulus,`E=(Fl)/(Adeltal) = F/A xx l/((Deltal)/l)`
`100/0.025xx4xx10^(4)=1.6xx10^(8) n/M^(2)`