A 20 N stone is suspended from a wire and its length changes by `1%`. If the Young's modulus of the material of wire is `2xx10^(11)N//m^(2)`, then the area of cross-section of the wire is `2xx10^(11)N//m^(2)`, then the area of cross-section of the wire will be

Options

A. `10^(-3)mm^(2)`
B. `10^(-2)mm^(2)`
C. `10^(-1)mm^(2)`
D. `1 mm^(2)`

Answer

B

Explanation

`A=(FL)/(yl)=(20xx100)/(2xx10^(11) xx 1)=10^(-8)m^(2)=10^(-2)mm^(2)`.