A steel wire of length 4.7 m and cross-sectional area 3.0 `xx` `10^(-5)"m"^(2)` stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 `xx 10^(-5)"m"^(2)` under a given load. What is the ratio of the ratio of the Young's modulus of steel to that of copper?

Options

A. `1.8`
B. `3.6`
C. `0.6`
D. `8.7`

Answer

A

Explanation

`Y=(F//A)/(/_\l//l)`
So, `(Y_(S))/(Y_(C))=(Fxxl_(s))/(A_(s)xx/_\l)-:F/A_(c)xxl_(c)/(/_\l)`
`(A_(c)l_(s))/(A_(s)lP_(c))=(4x10^(-6)xx4.7)/(3xx10^(-6)xx3.5)=1.8`