A steel wire of length `4.7 m` and cross-sectional area `3 xx 10^(-6) m^(2)` stretches by the same amount as a copper wire of length `3.5 m` and cross-sectional area of `4 xx 10^(-6) m^(2)` under a given load. The ratio of Young's modulus of steel to that of copper is

Explanation

Length of the steel wire, `L_(1)` = 4.7 m
Area of cross-section of the steel wire, `A_(1)=3.0xx10^(-5)"m"^(2)`
Length of the copper wire, `L_(2)`= 3.5 m
Area of cross-section of the copper wire, `A_(2)=4.0xx10^(-5)"m"^(2)`
Change in length = `DeltaL_(1)=DeltaL_(2)=DeltaL`
Force applied in both the casses = F
Young's modulus of the steel wire:
`Y_(1)=(F_(1))/(A_(1))=(L_(1))/(DeltaL)`
`" "(Fxx3.5)/(4.0xx10^(-5)xxDeltaL)" "...(ii)`
Dividing (i) by (ii), we get
`(Y_(1))/(Y_(2))=(4.7xx4.0xx10^(-5))/(3.0xx10^(-5)xx3.5)=1.79:1`
The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.