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

Given , For steel wire, `A_(1) = 3.0 xx 10^(-5)m^(2), l_(1)=4.7m, Delta l_(1)=Delta l , F_(1) = F`
for copper wire, `A_(2) = 4.0 xx 10^(-5)m^(2), l_(1)=3.5m, Delta l_(2)=Delta l , F_(2) = F`
Let `Y_(1), Y_(2)` be the Young's modulus of steel wire and copper wire respectively.
`:. Y_(1) = (F_1)/(A_1) xx (l_1)/(Delta l_(1)) = (F)/(3.0 xx 10^(-5)) xx (4.7)/(Delta l)`
and `Y_(2) = (F_2)/(A_2) xx (l_2)/(Delta l_(2)) = (F)/(4.0 xx 10^(-5)) xx (3.5)/(Delta l)` ...(ii)
`:. (Y_1)/(Y_2) = (4.7 xx 4 xx 10^(-5))/(3.5 xx 3.0 xx 10^(-5)) = 1.8`
Hence, `Y_(L):Y_(2) = 1:8:1`.