Question

A steel wire of length 4.7 m and cross-sectional area $3.0\times {10}^{-5}{m}^2$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0\times {10}^{-5}{m}^2$ under a given load. What is the ratio of the Youngs modulus of steel to that of copper ?

Answer 1

Young's Modulus for steel, $Y_s$,
$Y_s=\frac{F\times L_s}{A_s\times \delta l}$
Young's Modulus for copper, $Y_c$,
$Y_c=\frac{F\times L_c}{A_c\times \delta l}$
$\therefore \frac{Y_s}{Y_c}=\frac{L_s\times A_c}{L_c\times A_s}$ $\because$F and $\delta l$ are constant
$\therefore \frac{Y_s}{Y_c}=\frac{4.7\times 4.0\times {10}^{-5}}{3.5\times 3.0\times {10}^{-5}}$
$\therefore \frac{Y_s}{Y_c}=1.79:1$