Question

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

Answer 1

Given, length of steel wire, $L_1=4.7m$
Area of cross section of the steel wire, $A_1=3\times {10}^{-5}{m}^2$
Let change in length of steel wire be $\Delta L_1$ and force applied is $F$.
Young’s modulus of the steel wire,

$Y_1=\frac{\frac{F}{A_1}}{\frac{\Delta L_1}{L_1}}$

$=\frac{F}{A_1}\times \frac{L_1}{\Delta L_1}$

$=\frac{F}{\Delta L_1}\times \frac{4.7}{3\times {10}^{-5}}$ ...$\left(i\right)$

Given, length of copper wire, $L_2=3.5m$
Area of cross section of the copper wire, $A_2=4\times {10}^{-5}{m}^2$
And both wire stretches by same amount, so change in length of copper wire will also be $\Delta L_1$.
Force applied in both the cases is the same,
So, Force applied on copper wire be $F$.
Young’s modulus of the copper wire,

$Y_2=\frac{\frac{F}{A_2}}{\frac{\Delta L_1}{L_2}}$

$=\frac{F}{A_2}\times \frac{L_2}{\Delta L_1}$

$=\frac{F}{\Delta L_1}\times \frac{3.5}{4\times {10}^{-5}}$ ...$\left(ii\right)$
From equation (i) and (ii)
Therefore, $\frac{Y_1}{Y_2}=1.79\approx 1.8$