Question

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

• A. $1.5:2$
• B. $1.8:2$
• C. $1.5:1$
• D. $1.8:1$

Answer 1

The correct option is D $1.8:1$

For steel wire
$A_1=3\times {10}^{-5}{\text{m}}^2,l_1=4.7\text{m},\Delta l_1=\Delta l,F_1=F$
For copper wire,
$A_2=4\times {10}^{-5}{\text{m}}^2,l_2=3.5\text{m},\Delta l_2=\Delta l,F_2=F$
Let $Y_1$ and $Y_2$ be the Young's modulus of the steel wire and copper wire, respectively.
Then, $Y_1=\frac{F_1}{A_1}\times \frac{l_1}{\Delta l_1}=\frac{F}{3\times {10}^{-5}}\times \frac{4.7}{\Delta l}....\left(1\right)$
and
$Y_2=\frac{F_2\times l_2}{A_2\times \Delta l_2}=\frac{F\times 3.5}{4\times {10}^{-5}\times \Delta l}....\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we get
$\frac{Y_1}{Y_2}=\frac{4.7\times 4.0\times {10}^{-5}}{3.5\times 3.0\times {10}^{-5}}=1.8$
So, $Y_1:Y_2=1.8:1$
Thus, option (d) is the correct answer.