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A steel wire of length $4.7 \mathrm{~m}$ and cross-section $3 \times 10^{-5} \mathrm{~m}^2$ stretches by the same amount as a copper wire of length $3.5 \mathrm{~m}$ and cross-section $4 \times 10^{-5} \mathrm{~m}^2$ under a given load. What is the ratio of the Young's modulus of steel to that of copper?
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Verified Answer
For steel wire, $A_1=3 \times 10^{-5} \mathrm{~m}^2, l_1=4.7 \mathrm{~m}$;
For copper wire, $A_2=4 \times 10^{-5} \mathrm{~m}^2, l_2=3.5 \mathrm{~m}$
As $\Delta l_1=\Delta l_2=\Delta l$ and $F_1=F_2=F$
$$
\begin{aligned}
&\therefore \quad Y_1=\frac{F_1 l_1}{A_1 \Delta l_1}=\frac{F}{3 \times 10^{-5}} \times \frac{4.7}{\Delta l} \\
&Y_2=\frac{F_2 l_2}{A_2 \Delta l_2}=\frac{F \times 3.5}{4 \times 10^{-5} \Delta l} \\
&\therefore \quad \frac{Y_1}{Y_2}=\frac{4.7 \times 4 \times 10^{-5}}{3.5 \times 3 \times 10^{-5}}=\frac{18.8}{10.5}=1.8 \\
&\therefore \quad Y_1: Y_2=\frac{18}{10}=\frac{9}{5}
\end{aligned}
$$
For copper wire, $A_2=4 \times 10^{-5} \mathrm{~m}^2, l_2=3.5 \mathrm{~m}$
As $\Delta l_1=\Delta l_2=\Delta l$ and $F_1=F_2=F$
$$
\begin{aligned}
&\therefore \quad Y_1=\frac{F_1 l_1}{A_1 \Delta l_1}=\frac{F}{3 \times 10^{-5}} \times \frac{4.7}{\Delta l} \\
&Y_2=\frac{F_2 l_2}{A_2 \Delta l_2}=\frac{F \times 3.5}{4 \times 10^{-5} \Delta l} \\
&\therefore \quad \frac{Y_1}{Y_2}=\frac{4.7 \times 4 \times 10^{-5}}{3.5 \times 3 \times 10^{-5}}=\frac{18.8}{10.5}=1.8 \\
&\therefore \quad Y_1: Y_2=\frac{18}{10}=\frac{9}{5}
\end{aligned}
$$
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