Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is $2 \,cm$, then how much is the elongation in steel and copper wire respectively? Given, $Y_{\text {steel }}=20 \times 10^{11} \,dyne / \ cm ^2$, $Y_{\text {copper }}=12 \times 10^{11} \,dy ne / \ cm ^2$
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Let us say that elongation in copper $=x$
Than elongation in steel $=2-x$
We know
$\frac{F L}{A Y}=\Delta x$
$\because F, A, L$ are same only material is different We can say
$\frac{1}{Y} \propto \Delta x$
$\frac{Y_2}{Y_1}=\frac{\Delta x_1}{\Delta x_2}$ $\left\{\begin{array}{l}\text { Where } \\ Y_2=Y_{\text {steel }} \\ Y_1=Y_{\text {copper }} \\ \Delta x_1=\text { elongation in copper }=x \\ \Delta x_2=2-x\end{array}\right.$
Substituting values
$\frac{20 \times 10^{11}}{12 \times 10^{11}}=\frac{x}{2-x}$
$x=1.25 \,cm$
So $\Delta x_{\text {copper }}=1.25 \,cm , \Delta x_{\text {sleel }}=0.75 \,cm$
Than elongation in steel $=2-x$
We know
$\frac{F L}{A Y}=\Delta x$
$\because F, A, L$ are same only material is different We can say
$\frac{1}{Y} \propto \Delta x$
$\frac{Y_2}{Y_1}=\frac{\Delta x_1}{\Delta x_2}$ $\left\{\begin{array}{l}\text { Where } \\ Y_2=Y_{\text {steel }} \\ Y_1=Y_{\text {copper }} \\ \Delta x_1=\text { elongation in copper }=x \\ \Delta x_2=2-x\end{array}\right.$
Substituting values
$\frac{20 \times 10^{11}}{12 \times 10^{11}}=\frac{x}{2-x}$
$x=1.25 \,cm$
So $\Delta x_{\text {copper }}=1.25 \,cm , \Delta x_{\text {sleel }}=0.75 \,cm$
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