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The radii and Young's moduli of two uniform wires $A$ and $B$ are in the ratio $2: 1$ and $1: 2$ respectively. Both wires are subjected to the same longitudinal force. If the increase in length of the wire $A$ is one percent, the percentage increase in length of the wire $B$ is
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The correct answer is:
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Increase in length, $\Delta l=\frac{F l}{A Y}$
or $\Delta l=\frac{F l}{\pi r^2 y}$
Percent increase in length,
$\Delta x=\frac{\Delta l}{l} \times 100=\frac{F}{\pi r^2 y}$
Here, same longitudinal force is applied.
So,
$\begin{aligned}
& \frac{\Delta x_1}{\Delta x_2}=\left(\frac{r_2}{r_1}\right)^2 \cdot\left(\frac{y_2}{y_1}\right) \\
& \frac{1}{\Delta x_2}=\left(\frac{1}{2}\right)^2 \cdot\left(\frac{2}{1}\right)=\frac{1}{4} \times \frac{2}{1} \\
& \frac{1}{\Delta x_2}=\frac{1}{2} \\
& \Delta x_2=1 \times 2=2 \%
\end{aligned}$
or $\Delta l=\frac{F l}{\pi r^2 y}$
Percent increase in length,
$\Delta x=\frac{\Delta l}{l} \times 100=\frac{F}{\pi r^2 y}$
Here, same longitudinal force is applied.
So,
$\begin{aligned}
& \frac{\Delta x_1}{\Delta x_2}=\left(\frac{r_2}{r_1}\right)^2 \cdot\left(\frac{y_2}{y_1}\right) \\
& \frac{1}{\Delta x_2}=\left(\frac{1}{2}\right)^2 \cdot\left(\frac{2}{1}\right)=\frac{1}{4} \times \frac{2}{1} \\
& \frac{1}{\Delta x_2}=\frac{1}{2} \\
& \Delta x_2=1 \times 2=2 \%
\end{aligned}$
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