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One end of a metal wire is fixed to a ceiling and a load of $2 \mathrm{~kg}$ hangs from the other end. A similar wire is attached to the bottom of the load and another load of $1 \mathrm{~kg}$ hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be____________.
[Area of cross section of wire $=0.005 \mathrm{~cm}^2$, $\mathrm{Y}=2 \times 10^{11}\ \mathrm{Nm}^{-2}$ and $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$
$5$
$10$
$8$
$3$
Solution

$\Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}}$
$\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\mathrm{F}}{\mathrm{AY}}$
$\frac{\frac{\Delta \mathrm{L}_1}{\mathrm{~L}_1}}{\frac{\Delta \mathrm{L}_2}{\mathrm{~L}_2}}=\frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{30}{10}=3$