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Two metallic wires $P$ and $Q$ have same volume and are made up of same material. If their area of cross sections are in the ratio $4: 1$ and force $F_1$ is applied to $\mathrm{P}$, an extension of $\Delta l$ is produced. The force which is required to produce same extension in $Q$ is $\mathrm{F}_2$.The value of $\frac{\mathrm{F}_1}{\mathrm{~F}_2}$ is__________.
$16$
$14$
$20$
$50$
Solution
$ \mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell} $
$ \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} $
$ \mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}} $
$ \Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^2 \mathrm{Y}}$
$Y$ & $V$ is same for both the wires
$ \Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^2} $
$ \frac{\Delta \ell_1}{\Delta \ell_2}=\frac{\mathrm{F}_1}{\mathrm{~A}_1^2} \times \frac{\mathrm{A}_2^2}{\mathrm{~F}_2} $
$ \Delta \ell_1=\Delta \ell_2 $
$ \mathrm{~F}_1 \mathrm{~A}_2^2=\mathrm{F}_2 \mathrm{~A}_1^2 $
$ \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{A}_1^2}{\mathrm{~A}_2^2}=\left(\frac{4}{1}\right)^2=16$
