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 _______.

\( \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \)

\( \mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}} \)

\( \Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^2 \mathrm{Y}}\)

\(Y\) & \(V\) is same for both the wires

\( \Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^2} \)

\( \frac{\Delta \ell_1}{\Delta \ell_2}=\frac{\mathrm{F}_1}{\mathrm{~A}_1^2} \times \frac{\mathrm{A}_2^2}{\mathrm{~F}_2} \)

\( \Delta \ell_1=\Delta \ell_2 \)

\( \mathrm{~F}_1 \mathrm{~A}_2^2=\mathrm{F}_2 \mathrm{~A}_1^2 \)

\( \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{A}_1^2}{\mathrm{~A}_2^2}=\left(\frac{4}{1}\right)^2=16\)