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 $P$, 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 $\frac{F_1}{F_2}$ is $\underline{\hspace{2cm}}$.