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 $∆l$ is produced. The force which is required to produce same extension in $\mathrm{Q}$ is $F_2$. The value of $\frac{F_1}{ F_2}$ is ______.

The formula for Young's modulus is given by

$\begin{array}{rcl}Y& =& \frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{∆l}{l}}}\\ & =& \frac{Fl}{A∆l}\end{array}$

From above equation, it follows that

$\Delta l=\frac{Fl}{AY} ...\left(1\right)$

Again, volume of the wire is given by

$$\begin{gathered} V=Al \\ \Rightarrow l=\frac{V}{A} \end{gathered}$$

Hence, from equation (1),

$∆l=\frac{FV}{A^{2}Y} ...\left(2\right)$

As, $Y$ and $V$ is the same for both the wires,

$∆l\propto \frac{F}{ {\mathrm{A}}^2}$

Hence, for two wires, it can be written that

$\frac{∆l_1}{∆l_2}=\frac{F_1}{ A_1^2}\times \frac{A_2^2}{ F_2} ...\left(3\right)$

As $∆l_1=∆l_2$, from equation (3), it follows that

$$\begin{gathered} F_1 A_2^2=F_2 A_1^2 \\ \Rightarrow \frac{F_1}{ F_2}=\frac{A_1^2}{ A_2^2} \\ ={\left(\frac{4}{1}\right)}^2 \\ =16 \end{gathered}$$