Two metallic wires P and Q have same volume and are made up of same...

The formula for Young's modulus is given by

$\begin{cases}Y & = & \dfrac{\dfrac{F}{A}}{\dfrac{∆ l}{l}} \\ & = & \dfrac{Fl}{A ∆ l}\end{cases}$

From above equation, it follows that

$\Deltal = \dfrac{Fl}{AY} . . . \left(1\right)$

Again, volume of the wire is given by

$V = Al ⇒ l = \dfrac{V}{A}$

Hence, from equation (1),

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

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

$∆ l ∝ \dfrac{F}{{A}^{2}}$

Hence, for two wires, it can be written that

$\dfrac{∆ {l}_{1}}{∆ {l}_{2}} = \dfrac{{F}_{1}}{{A}_{1}^{2}} \times {A}_{2}^{2}{F}_{2} . . . \left(3\right)$

As $∆ {l}_{1} = ∆ {l}_{2}$, from equation (3), it follows that

${F}_{1}{A}_{2}^{2} = {F}_{2}{A}_{1}^{2} ⇒ \dfrac{{F}_{1}}{{F}_{2}} = {A}_{1}^{2}{A}_{2}^{2} = {\left(\dfrac{4}{1}\right)}^{2} = 16$