In the ratio of diameters, lengths and Young's modulus of steel and...

We know that $\text{ Young's modulus }$$\begin{aligned} \gamma& =\frac{\mathrm{FL}}{\Delta 4 \mathrm{~L}}\\ & =\frac{4 \mathrm{FL}}{\pi \mathrm{D}^2 4 \mathrm{~L}}\\ \Rightarrow \quad \Delta \mathrm{L}& =\frac{4 \mathrm{FL}}{\pi \mathrm{D}^2 \mathrm{~L}} \end{aligned}$$$According to questions.$$\begin{aligned}& \frac{\Delta \mathrm{L}_{\mathrm{S}}}{\Delta \mathrm{L}_{\mathrm{C}}}=\frac{\mathrm{F}_{\mathrm{S}}}{\mathrm{F}_{\mathrm{C}}} \cdot \frac{\mathrm{L}_{\mathrm{S}}}{\mathrm{L}_{\mathrm{C}}} \cdot \frac{\mathrm{D}_{\mathrm{C}}^2 \mathrm{~L}_{\mathrm{S}}}{\mathrm{D}_{\mathrm{S}}^2 \mathrm{~L}_{\mathrm{C}}}\\ & \text{ Here } \mathrm{F}_{\mathrm{S}}=(5 \mathrm{~m}+2 \mathrm{~m}) \mathrm{g}=7 \mathrm{mg}\\ & \mathrm{F}_{\mathrm{C}}=5 \mathrm{mg},=?, \frac{\mathrm{D}_{\mathrm{S}}}{\mathrm{D}_{\mathrm{C}}}=P, \\ & \frac{\mathrm{L}_{\mathrm{S}}}{\mathrm{L}_{\mathrm{C}}}=\mathrm{S} \end{aligned}$$Putting in eq. (ii)$$\begin{aligned}\frac{\Delta \mathrm{L}_{\mathrm{S}}}{\Delta \mathrm{L}_{\mathrm{C}}}& =\frac{7 \mathrm{mg}}{5 \mathrm{mg}} \timesq+\left(\frac{1}{\mathrm{P}}\right)^2\left(\frac{1}{\mathrm{~S}}\right)\\ & =\frac{7 q}{5 p^2 \mathrm{~S}} \end{aligned}$$