Two wires A and B are of the same materials. Their lengths are in...

Force $F=\frac{Y A \Delta l}{l}$ Substituting area $A=\pi$\left(\frac{d}{2}\right)^{2}=\frac{\pi d^{2}}{4}$So, we have$\quad F=\frac{Y \pi d^{2} \Delta l}{4 l}$F \propto$\frac{d^{2} \Delta l}{l} \left[\because \frac{Y \pi}{4}=\text { constant }\right]$Hence, using Eq. (i), we have$$\begin{array}{l}F_{A} \propto \frac{d_A^{2} \Delta l_{A}}{l_{A}}\\ F_{B} \propto \frac{d_B^{2} \Delta l_{B}}{l_{B}} \end{array}$$From Eqs. (i) and (ii), we get$\frac{F_{A}}{F_{B}}=\frac{d_A^{2} \Delta l_{A}}{l_{A}} \times \frac{l_{B}}{d_B^{2} \Delta l_{B}}$as \Delta l_{A}=\Delta l_{B}=\Delta l Therefore$\frac{F_{A}}{F_{B}}=\frac{d_A^{2}}{d_B^{2}} \times \frac{l_{B}}{l_{A}}$Putting$\frac{d_{A}}{d_{B}}=\frac{2}{1}$and$\frac{l_{B}}{l_{A}}=\frac{2}{1}$we get$\quad F_{A}: F_{B}=8: 1