Two wires A and B are made of same material having ratio of lengths L_A L_B= 1 3 and their diameters ratio d_A d_B=2. If both the wires are stretched…

$\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}=\frac{1}{3}$LB​LA​​=31​ and $\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}=2$dB​dA​​=2

$\Delta \mathrm{L}_{\mathrm{A}}=\frac{\mathrm{F}_{\mathrm{A}} \mathrm{L}_{\mathrm{A}}}{\mathrm{A}_{\mathrm{A}} \mathrm{Y}_{\mathrm{A}}}$ΔLA​=AA​YA​FA​LA​​ and $\Delta \mathrm{L}_{\mathrm{B}}=\frac{\mathrm{F}_{\mathrm{B}} \mathrm{L}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{B}} \mathrm{Y}_{\mathrm{B}}}$ΔLB​=AB​YB​FB​LB​​

Given, $F_A=F_B$FA​=FB​ and $Y_A=Y_B$YA​=YB​

$$\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\frac{\frac{\mathrm{F}_{\mathrm{A}} \mathrm{L}_{\mathrm{A}}}{\mathrm{A}_{\mathrm{A}} \mathrm{Y}_{\mathrm{A}}}}{\frac{\mathrm{F}_{\mathrm{B}} \mathrm{L}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{B}} \mathrm{Y}_{\mathrm{B}}}}=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\mathrm{A}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{A}}}\right)$$ΔLB​ΔLA​​=AB​YB​FB​LB​​AA​YA​FA​LA​​​=(LB​LA​​)(AA​AB​​)

$$\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\frac{\pi}{4} \mathrm{~d}_{\mathrm{B}}^2}{\frac{\pi}{4} \mathrm{~d}_{\mathrm{A}}^2}\right)=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\mathrm{d}_{\mathrm{B}}}{\mathrm{d}_{\mathrm{A}}}\right)^2$$ΔLB​ΔLA​​=(LB​LA​​)(4π​ dA2​4π​ dB2​​)=(LB​LA​​)(dA​dB​​)2

$$\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)^2=\frac{1}{12}$$ΔLB​ΔLA​​=(31​)(21​)2=121​