Two wires A and B are made of same material having ratio of lengths $\frac{L_A}{L_B}=\frac{1}{3}$ and their…
- $1: 6$
- $1: 12$
- $3: 4$
- $1: 3$
Solution
$\Delta \mathrm{L}_{\mathrm{A}}=\frac{\mathrm{F}_{\mathrm{A}} \mathrm{L}_{\mathrm{A}}}{\mathrm{A}_{\mathrm{A}} \mathrm{Y}_{\mathrm{A}}}$ and $\Delta \mathrm{L}_{\mathrm{B}}=\frac{\mathrm{F}_{\mathrm{B}} \mathrm{L}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{B}} \mathrm{Y}_{\mathrm{B}}}$
Given, $F_A=F_B$ and $Y_A=Y_B$
$\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\frac{\frac{\mathrm{F}_{\mathrm{A}} \mathrm{L}_{\mathrm{A}}}{\mathrm{A}_{\mathrm{A}} \mathrm{Y}_{\mathrm{A}}}}{\frac{\mathrm{F}_{\mathrm{B}} \mathrm{L}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{B}} \mathrm{Y}_{\mathrm{B}}}}=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\mathrm{A}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{A}}}\right)$
$\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\frac{\pi}{4} \mathrm{~d}_{\mathrm{B}}^2}{\frac{\pi}{4} \mathrm{~d}_{\mathrm{A}}^2}\right)=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\mathrm{d}_{\mathrm{B}}}{\mathrm{d}_{\mathrm{A}}}\right)^2$
$\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)^2=\frac{1}{12}$
Asked in: JEE Main 2025 (07 Apr Shift 1)
Practice more Mechanical Properties of Solids questions on Aicharya