JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2025
- Easy
- MCQ
Two wires A and B are made of same material having ratio of lengths \(\frac{L_A}{L_B}=\frac{1}{3}\) and their diameters ratio \(\frac{d_A}{d_B}=2\). If both the wires are stretched using same force, what would be the ratio of their respective elongations?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(1: 12\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}=\frac{1}{3}\) and \(\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}=2\)
\(\Delta \mathrm{L}_{\mathrm{A}}=\frac{\mathrm{F}_{\mathrm{A}} \mathrm{L}_{\mathrm{A}}}{\mathrm{A}_{\mathrm{A}} \mathrm{Y}_{\mathrm{A}}}\) 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 \(Y_A=Y_B\)
\(\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\frac{\frac{\mathrm{F}_{\mathrm{A}} \mathrm{L}_{\mathrm{A}}}{\mathrm{A}_{\mathrm{A}} \mathrm{Y}_{\mathrm{A}}}}{\frac{\mathrm{F}_{\mathrm{B}} \mathrm{L}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{B}} \mathrm{Y}_{\mathrm{B}}}}=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\mathrm{A}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{A}}}\right)\)
\(\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\frac{\pi}{4} \mathrm{~d}_{\mathrm{B}}^2}{\frac{\pi}{4} \mathrm{~d}_{\mathrm{A}}^2}\right)=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\mathrm{d}_{\mathrm{B}}}{\mathrm{d}_{\mathrm{A}}}\right)^2\)
\(\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)^2=\frac{1}{12}\)
\(\Delta \mathrm{L}_{\mathrm{A}}=\frac{\mathrm{F}_{\mathrm{A}} \mathrm{L}_{\mathrm{A}}}{\mathrm{A}_{\mathrm{A}} \mathrm{Y}_{\mathrm{A}}}\) 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 \(Y_A=Y_B\)
\(\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\frac{\frac{\mathrm{F}_{\mathrm{A}} \mathrm{L}_{\mathrm{A}}}{\mathrm{A}_{\mathrm{A}} \mathrm{Y}_{\mathrm{A}}}}{\frac{\mathrm{F}_{\mathrm{B}} \mathrm{L}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{B}} \mathrm{Y}_{\mathrm{B}}}}=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\mathrm{A}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{A}}}\right)\)
\(\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\frac{\pi}{4} \mathrm{~d}_{\mathrm{B}}^2}{\frac{\pi}{4} \mathrm{~d}_{\mathrm{A}}^2}\right)=\left(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}\right)\left(\frac{\mathrm{d}_{\mathrm{B}}}{\mathrm{d}_{\mathrm{A}}}\right)^2\)
\(\frac{\Delta \mathrm{L}_{\mathrm{A}}}{\Delta \mathrm{L}_{\mathrm{B}}}=\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)^2=\frac{1}{12}\)
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