Mechanical Properties of Solids

JEE Main level, single correct, medium

118 of 118

Question 118 of 118

Two wires A and B are made of same material having ratio of lengths LALB=13\frac{L_A}{L_B}=\frac{1}{3} and their diameters ratio dAdB=2\frac{d_A}{d_B}=2. If both the wires are stretched using same force, what would be the ratio of their respective elongations?

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Answer: (B) 1:121: 12

LALB=13\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}=\frac{1}{3} and dAdB=2\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}=2
ΔLA=FALAAAYA\Delta \mathrm{L}_{\mathrm{A}}=\frac{\mathrm{F}_{\mathrm{A}} \mathrm{L}_{\mathrm{A}}}{\mathrm{A}_{\mathrm{A}} \mathrm{Y}_{\mathrm{A}}} and ΔLB=FBLBABYB\Delta \mathrm{L}_{\mathrm{B}}=\frac{\mathrm{F}_{\mathrm{B}} \mathrm{L}_{\mathrm{B}}}{\mathrm{A}_{\mathrm{B}} \mathrm{Y}_{\mathrm{B}}}
Given, FA=FBF_A=F_B and YA=YBY_A=Y_B
ΔLAΔLB=FALAAAYAFBLBABYB=(LALB)(ABAA)\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)
ΔLAΔLB=(LALB)(π4 dB2π4 dA2)=(LALB)(dBdA)2\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
ΔLAΔLB=(13)(12)2=112\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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