Mechanical Properties of Solids

JEE Main level, single correct, medium

52 of 118

Question 52 of 118

A wire of length LL is hanging from a fixed support. The length changes to L1{L}_{1} and L2{L}_{2} when masses 1kg1kg and 2kg2kg are suspended respectively from its free end. Then the value of LL is equal to

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Answer: (C) 2L1−L22{L}_{1}-{L}_{2}

From Hooke's law, ∆L=FLAY⇒∆L∝F∆L=\dfrac{FL}{AY}\Rightarrow ∆L\propto F. Therefore,
L2−LL1−L=m2gm1g=21\dfrac{{L}_{2}-L}{{L}_{1}-L}=\dfrac{{m}_{2}g}{{m}_{1}g}=\dfrac{2}{1} ⇒L=2L1−L2\Rightarrow L=2{L}_{1}-{L}_{2}

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