Mechanical Properties of Solids

JEE Main level, single correct, medium

40 of 118

Question 40 of 118

A force is applied to a steel wire AA, rigidly clamped at one end. As a result elongation in the wire is 0.2mm0.2mm. If same force is applied to another steel wire BB of double the length and a diameter 2.42.4 times that of the wire AA, the elongation in the wire BB will be (wires having uniform circular cross sections)

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Answer: (D) 6.9×10–2mm6.9\times {10}^{–2}mm

As we know, Y=FA∆llY=\dfrac{F}{\dfrac{A}{\dfrac{∆l}{l}}}
⇒F=YAl∆l\Rightarrow F=\dfrac{YA}{l}∆l
For both cases force applied is the same, therefore
(YA∆ll)1=(YA∆ll)2{\left(\dfrac{YA∆l}{l}\right)}_{1}={\left(\dfrac{YA∆l}{l}\right)}_{2}
⇒∆l2∆l1=A1A2×l2l1\Rightarrow \dfrac{∆{l}_{2}}{∆{l}_{1}}=\dfrac{{A}_{1}}{{A}_{2}}\times \dfrac{{l}_{2}}{{l}_{1}}
⇒∆l20.2=πR2π(2.4R)2×21\Rightarrow \dfrac{∆{l}_{2}}{0.2}=\dfrac{\pi {R}^{2}}{\pi {\left(2.4R\right)}^{2}}\times \dfrac{2}{1}
⇒∆l2=6.9×10−2mm\Rightarrow ∆{l}_{2}=6.9\times {10}^{-2}mm

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