Mechanical Properties of Solids

JEE Main level, single correct, medium

46 of 118

Question 46 of 118

The force required to stretch a wire of cross-section 1cm21{cm}^{2} to double its length will be :
(Given Yong's modulus of the wire =2×1011Nm−2=2\times {10}^{11}N{m}^{-2})

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Answer: (C) 2×107N2\times {10}^{7}N

The wire is stretched to double its length. Therefore, if the length of the wore was ll, change in the length ∆l=l∆l=l.
Applying Hooke's law,
Y=stressstrain=FlA∆l⇒F=YA∆ll=YA⇒F=2×1011×1×10−4=2×107NY=\dfrac{\text{stress}}{\text{strain}}=\dfrac{Fl}{A∆l} \Rightarrow F=\dfrac{YA∆l}{l}=YA \Rightarrow F=2\times {10}^{11}\times 1\times {10}^{-4}=2\times {10}^{7}N
=2×1011×10−4(2l−ll)=2\times {10}^{11}\times {10}^{-4}\left(\dfrac{2l-l}{l}\right)
=2×107N=2\times {10}^{7}N

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