Mechanical Properties of Solids

JEE Main level, numerical answer, medium

37 of 118

Question 37 of 118

A steel rod has a radius of 20mm20mm and a length of 2.0m.2.0m. A force of 62.8kN62.8kN stretches it along its length. Young's modulus of steel is 2.0×1011Nm−22.0\times {10}^{11}N{m}^{-2}. The longitudinal strain produced in the wire is _________\times {10}^{-5}.

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Answer: 25

Young's modulus is given as
Y=Y=StressStrain\dfrac{Stress}{Strain}=FA∆LL=\dfrac{\dfrac{F}{A}}{\dfrac{∆L}{L}}
The given data is
Stress=62.8kN=62.8kN
r=20×10−3mr=20\times {10}^{-3}m
Y=2×1011Nm−2Y=2\times {10}^{11}N{m}^{-2}
Thus,
Strain=FA×Y⇒∆LL=62.8×103Nπ×(20×10−3m)2×(2×1011Nm−2)=25×10−5NStrain=\dfrac{F}{A\times Y} \Rightarrow \dfrac{∆L}{L}=\dfrac{62.8\times {10}^{3}N}{\pi \times {\left(20\times {10}^{-3}m\right)}^{2}\times \left(2\times {10}^{11}N{m}^{-2}\right)}=25\times {10}^{-5}N

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