Mechanical Properties of Solids

JEE Main level, numerical answer, medium

111 of 118

Question 111 of 118

The elastic potential energy stored in a steel wire of length 20m20m stretched through 2cm2cm is 80J.80J. The cross sectional area of the wire is _____ mm2.{mm}^{2}. (Given, Y=2.0×1011Nm–2Y=2.0\times {10}^{11}N{m}^{–2})

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Answer: 40

The formula to calculate the Young's modulus of the material is given by
Y=stressstrain=stress∆lL...(1)\begin{matrix}Y & = & \dfrac{stress}{strain} \\ & = & \dfrac{stress}{\dfrac{∆l}{L}}...\left(1\right)\end{matrix}
The formula to calculate the potential energy (U)\left(U\right) stored into the wire is given by
U=12×stress×strain×volume=12×stress×∆lL×A×L=12×stress×A×∆l...(2)\begin{matrix}U & = & \dfrac{1}{2}\times stress\times strain\times volume \\ & = & \dfrac{1}{2}\times stress\times \dfrac{∆l}{L}\times A\times L \\ & = & \dfrac{1}{2}\times stress\times A\times ∆l...\left(2\right)\end{matrix}
From equation (1) and equation (2), it can be written that
U=12×Y×(∆l)2L×A...(3)U=\dfrac{1}{2}\times Y\times \dfrac{{\left(∆l\right)}^{2}}{L}\times A...\left(3\right)
Substitute the values of the known parameters into equation (3) and solve to calculate the required cross-sectional area of the wire.
80J=12×2.0×1011Nm−2×(0.02m)220m×A⇒A=2×80J×202.0×1011×4×10−4Nm−1=4×10−5m2×106mm21m2=40mm2\begin{matrix}80J & = & \dfrac{1}{2}\times 2.0\times {10}^{11}{Nm}^{-2}\times \dfrac{{\left(0.02m\right)}^{2}}{20m}\times A \\ & \Rightarrow & A=\dfrac{2\times 80J\times 20}{2.0\times {10}^{11}\times 4\times {10}^{-4}{Nm}^{-1}} \\ & = & 4\times {10}^{-5}{m}^{2}\times \dfrac{{10}^{6}{mm}^{2}}{1{m}^{2}} \\ & = & 40{mm}^{2}\end{matrix}

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