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JEE Advanced 2023 Paper 2 · Physics · Fluid Mechanics, Elasticity and Surface Tension · medium

A steel wire of length 0.50 m is stretched between two rigid clamps at a temperature at which it has exactly zero tension. It is then cooled…

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A steel wire of length 0.50 m0.50\ \mathrm{m} is stretched between two rigid clamps at a temperature at which it has exactly zero tension. It is then cooled uniformly by ΔT\Delta T kelvin, and the clamps stay fixed so the wire becomes taut. For steel take Young's modulus Y=2.0×1011 N m−2Y=2.0\times10^{11}\ \mathrm{N\,m^{-2}}, coefficient of linear expansion α=1.2×10−5 K−1\alpha=1.2\times10^{-5}\ \mathrm{K^{-1}} and density ρ=8000 kg m−3\rho=8000\ \mathrm{kg\,m^{-3}} (assume the wire's length, area and density are unchanged by the small strain). The fundamental frequency of transverse vibration of the wire is measured to be 60 Hz60\ \mathrm{Hz}. Find ΔT\Delta T in kelvin.
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Answer: 12

Because the clamps are rigid, the thermal contraction αΔT L\alpha\Delta T\,L is prevented by an elastic extension, so the strain is αΔT\alpha\Delta T and the tension is T=YAαΔTT=YA\alpha\Delta T. Linear mass density is μ=ρA\mu=\rho A, so the wave speed is v=T/μ=YαΔTρv=\sqrt{T/\mu}=\sqrt{\dfrac{Y\alpha\Delta T}{\rho}}, independent of the cross-sectional area. The fundamental is f=v2Lf=\dfrac{v}{2L}, so v=2Lf=2×0.50×60=60 m s−1v=2Lf=2\times0.50\times60=60\ \mathrm{m\,s^{-1}}. Then ΔT=ρv2Yα=8000×36002.0×1011×1.2×10−5=2.88×1072.4×106=12 K\Delta T=\dfrac{\rho v^2}{Y\alpha}=\dfrac{8000\times3600}{2.0\times10^{11}\times1.2\times10^{-5}}=\dfrac{2.88\times10^{7}}{2.4\times10^{6}}=12\ \mathrm{K}.
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