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PhysicsInteger+4Question 43 of 69

IIT-JEE 2011 · Paper 1, Physics question 43

Steel wire of length 'L' at 40 °C is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from 40 °C to 30 °C to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is 10−510^{-5}/°C, Young's modulus of steel is 1011 N/m210^{11}\ \mathrm{N/m^2} and radius of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of 'm' in kg is nearly

Answer and worked solution

Official answer: 3

Equal and opposite changes. Cooling shortens the wire by LαΔT=10−4LL\alpha\Delta T=10^{-4}L, which must equal the elastic extension mgLAY\dfrac{mgL}{AY}:
mg=10−4×π(10−3)2×1011=31.4 N,m≈3.1 kg.mg=10^{-4}\times\pi(10^{-3})^2\times10^{11}=31.4\ \text{N},\qquad m\approx3.1\ \text{kg}.

Answer: 3

Topics:Thermal stressHooke's law and elastic moduli
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