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QPages 12-14: Q9: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q9 Solution: A resistance thermometer reads , , and at the ice point 0�C),…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often read $R_\theta = R_0 1 + \alpha\theta + \beta\theta^2$ as $R_\theta = R_0 + \alpha\theta + \beta\theta^2$, forgetting that $R_0$ multiplies the entire bracket. Use the correct form $R_\theta = R_0(1+\alpha\theta+\beta\theta^2)$ before setting up the equations.

A resistance thermometer reads R=20.0 ΩR = 20.0 \ \Omega, 27.5 Ω27.5 \ \Omega, and 50.0 Ω50.0 \ \Omega at the ice point 0�C), the steam point 100�C) and the zinc point 420�C), respectively. Assuming that the resistance varies with temperature as Rθ=R01+αθ+βθ2)R_\theta = R_0 1 + \alpha\theta + \beta\theta^2), find the values of R0R_0, α\alpha and β\beta. Here θ\theta represents the temperature on the Celsius scale.

Step-by-Step Verified Solution

**Answer:** R0=20.0 ΩR_0 = 20.0\ \Omega, α=3.8×10−3\alpha = 3.8 \times 10^{-3} /�C, β=−5.6×10−7\beta = -5.6 \times 10^{-7} /�C_ **Step-by-Step Explanation:** Given: R(0)=20.0R(0) = 20.0, R(100)=27.5R(100) = 27.5, R(420)=50.0R(420) = 50.0. From R(0)R(0): R0=20.0 ΩR_0 = 20.0\ \Omega. At 100�C: 27.5=20(1+100α+10000β)27.5 = 20(1 + 100\alpha + 10000\beta) At 420�C: 50.0=20(1+420α+176400β)50.0 = 20(1 + 420\alpha + 176400\beta) Solving the two equations: From first: 0.375=100α+10000β0.375 = 100\alpha + 10000\beta From second: 1.5=420α+176400β1.5 = 420\alpha + 176400\beta Multiply first by 4.2: 1.575=420α+42000β1.575 = 420\alpha + 42000\beta Subtract from second: −0.075=134400β⇒β=−5.58×10−7-0.075 = 134400\beta \Rightarrow \beta = -5.58 \times 10^{-7} Then 100α=0.375−10000(−5.58×10−7)=0.375+0.00558=0.3806⇒α=3.806×10−3100\alpha = 0.375 - 10000(-5.58 \times 10^{-7}) = 0.375 + 0.00558 = 0.3806 \Rightarrow \alpha = 3.806 \times 10^{-3}

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