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QPages 12-14: Q8: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q8 Solution: A platinum resistance thermometer reads 0� when its resistance…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently plug $T=100\frac{R-R_0}{R_{100}-R_0}$ into any resistance‑temperature problem, assuming the relation is perfectly linear; however, platinum’s $R$‑$T$ curve is only approximately linear, so the formula is valid only when the problem explicitly states a linear calibration between the two given points.

A platinum resistance thermometer reads 0� when its resistance is 80 Ω\Omega and 100� when its resistance is 90 Ω\Omega. Find the temperature at the platinum scale at which the resistance is 86 Ω\Omega.

Step-by-Step Verified Solution

**Answer:** TPt=60�PtT_{Pt} = 60�\text{Pt} **Step-by-Step Explanation:** R0=80 ΩR_0 = 80\ \Omega, R100=90 ΩR_{100} = 90\ \Omega, R=86 ΩR = 86\ \Omega.
TPt=100×R−R0R100−R0=100×86−8090−80=100×610=60�PtT_{Pt} = 100 \times \frac{R - R_0}{R_{100} - R_0} = 100 \times \frac{86 - 80}{90 - 80} = 100 \times \frac{6}{10} = 60�\text{Pt}

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