NECO 2003 Physics · Question 17

A platinum-resistance thermometer has a resistance of 4Ω at 0°C and 12Ω at 100°C. Assuming that the resistance…

Question

A platinum-resistance thermometer has a resistance of 4Ω at 0°C and 12Ω at 100°C. Assuming that the resistance changes uniformly with temperature. Calculate the resistance of the thermometer when the temperature is 45°C.

A.6.0Ω
B.6.42Ω
C.7.6Ω
D.8.4Ω
E.16.0Ω

Answer: C. 7.6Ω

Explanation written with AI and checked against the answer key

Analyze the question The resistance changes uniformly (linearly) with temperature. We know R at 0°C and at 100°C, and we need R at 45°C. Concept For a linear change, R_θ = R₀(1 + αθ), or equivalently, the resistance rises by the same amount for each degree. Solution Change in resistance over the fixed interval: 12 Ω − 4 Ω = 8 Ω over 100°C Rate of change = 8/100 = 0.08 Ω per °C Rise from 0°C to 45°C = 0.08 × 45 = 3.6 Ω R₄₅ = 4 + 3.6 = 7.6 Ω Check using the ratio formula: (R₄₅ − R₀)/(R₁₀₀ − R₀) = 45/100 (R₄₅ − 4)/8 = 0.45 R₄₅ = 4 + 3.6 = 7.6 Ω ✓ Answer: C. 7.6 Ω Exam tip For "uniform change" questions, use the fraction method: (R − R₀)/(R₁₀₀ − R₀) = θ/100. Compute the fraction of the temperature interval (45/100), then apply the same fraction to the resistance interval. Be careful not to forget to add back R₀. Forgetting it gives 3.6 Ω, which isn't an option, so that's a good self-check.

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