Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# A Resistance Thermometer Reads R = 20.0 ω, 27.5 ω, and 50.0 ω at the Ice Point (0°C), the Steam Point (100°C) and the Zinc Point (420°C), Respectively. - Physics

A resistance thermometer reads R = 20.0 Ω, 27.5 Ω, and 50.0 Ω 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 = R0 (1 + αθ + βθ2), find the values of R0, α and β. Here θ represents the temperature on the Celsius scale.

#### Solution

Given:
Reading on resistance thermometer at ice point, R0 = 20 Ω
Reading on resistance thermometer at steam point, R100 = 27.5 Ω
Reading on resistance thermometer at zinc point, R420 = 50 Ω
The variation of resistance with temperature in Celsius scale,θ, is given as:

$R_{100} = R_0 \left( 1 + \alpha \theta + \beta \theta^2 \right)$

$\Rightarrow R_{100} = R_0 + R_0 \alpha\theta + R_0 \beta \theta^2$

$\Rightarrow R_{100} = R_0 + R_0 \alpha\theta + R_0 \beta \theta^2$

$\Rightarrow \frac{\left( R_{100} - R_0 \right)}{R_0} = \alpha\theta + \beta \theta^2$

$\Rightarrow \frac{27 . 5 - 20}{20} = \alpha\theta + \beta \theta^2$

$\Rightarrow \frac{7 . 5}{20} = \alpha \times 100 + \beta \times 10000 . . . \left( i \right)$

$Also, R_{420} = R_0 \left( 1 + \alpha\theta + \beta \theta^2 \right)$

$\Rightarrow R_{420} = R_0 + R_0 \alpha\theta + R_0 \beta \theta^2$

$\Rightarrow \frac{R_{420} - R_0}{R_0} = \alpha\theta + \beta \theta^2$

$\Rightarrow \frac{50 - 20}{20} = 420\alpha + 176400 \beta$

$\Rightarrow \frac{3}{2} = 420\alpha + 176400 \beta . . . \left( ii \right)$

Solving (i) and (ii), we get:
α = 3.8 ×10–3°C​-1
β = –5.6 ×10–7°C-1
Therefore, resistance Ris 20 Ω and the value of α is 3.8 ×10–3°C​-1  and that of β is –5.6 ×10–7°C-1.

Concept: Temperature and Heat
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 1 Heat and Temperature
Q 9 | Page 13