At the ice point ( 0^ ^ ^ C ), the steam point ( 100^ ^ ^ C ), and the zinc point ( 420^ ^ ^ C ), a resistance thermometer reads R = 20.0 , 27.5 , and 50.0 respectively. Given that the resistance changes with

Correct answer: A. 20.0 , 3.8 × 10^ -3 ^ ^ ^ ^ -1 , -5.6 × 10^ -7 ^ ^ ^ ^ -2 Given the relation R_θ = R_0(1 + αθ + βθ^2) At θ = 0^ , R_0 = 20.0 At θ = 100^ , R_ 100 = 27.5 27.5 = 20.0(1 + 100α + 10000β) 100α + 10000β = (27.5)/(20.0) - 1 = 0.375 α + 100β = 0.00375 --- (i) At θ = 420^ , R_ 420 = 50.0 50.0 = 20.0(1 + 420α + 176400β) 420α + 176400β = (50.0)/(20.0) - 1 = 1.5 α + 420β = (1.5)/(420) = (1)/(280) --- (ii) Subtracting equation (i) from equation (ii): 320β = (1)/(280) - (3)/(800) 320β = (20 - 21)/(5600) = -(1)/(5600) β = -(1)/(320 × 5600) -5.6 × 10^ -7 ^ ^ -2 Substituting the value of β in equation (i): α = 0.00375 - 100 (-(1)/(1792000)) α = 0.00375 + 0.0000558 3.8 × 10^ -3 ^ ^ -1 Thus, R_0 = 20.0 , α = 3.8 × 10^ -3 ^ ^ -1 , and β = -5.6 × 10^ -7 ^ ^ -2 .

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Heat and Temperature

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat and Temperature

At the ice point ( 0^ ^ ^ C ), the steam point ( 100^ ^ ^ C ), and...

At the ice point (

0^{}^{{}^{\circ}} \text{C}

), the steam point (

100^{}^{{}^{\circ}} \text{C}

), and the zinc point (

420^{}^{{}^{\circ}} \text{C}

), a resistance thermometer reads

R = 20.0 \Omega

27.5 \Omega

, and

50.0 \Omega

respectively. Given that the resistance changes with temperature according to the relation

R_\theta = R_0(1 + \alpha\theta + \beta\theta^2)

, determine the values of

R_0

\alpha

, and

\beta

. Here

\theta

represents the temperature on the Celsius scale.

Options

A $20.0 \Omega$ , $3.8 \times 10^{-3} ^{}^{{}^{\circ}}\textC^{-1}$ , $-5.6 \times 10^{-7} ^{}^{{}^{\circ}}\textC^{-2}$
B $27.5 \Omega$ , $3.8 \times 10^{-3} ^{}^{{}^{\circ}}\textC^{-1}$ , $-5.6 \times 10^{-7} ^{}^{{}^{\circ}}\textC^{-2}$
C $20.0 \Omega$ , $5.6 \times 10^{-3} ^{}^{{}^{\circ}}\textC^{-1}$ , $-3.8 \times 10^{-7} ^{}^{{}^{\circ}}\textC^{-2}$
D $20.0 \Omega$ , $3.8 \times 10^{-3} ^{}^{{}^{\circ}}\textC^{-1}$ , $5.6 \times 10^{-7} ^{}^{{}^{\circ}}\textC^{-2}$

Correct answer

A. $20.0 \Omega$ , $3.8 \times 10^{-3} ^{}^{{}^{\circ}}\textC^{-1}$ , $-5.6 \times 10^{-7} ^{}^{{}^{\circ}}\textC^{-2}$

Step-by-step solution

Given the relation

R_\theta = R_0(1 + \alpha\theta + \beta\theta^2)

\theta = 0^\circ\text{C}

R_0 = 20.0 \Omega

\theta = 100^\circ\text{C}

R_{100} = 27.5 \Omega

27.5 = 20.0(1 + 100\alpha + 10000\beta)

100\alpha + 10000\beta = \dfrac{27.5}{20.0} - 1 = 0.375

\alpha + 100\beta = 0.00375

--- (i)

At

\theta = 420^\circ\text{C}

R_{420} = 50.0 \Omega

50.0 = 20.0(1 + 420\alpha + 176400\beta)

420\alpha + 176400\beta = \dfrac{50.0}{20.0} - 1 = 1.5

\alpha + 420\beta = \dfrac{1.5}{420} = \dfrac{1}{280}

--- (ii)

Subtracting equation (i) from equation (ii):

320\beta = \dfrac{1}{280} - \dfrac{3}{800}

320\beta = \dfrac{20 - 21}{5600} = -\dfrac{1}{5600}

\beta = -\dfrac{1}{320 \times 5600} \approx -5.6 \times 10^{-7} ^\circ\textC^{-2}

Substituting the value of

\beta

in equation (i):

\alpha = 0.00375 - 100

\left(-\dfrac{1}{1792000}\right)

\alpha = 0.00375 + 0.0000558 \approx 3.8 \times 10^{-3} ^\circ\textC^{-1}

Thus,

R_0 = 20.0 \Omega

\alpha = 3.8 \times 10^{-3} ^\circ\textC^{-1}

, and

\beta = -5.6 \times 10^{-7} ^\circ\textC^{-2}

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