The resistance of a thermometer is 100 Omega | vedclass

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(D) Let the resistance be a linear function of temperature: $R_T = R_0 + \alpha T$.Given: $R_1 = 100 \ \Omega$ at $T_1 = 273 \ K$ and $R_2 = 300 \ \Om

Vedclass · Accepted Answer

$573$

Vedclass · Answer

(D) Let the resistance be a linear function of temperature: $R_T = R_0 + \alpha T$.Given: $R_1 = 100 \ \Omega$ at $T_1 = 273 \ K$ and $R_2 = 300 \ \Omega$ at $T_2 = 873 \ K$.The slope of the resistance-temperature graph is $m = \frac{R_2 - R_1}{T_2 - T_1} = \frac{300 - 100}{873 - 273} = \frac{200}{600} = \frac{1}{3} \ \Omega/K$.We want to find the temperature $T$ where $R = 200 \ \Omega$.Using the equation of a line: $R - R_1 = m(T - T_1)$.$200 - 100 = \frac{1}{3}(T - 273)$.$100 = \frac{1}{3}(T - 273)$.$300 = T - 273$.$T = 300 + 273 = 573 \ K$.

The resistance of a thermometer is $100 \Omega$ at the triple point of water $(273 \ K)$ and is $300 \Omega$ at the melting point of gold $(873 \ K)$. The temperature at which the resistance of the thermometer is $200 \Omega$ is: (in $K$)

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