(D) Given: Resistance at ice point $(R_0) = 5\; \Omega$ Resistance at steam point $(R_{100}) = 5.39\; \Omega$ Resistance in hot bath $(R_t) = 5.795

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(D) Given: Resistance at ice point $(R_0) = 5\; \Omega$ Resistance at steam point $(R_{100}) = 5.39\; \Omega$ Resistance in hot bath $(R_t) = 5.795\; \Omega$ The formula for temperature $(t)$ in a platinum resistance thermometer is given by: $t = \frac{R_t - R_0}{R_{100} - R_0} \times 100$ Substituting the values: $t = \frac{5.795 - 5}{5.39 - 5} \times 100$ $t = \frac{0.795}{0.39} \times 100$ $t = 2.03846 \times 100$ $t \approx 203.85^{\circ}C$

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Temperature and Temperature scales

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The resistance of the platinum wire of a platinum resistance thermometer at the ice point is $5\; \Omega$ and at steam point is $5.39\; \Omega$. When the thermometer is inserted in a hot bath,the resistance of the platinum wire is $5.795\; \Omega$. Calculate the temperature of the bath in $^oC$. (in $^{\circ}C$)

A
$279.64$
B
$462.95$
C
$593.24$
D
$203.85$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Temperature and Temperature scales

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