Question

The resistance of a resistance thermometer has values $\;2.71$ and $3.70$ ohms at ${10^0}C$ and ${100^0}C$ respectively. The temperature at which the resistance is $3.26$ohm is-
A) ${40^0}C$
B) ${50^0}C$
C) ${60^0}C$
D) ${70^0}C$

Answer 1

Hint
We know the resistances at two temperatures. Substitute it in the expression ${R_2} = {R_1}(1 + \alpha ({t_2} - {t_1}))$. Now, using the same conditions substitute the resistance of the temperature required and ant one of the other two resistances. Solving the first equation we get the value of $\alpha $. Now substitute the value of $\alpha $ in the second equation to calculate the required temperature.

Complete step-by-step answer
Resistance of the conductor is directly proportional to the temperature.
$R \propto T$
Consider ${R_1} $ and ${R_2}$ as the resistances at temperatures ${T_1}$ and ${T_2}$. Where, $\alpha $ is the coefficient of resistance.
${R_2} = {R_1}(1 + \alpha ({t_2} - {t_1}))$,
It is given that,
${R_1} = 2.71\Omega $
${R_2} = 3.70\Omega $
${T_1} = {10^0}C$
${T_2} = {100^0}C$
Substitute in the resistance temperature relation to know the value of $3.70 = 2.71[1 + \alpha (100 - 10)] \to (1)$
Now, we need to calculate the temperature $T$ at which resistance is $R = 3.26\Omega $.
$3.26 = 2.71[1 + \alpha (T - 10)] \to (2)$
From equation $(1)$we get,
$\alpha = \dfrac{{3.70 - 2.71}}{{90 \times 2.71}}$
$\alpha = \dfrac{1}{{90 \times 2.71}}$
Substitute the value of $\alpha $ in $(2)$.
$3.26 = 2.71\left[ {1 + \dfrac{1}{{90 \times 2.71}} \times \left( {T - 10} \right)} \right]$
$\Rightarrow 3.26 = 2.71 + \dfrac{{T - 10}}{{90}}$
$\Rightarrow T - 10 = 90(3.26 - 2.71)$
$\Rightarrow T = 49.5 + 10$
$\Rightarrow T = 59.5 \approx {60^0}C$
Hence, the resistance of the conductor is $3.26\Omega $ when temperature is ${60^0}C$ .

The correct option is C.

Note
The resistance of the conductor also depends on other factors like length and area of cross section. Resistance is directly proportional to the length. Resistance of the conductor is inversely proportional to the area of cross section of the conductor. Resistance is given by,
$R = \dfrac{{\rho l}}{A}$
Here, $\rho $ is the resistivity of material of the conductor.