Question

A platinum resistance thermometer has a resistance of $50\Omega $ at ${20^0}C$. When dipped in a liquid the resistance becomes $76.8\Omega $. The temperature coefficient of resistance for platinum is $\alpha = 3.92 \times {10^{ - 3}}/{}^0C$. The temperature of the liquid is
A. ${100^0}C$
B. ${137^0}C$
C. ${167^0}C$
D. ${200^0}C$

Answer 1

Hint: First try to find the relation between electric resistance and the temperature of the thermometer of the given material. Then find the ratio of resistance of the thermometer in both the cases given in the question. Then apply the same ratio to find the temperature of the liquid when the thermometer is dipped into it.

Formula used:
Resistance, $R_t = R_o(1 + \alpha t)$
Where, $\alpha $ is temperature coefficient, $t$ is temperature, $R_o$ is the resistance at normal conditions and $R_t$ is the resistance at $t$ temperature.

Complete step by step solution:
First start with the given information from the question:
Resistance of thermometer in normal case, ${R_1} = 50\Omega $.
Resistance of thermometer after dipped in liquid, ${R_2} = 76.8\Omega $.
Temperature in normal case without the liquid, ${t_1} = {20^0}C$.
Now, we know that resistance is given by:
$R_t = R_o(1 + \alpha t)$
Where, $\alpha = 3.92 \times {10^{ - 3}}$

Then the ratio of both resistances will be given by:
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\left( {1 + \alpha {t_1}} \right)}}{{\left( {1 + \alpha {t_2}} \right)}}$
$\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\left( {1 + 3.92 \times {{10}^{ - 3}\times 20}} \right)}}{{\left( {1 + 3.92 \times {{10}^{ - 3}}{t_2}} \right)}} \\
\Rightarrow \dfrac{{50}}{{76.8}}= \dfrac{{\left( {1 + 3.92 \times {{10}^{ - 3}\times 20}} \right)}}{{\left( {1 + 3.92 \times {{10}^{ - 3}}{t_2}} \right)}} \\ $
$\Rightarrow 50\left( {1 + 3.92 \times {{10}^{ - 3}}{t_2}} \right)= 76.8\left( {1 + 3.92 \times {{10}^{ - 3}\times 20}} \right)\\ $
$\Rightarrow 50 \times 3.92 \times {10}^{ - 3} \times {t_2} = 76.8 -50 + 76.8 \times 3.92 \times {10}^{ - 3}\times 20 \\ $
$\Rightarrow t_2 = \dfrac{26.8 + 6.021}{0.196}= \dfrac{32.821}{0.196} \\ $

After solving above ratio, we get:
$\therefore {t_2} = {167^0}C$

Hence, the correct answer is option C.

Note: Here the resistance of the thermometer in both the cases that is with liquid and without liquid is already given in the question so we need to find only the ratio of both resistances in order to get the temperature of liquid but any of the value if is not present then we do not get the exact value but one value in terms of another.