Question

The reading of the Centigrade thermometer coincides with that of the Fahrenheit thermometer in a liquid. The temperature of the liquid is:
A) $-{40}^{\circ }C$
B) $0^{\circ }C$
C) ${100}^{\circ }C$
D) ${300}^{\circ }C$

Answer 1

Hint: Centigrade and Fahrenheit scales are both temperatures measuring scales.
In centigrade the freezing point of water is taken as the reference point for centigrade scale the freezing point of water is taken as $0^{\circ }C$ and boiling point of water is taken as ${100}^{\circ }C$ and the distance between these two points is divided into 100 equal divisions
In the Fahrenheit scale, the freezing point of water is labelled as ${32}^{\circ }F$ and boiling point of water is taken as ${212}^{\circ }F$ and the distance between these two points is divided into 180 equal divisions.

Complete step by step solution:
As we know that in the Centigrade scale the distance between the freezing and boiling point of water is divided into 100 equal units and in the Fahrenheit scale the same distance is divided into 180 equal units
So we can say that the in one scale same temp difference is divided into 100 units and on other it is divided into 180 equal units
So it can be written as
$$\begin{gathered} {180}^{\circ }F={100}^{\circ }C \\ or,{\displaystyle \frac{{180}^{\circ }F}{{100}^{\circ }C}}=1 \\ \Rightarrow {\displaystyle \frac{9^{\circ }F}{5^{\circ }C}}=1 \end{gathered}$$
From here it is clear that a temperature increase of 9 Fahrenheit degrees is equivalent to an increase of 5 Celsius degrees.
Now as the freezing point of water in Centigrade scale is taken as $0^{\circ }C$ and ${32}^{\circ }F$ in Fahrenheit scale we can write its as
$$\begin{gathered} 5^{\circ }F=9^{\circ }C+32 \\ {\Rightarrow }^{\circ }F={\displaystyle \frac{9}{5}}\left({}^{\circ }C\right)+32 \end{gathered}$$
Now this comes out to be the relation between both scales.
In the question, it is asked as to when the reading of two scales will coincide or in other words when the two thermometers will show the same temperature reading.
For that let the temperature reading on both the scale be $x$.
$$\begin{gathered} \Rightarrow x={\displaystyle \frac{9}{5}}x+32 \\ \Rightarrow 5x=9x+160 \\ \Rightarrow 5x-9x=160 \\ \therefore x=-{40}^{\circ }C \end{gathered}$$
Final answer is (A), The reading of both thermometers will coincide at $-{40}^{\circ }C$.

Note: While solving the equation the signs are to be taken care of.
All the temperature scales are made for temperature measurement and take various reference points for their calibration.
The SI unit of temperature measurement is Kelvin (K).