Question

Boiling point of water on the Fahrenheit scale is
$\begin{align}
 & A.\text{ }100{}^\circ F \\
 & B.\text{ }112{}^\circ F \\
 & C.\text{ }212{}^\circ F \\
 & D.\text{ }32{}^\circ F \\
\end{align}$

Answer 1

Hint: In order to convert one temperature scale to another
We use
$\dfrac{\operatorname{Re}ading-freezing\text{ }po\operatorname{int}}{Boiling\text{ point-freezing point}}=constant $

Complete step-by-step answer:
As temperature is linearly varies to height of mercury level
Now it is given from question
For Celsius scale
Freezing point$={{0}^{0}}\text{ C}$
Boiling point $={{100}^{0}}C $
For Fahrenheit scale
 Freezing point$={{32}^{0}}\text{ F}$
Boiling point $={{212}^{0}}F$
Let us assume boiling point on Fahrenheit scale is x⁰ F,
So using the relation
$\dfrac{\operatorname{Re}ading-freezing\text{ }po\operatorname{int}}{Boiling\text{ point-freezing point}}=constant$
So we can write
$\begin{align}
  & \dfrac{x{}^\circ F-32{}^\circ F}{212{}^\circ F-32{}^\circ F}=\dfrac{100{}^\circ C-0{}^\circ C}{100{}^\circ C-0{}^\circ C} \\
 & \Rightarrow \dfrac{x{}^\circ F-32{}^\circ F}{180{}^\circ F}=\dfrac{100{}^\circ C}{100{}^\circ C} \\
 & \Rightarrow \dfrac{x{}^\circ F-32{}^\circ F}{180{}^\circ F}=1 \\
 & \Rightarrow x{}^\circ F-32{}^\circ F=180{}^\circ F \\
 & transposing\text{ 32}{}^\circ \text{F on right hand side} \\
 & \Rightarrow x=180{}^\circ F+32{}^\circ F \\
 & \Rightarrow x=212{}^\circ F \\
 & \\
\end{align}$
Hence boiling point on Fahrenheit scale is ${{212}^{0}}F$ , so option C is correct.
Note: During calculation it must be written steps in proper units , also expansion of mercury level is a linear function of temperature. We can also use the relation $F=32+\dfrac{9}{5}C$ to change directly celsius reading to Fahrenheit reading.