Question

The steam point and the ice point for a mercury thermometer are marked as \[80{}^\circ \] and $20{}^\circ $. What will be the temperature in centigrade mercury scale when this thermometer reads \[32{}^\circ \]
A. $20{}^\circ $
B. $30{}^\circ $
C. $10{}^\circ $
D. $0{}^\circ $

Answer 1

Hint: temperature to the temperature measured relationship of a mercury thermometer will be helpful for solving the question. The steam point and ice point are the key fetchers for the calculation.
As per the question,
Steam point of the mercury thermometer is $80{}^\circ $
Ice point of the mercury thermometer is $20{}^\circ $

Formula used:
Temperature corresponding in centigrade to a temperature measured by mercury thermometer;
${{T}_{c}}=\dfrac{T-{{T}_{\circ }}}{{{T}_{100}}-{{T}_{\circ }}}\times 100$

Complete answer:
A mercury thermometer is a device which is used to measure the temperature corresponding to its expansion and contraction due to temperature. Mercury filled in the capillary tube has its own ice point and stem point according to which a measurement scale of temperature is calculated and marked on the thermometer.
As per the question it is said that the steam point of the mercury thermometer is $80{}^\circ $ and the ice point is $20{}^\circ $. By the standard temperature to a temperature measured by mercury formula, which is ${{T}_{c}}=\dfrac{T-{{T}_{\circ }}}{{{T}_{100}}-{{T}_{\circ }}}\times 100$

Where,
$\begin{align}
  & {{T}_{c}}=temperature\ in\ centigrade \\
 & T=temperature\ measured \\
 & {{T}_{\circ }}=\ ice\ po\operatorname{int} \\
 & {{T}_{100}}=steam\ po\operatorname{int} \\
\end{align}$

When, the measured temperature is 32, actual temperature in centigrade would be:
${{T}_{\circ }}=\dfrac{32-20}{80-20}\times 100$
${{T}_{\circ }}=20{}^\circ C$

So, the correct answer is “Option A”.

Note:
In this question mercury scale reading is not the same as that of regular temperature scale. This is due to the reason that the ice point is $20{}^\circ $ instead of $0{}^\circ $ and steam point is $80{}^\circ $ instead of $100{}^\circ $. If in case the steam and ice point where $100{}^\circ $ and $0{}^\circ $ then there will be no difference between the measured temperature and temperature in case of mercury thermometer.