Question

A mercury Celsius thermometer is transferred from melting ice to a hot liquid. The mercury rises to $\frac{9}{10}$th of the distance between the two fixed points. Find the temperature of the liquid in Fahrenheit scale.

• A. ${162}^{\circ }\text{F}$
• B. ${194}^{\circ }\text{F}$
• C. ${90}^{\circ }\text{F}$
• D. ${112}^{\circ }\text{F}$

Answer 1

The correct option is B ${194}^{\circ }\text{F}$

According to the question mercury rises to $\frac{9}{10}$ th of the distance

So,

We know, for Celsius scale
UFP $={100}^{\circ }\text{C}$, LFP $=0^{\circ }\text{C}$

From the question we can write
$\therefore$ Temperature of liquid given $=\frac{9}{10}\times (UFP-LFP)=\frac{9}{10}\times (100-0{)}^{\circ }\text{C}={90}^{\circ }\text{C}$
Now, the temperature in Fahrenheit scale is given by
$\frac{9}{5}C=F-32$
$\Rightarrow \frac{9}{5}\times 90=F-32$
$\Rightarrow F={194}^{\circ }\text{F}$

Hence the temperature is ${194}^{0}F$