Question

If a thermometer reads freezing point of water as ${20}^{\circ }\text{C}$ and boiling point as ${150}^{\circ }\text{C},$ how much thermometer would read when the actual temperature is ${60}^{\circ }\text{C}$?

• A. ${98}^{\circ }\text{C}$
• B. ${110}^{\circ }\text{C}$
• C. ${40}^{\circ }\text{C}$
• D. ${60}^{\circ }\text{C}$

Answer 1

The correct option is A ${98}^{\circ }\text{C}$

Temperature on any scale can be converted into other scale by $\frac{x-L.F.P}{U.F.P-L.F.P}=$ Constant for all scales.
So, $\frac{x-20}{150-20}=\frac{60-0}{100-0}$
$\Rightarrow x={98}^{\circ }\text{C}$