Question

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

• A. ${100}^{\circ }$
• B. ${98}^{\circ }$
• C. ${110}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is B ${98}^{\circ }$

For first thermometer,
$L.F.P=\text{Freezing point}={20}^{\circ }$
$U.F.P=\text{Boiling point}={150}^{\circ }$

For actual thermometer,
$L.F.P=0^{\circ }C$
$U.F.P={100}^{\circ }C$

We know,
$\frac{\text{Reading on any scale}-L.F.P}{U.F.P-L.F.P}=\text{constant for all scales}$
$\therefore \frac{x-{20}^{\circ }}{{150}^{\circ }-{20}^{\circ }}=\frac{{60}^{\circ }-0^{\circ }}{{100}^{\circ }-0^{\circ }}$
$\Rightarrow x={98}^{\circ }$