Question

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

• A. $98C$
• B. $110C$
• C. $40C$
• D. $60C$

Answer 1

The correct option is A $98C$

Temperaturee on any scale can be converted into other scale by applying,
$\Rightarrow \frac{x-LFP}{UFP-LFP}=\text{constant}...\left(i\right)$
where $UFP=$Upper fixed point (boiling point)
$LFP=$Lower fixed point (freezing point)
$x=$ temperature on the given scale
As per question on $C-$scale
$LFP=20C\&UFP=150C$
$\Rightarrow$let the measured temperature on $C-$ scale is $x$ corresponding to ${60}^{\circ }\text{C}$ on centigrade scale for which $LFP=0^{\circ }\text{C},UFP={100}^{\circ }\text{C}$
Hence from Eq.$\left(i\right)$,
$\frac{x-20}{150-20}=\frac{60-0}{100-0}$
$\therefore x=98$
Hence measurement by thermometer will be $98\text{C}$