Question

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

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

Answer 1

Answer: (A)

Let temperature of thermometer be $\theta^{\circ} C$ at $60^{\circ} C$ then
$ 100-60 =150-\theta $
$ \Rightarrow 40 =150-\theta$ ... (i)
Also $60-0 =\theta-20 $
$ 60 =\theta-20$ ... (ii)
Dividing Eq. (i) by Eq. (ii), we get
$ \frac{40}{60}=\frac{150-\theta}{\theta-20} $
$\Rightarrow \frac{2}{3}=\frac{150-\theta}{\theta-20}$
$\Rightarrow 50=490 $
$\Rightarrow \theta=98^{\circ} C$