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)

$98{}^\circ C$

Answer 2

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 E (i) by E (ii), we get $\frac{40}{60}=\frac{150-\theta}{\theta-20}$ $\Rightarrow \frac{2}{3}=\frac{150-\theta}{\theta-20}$ $\Rightarrow 5\theta=490$ $\Rightarrow \theta= 98^{\circ}C$