If a thermometer reads freezing point of water as 20° C and boiling point as 150° C, how much thermometer read, when the actual temperature is 60° C ?

Question

If a thermometer reads freezing point of water as 20° C and boiling point as 150° C, how much thermometer read, when the actual temperature is 60° C ? (A) 98° C (B) 110° C (C) 40° C (D) 60° C

Tardigrade Admin · Accepted Answer

Let temperature of thermometer be θ° C at 60° C then 100-60 =150-θ ⇒ 40 =150-θ ... (i) Also 60-0 =θ-20 60 =θ-20 ... (ii) Dividing Eq. (i) by Eq. (ii), we get (40/60)=(150-θ/θ-20) ⇒ (2/3)=(150-θ/θ-20) ⇒ 50=490 ⇒ θ=98° C

Q. If a thermometer reads freezing point of water as $2 0^{\circ} C$ and boiling point as $15 0^{\circ} C$ , how much thermometer read, when the actual temperature is $6 0^{\circ} C$ ?

$9 8^{\circ} C$

$11 0^{\circ} C$

$4 0^{\circ} C$

$6 0^{\circ} C$

Q. If a thermometer reads freezing point of water as 20∘C and boiling point as 150∘C, how much thermometer read, when the actual temperature is 60∘C ?

98∘C

110∘C

40∘C

60∘C

Let temperature of thermometer be θ∘C at 60∘C then 100−60=150−θ ⇒40=150−θ ... (i) Also 60−0=θ−20 60=θ−20 ... (ii) Dividing Eq. (i) by Eq. (ii), we get 6040​=θ−20150−θ​ ⇒32​=θ−20150−θ​ ⇒50=490 ⇒θ=98∘C

Q. If a thermometer reads freezing point of water as $2 0^{\circ} C$ and boiling point as $15 0^{\circ} C$ , how much thermometer read, when the actual temperature is $6 0^{\circ} C$ ?

$9 8^{\circ} C$

$11 0^{\circ} C$

$4 0^{\circ} C$

$6 0^{\circ} C$