Question

Centigrade and Fahrenheit thermometers are dipped in boiling water and the water temperature is lowered until the Fahrenheit thermometer registers 140. The fall in temperature as registered by the centigrade thermometer is........ .
A) $30^\circ $
B) $40^\circ $
C) $60^\circ $
D) $80^\circ $

Answer 1

Hint: We know the relation between the centigrade and Fahrenheit scale so we can use that relation in this question.
We know the centigrade scale have $100^\circ c$ as the boiling point of water and in Fahrenheit scale $212^\circ F$

Step by step solution:
We know that in centigrade scale freezing point of water is $0^\circ c$ and boiling point $100^\circ c$ and it divide into 100 equal parts
And in Fahrenheit scale it start from $32^\circ F$ to $212^\circ F$ and divided into 180 equal parts
So we can write
$ \Rightarrow \dfrac{{\Delta {T_c}}}{{100}} = \dfrac{{\Delta {T_F}}}{{180}}$
This is the relation between centigrade and Fahrenheit scale
Here temperature fall down in Fahrenheit scale 212 to 140 so $\Delta {T_F} = 212 - 140$
Put in above equation
$ \Rightarrow \dfrac{{\Delta {T_c}}}{{100}} = \dfrac{{212 - 140}}{{180}}$
$ \Rightarrow \Delta {T_c} = \dfrac{{72}}{{180}} \times 100$
Solving this we get a fall down in centigrade scale.
$\therefore \Delta {T_c} = 40^\circ $

Hence option B is correct

Note:
We can solve this question by another method we know $\dfrac{c}{5} = \dfrac{{F - 32}}{{180}}$
Put F=140 $c = \left( {\dfrac{{140 - 32}}{{180}}} \right) \times 5$
Solving this we get $c = 60^\circ $
Since the temperature of boiling water is $100^\circ c$ so the require fall is $100 - 60 = 40^\circ c$