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Question: Centigrade and Fahrenheit thermometers are dipped in boiling water and the water temperature is lowe...

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) 3030^\circ
B) 4040^\circ
C) 6060^\circ
D) 8080^\circ

Explanation

Solution

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 100c100^\circ c as the boiling point of water and in Fahrenheit scale 212F212^\circ F

Step by step solution:
We know that in centigrade scale freezing point of water is 0c0^\circ c and boiling point 100c100^\circ c and it divide into 100 equal parts
And in Fahrenheit scale it start from 32F32^\circ F to 212F212^\circ F and divided into 180 equal parts
So we can write
ΔTc100=ΔTF180\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 ΔTF=212140\Delta {T_F} = 212 - 140
Put in above equation
ΔTc100=212140180\Rightarrow \dfrac{{\Delta {T_c}}}{{100}} = \dfrac{{212 - 140}}{{180}}
ΔTc=72180×100\Rightarrow \Delta {T_c} = \dfrac{{72}}{{180}} \times 100
Solving this we get a fall down in centigrade scale.
ΔTc=40\therefore \Delta {T_c} = 40^\circ

Hence option B is correct

Note:
We can solve this question by another method we know c5=F32180\dfrac{c}{5} = \dfrac{{F - 32}}{{180}}
Put F=140 c=(14032180)×5c = \left( {\dfrac{{140 - 32}}{{180}}} \right) \times 5
Solving this we get c=60c = 60^\circ
Since the temperature of boiling water is 100c100^\circ c so the require fall is 10060=40c100 - 60 = 40^\circ c