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Physics Question on thermal properties of matter

A Centigrade and Fahrenheit thermometers are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers a temperature of 140?. The fall of temperature as registered by the Centigrade thermometer is:

A

80∘C80{}^\circ C

B

40∘C40{}^\circ C

C

50∘C50{}^\circ C

D

90∘C90{}^\circ C

Answer

40∘C40{}^\circ C

Explanation

Solution

The relation between Centigrate ∘^{\circ} C scale of temperature and Fahrenheit ∘^{\circ} F scale is C5=F−329\frac{C}{5}=\frac{F - 32}{9} Fahrenheit F = 140 ∘^{\circ} F = C5=140−329\frac{C}{5}=\frac{140 - 32}{9} =1809=12={180}{9}=12 ⇒C=12×5=60∘C\Rightarrow C = 12\times5 = 60^{\circ}C ∴\therefore Fall in temperature in centigrade scale = initial temperature - final temperature = I00 ∘^{\circ} C - 60 ∘^{\circ} C = 40 ∘^{\circ} C