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:

Question

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° C  (B)  40° C  (C)  50° C  (D)  90° C

Tardigrade Admin · Accepted Answer

The relation between Centigrate ° C scale of temperature and Fahrenheit ° F scale is (C/5)=(F - 32/9) Fahrenheit F = 140 ° F = (C/5)=(140 - 32/9) =1809=12 ⇒ C = 12×5 = 60°C ∴ Fall in temperature in centigrade scale = initial temperature - final temperature = I00 ° C - 60 ° C = 40 ° C

Q. 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:

$80^{\circ} C$

$40^{\circ} C$

$50^{\circ} C$

$90^{\circ} C$

Q. 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:

80∘C

40∘C

50∘C

90∘C

The relation between Centigrate ∘ C scale of temperature and Fahrenheit ∘ F scale is 5C​=9F−32​ Fahrenheit F = 140 ∘ F = 5C​=9140−32​ =1809=12 ⇒C=12×5=60∘C ∴ Fall in temperature in centigrade scale = initial temperature - final temperature = I00 ∘ C - 60 ∘ C = 40 ∘ C

$80^{\circ} C$

$40^{\circ} C$

$50^{\circ} C$

$90^{\circ} C$