The sream poinr and rht ice point of a mercuty thermometer are marked as `80^0 and 20^0`. What will be the temperature in centigrade mercury scale when this thermometer reads `32^0`?

Explanation

Ice point `=20^0 (T_0)`, T_1 = 32^0 C`
steam point `=80^0 (T_(100))`
`T= (T_1 - T_0)/(T_100 - T_0) xx 100`
`= (32 - 20)/(80 - 20) xx 100`
`=12/60 xx 100 = 120/6 = 20^0 C`