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Q. The steam point and ice point of a mercury thermometer are marked as $80^{o}$ and $10^{o}$ . At what temperature on centigrade scale the reading of this thermometer will be $59^{o}$ ?

NTA AbhyasNTA Abhyas 2020

Solution:

$\frac{T ′ - 10}{80 - 10}=\frac{T_{C}}{100}$
$\frac{59 - 10}{80 - 10}=\frac{T_{C}}{100}$
$T_{C}=70^{o}C$