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Q. A uniform thermometre scale is at steady state with its $0 \,cm$ mark at $20^{\circ} C$ and $100\, cm$ mark at $100^{\circ} C$. Temperature of the $60 \,cm$ mark is

Thermal Properties of Matter

Solution:

$\frac{T-20}{100-20}=\frac{60-0}{100-0}$
$\frac{T-20}{80}=\frac{60}{100}$
$T=48+20$
$=68^{\circ} C$