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

The correct option is B ${68}^{\circ }\text{C}$
Given that,
$l_1=0\text{cm}$ mark is at temperature $T_1={20}^{\circ }\text{C}$
and $l_2=100\text{cm}$ mark is at temperature $T_2={100}^{\circ }\text{C}$.
Let the temperature reading of the thermometer at $l=60\text{cm}$ be $T$.
By using the formula $l_2-l_1=\alpha (T_2-T_1)$, we can write that
$\frac{T-T_1}{T_2-T_1}=\frac{l-l_1}{l_2-l_1}.........\left(1\right)$

Using($1)$, we get
$\frac{T-20}{100-20}=\frac{60-0}{100-0}$
$\Rightarrow \frac{T-20}{80}=\frac{60}{100}$
$\Rightarrow T=48+20={68}^{\circ }\text{C}$
Hence, option (b) is the correct answer.