MCQ
The graph between two temperature scales $A$ and $B$ is shown in between upper fixed point and lower fixed point there are $150$ equal division on scale $A$ and $100$ on scale $B$. The relationship for conversion between the two scales is given by:
  • A
    $\frac{\text{t}_\text{A}-180}{100}=\frac{\text{t}_\text{B}}{150}$
  • $\frac{\text{t}_\text{A}-30}{150}=\frac{\text{t}_\text{B}}{100}$
  • C
    $\frac{\text{t}_\text{B}-180}{150}=\frac{\text{t}_\text{A}}{100}$
  • D
    $\frac{\text{t}_\text{B}-40}{100}=\frac{\text{t}_\text{A}}{180}$

Answer

Correct option: B.
$\frac{\text{t}_\text{A}-30}{150}=\frac{\text{t}_\text{B}}{100}$
Key concept: Temperature on one scale can be converted into other scale by using the following identity.
$($image$)$
Reading on any scale $\frac{\text{LfP}}{\text{UFP}-\text{LFP}}=$ Constant for all scales
where, $\text{LFP} \rightarrow$ Lower fixed point
$\text{UFP} \rightarrow$ Upper fixed point
From the graph it is clear that the lowest point for scale $A$ is $30^\circ$ and highest point for the scale $A$ is $180^\circ$. Lowest point for scale $B$ is $0^\circ$ and highest point for scale $B$ is $100^\circ$ .
Hence, the relation between the two scales $A$ and $B$ is given by
$\frac{\text{T}_\text{A}-(\text{LFP})_\text{A}}{(\text{UFP)}_\text{A}-(\text{LFP})_\text{A}}=\frac{\text{T}_\text{B}-(\text{LFP})_\text{B}}{(\text{UFP)}_\text{B}-(\text{LFP})_\text{B}}$
$\Rightarrow\frac{\text{T}_\text{A}-30}{180-30}=\frac{\text{T}_\text{B}-0}{100-0}$
$\Rightarrow\frac{\text{t}_\text{A}-30}{150}=\frac{\text{t}_\text{B}}{100}$

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