The graph between two temperature scales $A$ and $B$ is shown in figure. Between upper fixed point and lower fixed point, there are $150$ equal divisions on scale $A$ and $100$ on scale $B$ . The relationship for conversion between the two scales in given by:

\frac{t_{A} - 180}{100} = \frac{t_{B}}{150}

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\frac{t_{A} - 30}{150} = \frac{t_{B}}{100}

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\frac{t_{B} - 180}{150} = \frac{t_{A}}{100}

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\frac{t_{B} - 40}{100} = \frac{t_{A}}{180}

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Solution

The correct option is C $\frac{t_{A} - 30}{150} = \frac{t_{B}}{100}$
A correct answer is an option(B).
The temperature on one scale can be converted into other scales by using the following identity.
$\frac{Reading on any scale - L F P}{U F P - L F P}$ = Constant for all scales
where $L F P$ is a Lower fixed point and $U F P$ is Upper fixed point
From the graph, it is clear that the lowest point for the scale $A$ is $30 °$ and the highest point for scale $A$ is $180 °$ .
The lowest point for scale $B$ is $0 °$ and the highest point for scale $B$ is $100 °$ . Hence, the relation between the two scales $A$ and $B$ is given by
$\frac{t_{A} - (L F P)_{A}}{(U F P)_{A} - (L F P)_{A}} = \frac{t_{B} - (L F P)_{B}}{(U F P)_{B} - (L F P)_{B}}$

$\frac{t_{A} - 30}{180 - 30} = \frac{t_{B} - 0}{100 - 0}$

$\frac{t_{A} - 30}{150} = \frac{t_{B}}{100}$

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