The graph between two temperature scales `A` and `B` is shown in Fig. Between upper fixed point and lower fixed point there are `150` equal divisions on scales `A` and `100` on scale `B`. The relation between the temperature in two scales is given by_

Options

A. `(t_A -180 )/(100) = (t_B)/(150)`
B. `(t_A - 30)/(150) = (t_B)/(100)`
C. `(t_B-180)/(150) = (t_A)/(100)`
D. `(t_B - 40)/(100) = (t_A)/(180)`

Answer

B

Explanation

It is clear from the graph that lowest point for scale A is `30^@` and lowest point for scale B is `0^@` . Highest point for the scale A is `180^@` and for scale B is `100^@` Hence, correct relation is
`((t_A - (LFP)_A)/((UFP)_A - (LFP)_A) = (t_B - (LFP)_B)/((UFP)_B - (LFP)_B)`

`rArr (t_A - 30)/(180 - 30) = (t_B - 0)/(100-0)`
`rArr (t_A - 30)/(150) = (t_B)/(100)`
where `LFP rarr` Lower fixed point
`UFP rarr` Upper fixed point.