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 `100^(@)`. Hence, correct relation is
`(t_(A)-(LEP)_(A))/((UFP)_(A)-(LEP)_(A))= (t_(B)-(LEP)_(B))/((UFP)_(B)-(LEP)_(B))`

`rArr (t_(A)-30)/(180-30) = (t_(B)-0)/(100-0)`
`rArr (t_(A)-30)/(150) = (t_(B))/(100)`
where, `LEP rarr` Lower fixed point
`UFP rarr` Upper fixed point