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_(A)-180)/(150)=(t_(A))/(100)`
D. `(t_(B)-40)/(100)=(t_(A))/(180)`

Answer

C

Explanation

It is clerar form 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)-(LPF)_(A))=(t_(B)-(LFP)_(B))/((UFP)_(B)-(LFP)_(B))`

`rArr" "(t_(A)-30)/(180-30)=(t_(B)-O)/(100-0)rArr(t_(A)-30)/(150)=(t_(B))/(100)`
where, LFP `to` lower fixed poing
UFP `to` Upper fixed point