On a new scale of temperature (which is linear) and called the `W` scale. The freezing and boiling points of water are `39^(@)W` and `239^(@)W` respectively. What will be the temperature on the new scale, corresponding to a temperature of `39^(@)C` on the Celsius scale?

Options

A. `200^(@)W`
B. `139^(@)W`
C. `78^(@)W`
D. `117^(@)W`

Answer

D

Explanation

`(X-LFP)/(UFP-LFP)=` constant
where, X = Any given temperature on that scale LFP = Lower fixed point (Freezing point)
UFP = Upper fixed poing (Boiling point)
`(W-39)/(239-39)=(39-0)/(100-0) rArr(W-39)/(200)=(39)/(100)`
`rArr " "W=78+39 rArrW=117^(@)W`