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

W is the temperature on new scale corresponding, to `39^(@)C` on `.^(@)C` scale.
`"S,=o, "(C-0)/(100-0)=(W-39)/(239-39)`
`rArr" "(C)/(100)=(W-39)/(200)`
`"or "w=(C)/(100)xx200+39`
`=(39)/(100)xx200+39=78+39=117`
So, temperature on new scale is `117^(@)W` corresponding to `39^(@)C`