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

Answer

C

Explanation

For new scale,
Freezing point of water = `39^@W`
Boiling point of water = `239^@W`
For Celsius scale, boiling point `= 100^@C`
` therefore ("Freezing point ")/("boiling point") = (39^@C - 0)/(100^@C - 0) = (x - 39^@ W)/(239^@W - 39^@ W)`
` x = 117^@ W`