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?
A
`200^@W`
B
`139^@W`
C
`78^@W`
D
`117^@W`
Text Solution
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The correct Answer is:
D
The relation between Celsium scale and new scale `(39^@C-0^@C)/(100^@C - 0^@C) = (t = 39^@W)/(239^@W -39^@W) rArr t = 117^@W`
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