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In new temperature scale W, freezing poi...

In new temperature scale W, freezing point of water is `39^(@)W` and boiling point is `239^(@)W`. What would be the temperature on new temperature scale for `39^(@)C` temperature ?

A

`200^(@)W`

B

`139^(@)W`

C

`78^(@)W`

D

`117^(@)W`

Text Solution

Verified by Experts

The correct Answer is:
D
`(""^(@)C-0)/(100-0)=(W^(@)-39)/(239-39)`
`(39)/(100)=(W-39)/(200)`
`2xx39+39=W`
`:.W=39xx(200)/(100)+39`
`=39xx2+39`
`=78+39`
`=117^(@)W`
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