On a temperature scale Y, water freezes at `-160^@ ` Y and boils at `-50^@` Y . On this Y scale , a temperature of 340 K is

Options

A. `-160.3^@Y`
B. `-96.3^@Y`
C. `-86.3^@Y`
D. `-76.3^@Y`

Answer

C

Explanation

On Y scale ,
Boiling point =`-50^(@)Y`
Freezing point =`-160^(@)Y`
On kelvin scale ,
Boiling point = 373 K
Freezing point = -273 K
Therefore ,
`(Y + 160)/(-50 + 160) =(340-273)/(373-273)`
`implies Y = -86.3^(@)Y`