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. `-106.3^(@)Y`
B. `-96.3^(@)Y`
C. `-86.3^(@)Y`
D. `-76.3^(@)Y`

Answer

C

Explanation

Let `T_(Y)` be the temperature on Y scale corresponding to `T_(K)` on kelvin scale.
`:. (T_(Y) - (-160))/(-50-(-160)) = (T_(K) -273)/(373-273) or (T_(Y) + 160)/(-50 + 160) = (340-273)/(100)`
or `100 (T_(Y) + 160) = 110 (67) or T_(Y) = -86.3^(@)Y`