A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by `62^@C` , the efficiency of the engine is doubled. The temperatures of the source and sink are

Options

A. `80^(@)C, 37^(@)C`
B. `95^(@)C,28 ^(@)C`
C. `90^(@)C,37^(@)C`
D. `99^(@)C,37^(@)C`

Answer

D

Explanation

Initially `eta=(1-(T_2)/(T_1)) = W/Q = 1/6` ..(i)
Finally `eta' = (1-(T'_2)/(T_1)) = (1-(T_2-62)/(T_1))`
`=1-(T_2)/(T_1)+62/(T_1)`
`= eta + 62/(T_1)` ..(ii)
It is given that `eta' = 2 eta` . Hence solving equation (i) and (ii)
`implies T_(1) = 372K = 99^(@)C` and `T_(2)= 310K = 37^(@)C`.