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`
Finally, `eta. = (1-(T_2.)/(T_1))=(1-((T_2-62))/(T_(1)))`
`=eta+(62)/(T_(1)`
Since, `eta. = 2eta`
By solving-
`T_1=372 K=99^@C, T_2=310K = 37^@C`.