A Carnot reversible engine converts `1//6` of heat input into work . When the temperature of the sink is reduced by 62 K, the efficiency of carnot's cycle becomes `1//3` . The temperature of the source and sink will be

Options

A. 362 K
B. 372 K
C. 392 K
D. 412 K

Answer

B

Explanation

`eta=(dW)/(dQ)=1-(T_(2))/(T_(1))`
`(1)/(6)=1-(T_(2))/(T_(1)),(T_(2))/(T_(1))=(5)/(6),T_(2)=(5T_(1))/(6)`
`3T_(2)-186=2T_(1)`
`3xx(5T_(1))/(6)-186=2T_(1)(5T_(1))/(6)-186=2T_(1)`
`5T_(1)-372=4T`
`therefore T_(1)=372K`.