A Carnot reversible engine convert 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

The efficiency of heat engine is given by η=WQ=1−Q2Q1=1−T2T1 Where, T1  is temperature ofsource and T2  is temperature of sink.Given, η1=16,η2=13 ∴ 16=T1−T2T1 ⇒ T2T1=56 (i) and 13=T1−(T2−62)T1 ⇒ T2T1−62T1=23 (ii) Solving Eqs. (i) and (ii), we get56−62T1=23 ⇒ T1=372K  and  T2=310K