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JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2018
- Hard
- MCQ
Two Carnot engines \(A\) and \(B\) are operated in series. Engine \(A\) receives heat from a reservoir at \(600\,K\) and rejects heat to a reservoir at temperature \(T\). Engine \(B\) receives; heat rejected by engine \(A\) and in turn rejects it to a reservoir at \(100\,K\). If the efficiencies of the two engines \(A\) and \(B\) are represented by \({\eta _A}\) and \({\eta _B}\) respectively, then what is the value of \(\frac{{{\eta _A}}}{{{\eta _B}}}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\frac{{7}}{12}\)
Step-by-step Solution
Detailed explanation
Efficiency of engine \(A,{n_A} = \frac{{{T_1} - {T_2}}}{{{T_1}}}\)
and \({n_B} = \frac{{{T_2} - {T_3}}}{{{T_2}}};{T_2} = \frac{{{T_1} + {T_3}}}{2} = 350\,K\)
\(or\frac{{{n_A}}}{{{n_B}}} = \frac{{\frac{{600 - 350}}{{600}}}}{{\frac{{350 - 100}}{{350}}}} = \frac{7}{{12}}\)
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