question_answer

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is [AIEEE 2005]

A)            \[1/3\]      

B)            \[2/3\]      

C)            \[1/2\]

D)            \[1/4\]

Correct Answer: A

Solution :

                   \[{{Q}_{1}}={{T}_{0}}{{S}_{0}}+\frac{1}{2}{{T}_{0}}{{S}_{0}}=\frac{3}{2}{{T}_{0}}{{S}_{0}}\] \[{{Q}_{2}}={{T}_{0}}{{S}_{0}}\]and \[{{Q}_{3}}=0\] \[\eta =\frac{W}{{{Q}_{1}}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}\]    \[=1-\frac{{{Q}_{2}}}{{{Q}_{1}}}=1-\frac{2}{3}=\frac{1}{3}\]