• # question_answer The temperature-energy diagram of a reversible engine cycle is given in the figure. Its efficiency is: A)  $\frac{1}{2}$                                    B) $\frac{1}{4}$             C)   $\frac{1}{3}$             D)   $\frac{2}{3}$

According to the figure ${{Q}_{1}}={{T}_{o}}{{S}_{o}}+\frac{1}{2}{{T}_{o}}{{S}_{o}}=\frac{3}{2}{{T}_{o}}{{S}_{o}}$ ${{Q}_{2}}={{T}_{o}}(2{{S}_{o}}-{{S}_{o}})={{T}_{o}}{{S}_{o}}$ ${{Q}_{3}}=0$ $\therefore$$\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}$ Hence, the correction option is (c).