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)   \[\frac{2}{3}\] 

Correct Answer: C

Solution :

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).