Q. The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

A) $1/2$
B) $1/4$
C) $1/3$
D) $2/3$
Explanation

(c) According to the figure
$\ \ \ \ \ \ \ \ \ $
$Q_{1} = T_{0}S_{0} + \frac{1}{2}T_{0}S_{0} = \frac{3}{2}T_{0}S_{0}$
${\ \ \ \ \ \ \ \ \ Q}_{2} = T_{0}\left( 2S_{0} - S_{0} \right) = T_{0}S_{0}$
$\ \ \ \ \ \ \ \ \ Q_{3} = 0$
$\ \ \ \ \ \ \ \ \ \ \ \eta = \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{2}{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{3}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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