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

Option 1 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

Option 4 - <math> <mrow> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

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5 months ago

Correct Option - 3

Detailed Solution:

$η = \frac{W . D . b y g a s}{H e a t a b s o r b}$

$η = \frac{H e a t (N e t)}{H e a t (A b s o r b)}$

$η = \frac{A r e a u n d e r t h e c u r v e}{H e a t A b s o r b}$

$η = \frac{\frac{1}{2} * S_{0} T_{0}}{\frac{1}{2} (3 T_{0}) S_{0}} = \frac{1}{3}$

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