Q73 / 75 · JEE Main 2020 (06 Sep Shift 2) · Physics

An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The percentage efficiency of the engine is close to

Correct answer: 19

Solution

$W = 2 P_{0} v_{0}$

$Q_{+} = W_{AB} + W_{BC}$

$= n . c_{v} {ΔT}_{AB} + {nC}_{P} {ΔT}_{B} C$

$= \frac{3}{2} \times [3 P_{0} v_{0} - P_{0} v_{0}] + \frac{5}{2} [6 P_{0} V_{0} - 3 P_{0} V_{0}]$

$= 3 P_{0} V_{0} + \frac{15}{2} P_{0} V_{0}$

$= \frac{21}{2} P_{0} V_{0}$

$η = \frac{W}{Q_{+}} = \frac{2 P_{0} V_{0}}{\frac{21}{2} P_{0} V_{0}} = \frac{4}{21}$

$η % = \frac{400}{21} \approx 19$

Hence, the correct answer is 19.