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Question Number 34891 by arnabmaiti550@gmail.com last updated on 12/May/18
Answered by ajfour last updated on 12/May/18
$${Work}\:{done}={Area}\:{under}\:{ellipse} \\ $$$$\Rightarrow\:\:\:{W}=\pi{P}_{\mathrm{0}} {V}_{\mathrm{0}} \\ $$$${Heat}\:{received}\:=\:{Area}\:{under} \\ $$$${upper}\:{ellipse}\:{curve}\:{and}\:{V}\:{axes} \\ $$$${between}\:{V}=\frac{{V}_{\mathrm{0}} }{\mathrm{2}}\:\:{and}\:\:{V}=\frac{\mathrm{3}{V}_{\mathrm{0}} }{\mathrm{2}}\: \\ $$$$\:\:\:\:+\:\bigtriangleup\left({Internal}\:{energy}\:{while}\right. \\ $$$$\left.{expansion}\right) \\ $$$$\Rightarrow\:\:{Q}={P}_{\mathrm{0}} {V}_{\mathrm{0}} +\frac{\pi{P}_{\mathrm{0}} {V}_{\mathrm{0}} }{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}{P}_{\mathrm{0}} \left(\frac{\mathrm{3}{V}_{\mathrm{0}} }{\mathrm{2}}−\frac{{V}_{\mathrm{0}} }{\mathrm{2}}\right) \\ $$$${efficiency}\:=\:\frac{{work}\:{done}\:{by}\:{gas}}{{heat}\:{received}} \\ $$$$\Rightarrow\:\:\:\eta\:=\:\frac{\pi{P}_{\mathrm{0}} {V}_{\mathrm{0}} }{\frac{\mathrm{5}}{\mathrm{2}}{P}_{\mathrm{0}} {V}_{\mathrm{0}} +\frac{\pi{P}_{\mathrm{0}} {V}_{\mathrm{0}} }{\mathrm{2}}}\:=\:\frac{\mathrm{2}\pi}{\pi+\mathrm{5}}\:. \\ $$$${So}\:\:\left({a}\right),\:{and}\:\left({c}\right)\:. \\ $$