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(C) The efficiency $\eta$ of a heat engine is defined as the ratio of the net work done to the total heat supplied.$\eta = \frac{W}{Q_{	ext{in}}} = \

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$980$

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(C) The efficiency $\eta$ of a heat engine is defined as the ratio of the net work done to the total heat supplied.$\eta = \frac{W}{Q_{	ext{in}}} = \frac{Q_{	ext{net}}}{Q_{	ext{in}}}$Here,the net heat exchanged in one cycle is $Q_{	ext{net}} = 1915 - 40 + 125 - Q = 2000 - Q$.The heat supplied $(Q_{	ext{in}})$ is the sum of all positive heat exchanges: $Q_{	ext{in}} = 1915 + 125 = 2040\, J$.Given $\eta = 50.0\% = 0.5$,we have:$0.5 = \frac{2000 - Q}{2040}$$1020 = 2000 - Q$$Q = 2000 - 1020 = 980\, J$.

Column-$1$	Column-$2$
$P$: Process-$I$	$A$: Adiabatic
$Q$: Process-$II$	$B$: Isobaric
$R$: Process-$III$	$C$: Isochoric
$S$: Process-$IV$	$D$: Isothermal

$A$ heat engine is involved with an exchange of heat of $1915\, J,$ $-40\, J,$ $+125\, J,$ and $-Q\, J$ during one cycle,achieving an efficiency of $50.0 \%$. The value of $Q$ is ....... $J$.

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