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

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$1 / 3$
$1 / 3$

$\eta=\frac{\Delta \mathrm{W}}{\mathrm{Q}_{\mathrm{BC}}}=\frac{\frac{\mathrm{S}_0 \mathrm{~T}_0}{2}}{\frac{3 \mathrm{~S}_0 \mathrm{~T}_0}{2}}=1 / 3$
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