The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is
- $1 / 2$
- $1 / 4$
- $1 / 3$
- $2 / 3$
Solution
$\eta=\frac{\Delta \mathrm{W}}{\mathrm{Q}_{\mathrm{BC}}}=\frac{\frac{\mathrm{S}_0 \mathrm{~T}_0}{2}}{\frac{3 \mathrm{~S}_0 \mathrm{~T}_0}{2}}=1 / 3$
Asked in: JEE Main 2005