Use the $T$–$S$ diagram idea

For a reversible process on a temperature-entropy diagram,

So, the area under the curve gives the heat transferred.

For a reversible cyclic engine,

and the net work is the area enclosed by the cycle on the $T$–$S$ diagram.

Read the standard geometry of the given cycle

The given figure corresponds to a triangular cycle on the $T$–$S$ plane with:

• upper temperature $= 2T$
• lower temperature $= T$
• entropy change along the top side $= 2S$

Thus,

• heat absorbed is the area under the top horizontal line: $$Q_{\text{in}} = 2T \cdot 2S = 4TS$$
• net work done is the area of the triangle enclosed: $$W = \frac{1}{2}(2S)(2T-T)=\frac{1}{2}(2S)(T)=TS$$

Calculate efficiency

Match with the options

So the correct option is A.

Compare with stored answer

Stored correct answer is D: $\frac{1}{3}$.

This does not match the efficiency obtained from the $T$–$S$ area calculation. Hence the stored answer appears incorrect, assuming the figure is the standard triangular $T$–$S$ cycle described above.