We need the efficiency of the rectangular cycle made of two isochoric and two isobaric processes for helium (monoatomic ideal gas).

Since the figure is not visible here, the standard interpretation for such JEE problems is that the rectangle has corner states:

• $A(P_1,V_1)$
• $B(P_2,V_1)$
• $C(P_2,V_2)$
• $D(P_1,V_2)$ with the cycle run in the order $A\to B\to C\to D\to A$.

For helium (monoatomic ideal gas): $C_V=\frac{3R}{2},\qquad C_P=\frac{5R}{2}$

1. Temperatures at the four corners

Using ideal gas law: $T=\frac{PV}{nR}$ So, $T_A=\frac{P_1V_1}{nR},\quad T_B=\frac{P_2V_1}{nR},\quad T_C=\frac{P_2V_2}{nR},\quad T_D=\frac{P_1V_2}{nR}$

2. Net work done in one cycle

For a rectangular $PV$ cycle, net work done is area enclosed: $W=(P_2-P_1)(V_2-V_1)$

3. Heat absorbed

Heat is absorbed in the two heating branches:

• $A\to B$ : isochoric heating
• $B\to C$ : isobaric expansion

(i) Heat in $A\to B$

$Q_{AB}=nC_V(T_B-T_A)$ $=n\cdot \frac{3R}{2}\left(\frac{P_2V_1-P_1V_1}{nR}\right)$ $Q_{AB}=\frac{3}{2}(P_2-P_1)V_1$

(ii) Heat in $B\to C$

$Q_{BC}=nC_P(T_C-T_B)$ $=n\cdot \frac{5R}{2}\left(\frac{P_2V_2-P_2V_1}{nR}\right)$ $Q_{BC}=\frac{5}{2}P_2(V_2-V_1)$

Hence total heat absorbed is $Q_{in}=Q_{AB}+Q_{BC}$ $Q_{in}=\frac{3}{2}(P_2-P_1)V_1+\frac{5}{2}P_2(V_2-V_1)$

4. Efficiency formula

Efficiency, $\eta=\frac{W}{Q_{in}}$ Thus, $\eta=\frac{(P_2-P_1)(V_2-V_1)}{\frac{3}{2}(P_2-P_1)V_1+\frac{5}{2}P_2(V_2-V_1)}$

5. Using the values from the figure

The standard figure for this question corresponds to pressure increasing from $P_1$ to $2P_1$ and volume increasing from $V_1$ to $2V_1$. So, $P_2=2P_1,\qquad V_2=2V_1$

Then, $W=(2P_1-P_1)(2V_1-V_1)=P_1V_1$

Also, $Q_{AB}=\frac{3}{2}(P_1)V_1=\frac{3}{2}P_1V_1$ $Q_{BC}=\frac{5}{2}(2P_1)(V_1)=5P_1V_1$

So, $Q_{in}=\frac{3}{2}P_1V_1+5P_1V_1=\frac{13}{2}P_1V_1$

Hence, $\eta=\frac{P_1V_1}{\frac{13}{2}P_1V_1}=\frac{2}{13}$ $\eta\approx 0.1538=15.4\%$

6. Option check

• A: $15.4\%$ ✅
• B: $9.1\%$ ❌
• C: $10.5\%$ ❌
• D: $12.5\%$ ❌

So the correct option is A.

7. Comparison with stored answer

Stored correct answer: A

Our derived answer: A ($15.4\%$)

They match.