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(C) For adiabatic processes $BC, DE, FA$, we have $TV^{\gamma-1} = 	ext{constant}$.For $BC$: $T_1 V_B^{\gamma-1} = T_2 V_C^{\gamma-1} \implies \frac{

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$3$

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(C) For adiabatic processes $BC, DE, FA$, we have $TV^{\gamma-1} = 	ext{constant}$.For $BC$: $T_1 V_B^{\gamma-1} = T_2 V_C^{\gamma-1} \implies \frac{V_C}{V_B} = \left(\frac{T_1}{T_2}ight)^{\frac{1}{\gamma-1}}$For $DE$: $T_2 V_D^{\gamma-1} = T_3 V_E^{\gamma-1} \implies \frac{V_E}{V_D} = \left(\frac{T_2}{T_3}ight)^{\frac{1}{\gamma-1}}$For $FA$: $T_3 V_F^{\gamma-1} = T_1 V_A^{\gamma-1} \implies \frac{V_A}{V_F} = \left(\frac{T_3}{T_1}ight)^{\frac{1}{\gamma-1}}$Multiplying these: $\frac{V_C}{V_B} \cdot \frac{V_E}{V_D} \cdot \frac{V_A}{V_F} = \left(\frac{T_1}{T_2} \cdot \frac{T_2}{T_3} \cdot \frac{T_3}{T_1}ight)^{\frac{1}{\gamma-1}} = 1$Thus, $\frac{V_E}{V_F} = \frac{V_B}{V_A} \cdot \frac{V_D}{V_C} = 2 	imes 2 = 4$. Statement $1$ is correct.Work in isothermal $EF$: $W_{EF} = RT_3 \ln\left(\frac{V_F}{V_E}ight) = RT_3 \ln\left(\frac{1}{4}ight) = -2RT_3 \ln(2)$. Magnitude is $2RT_3 \ln(2)$. Statement $2$ is correct.Heat supplied in $AB$ is $Q_{AB} = RT_1 \ln\left(\frac{V_B}{V_A}ight) = RT_1 \ln(2)$. Heat rejected in $EF$ is $Q_{EF} = RT_3 \ln\left(\frac{V_E}{V_F}ight) = RT_3 \ln(4) = 2RT_3 \ln(2)$. Ratio $\frac{Q_{AB}}{Q_{EF}} = \frac{RT_1 \ln(2)}{2RT_3 \ln(2)} = \frac{T_1}{2T_3}$. Statement $3$ is incorrect.Net work $W = W_{AB} + W_{CD} + W_{EF} = RT_1 \ln(2) + RT_2 \ln(2) - 2RT_3 \ln(2) = (T_1 + T_2 - 2T_3) R \ln(2)$. Statement $4$ is correct.Total correct statements = $3$.

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