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**Hint:**To solve this question, we need to identify each of the four processes in the given T-V diagram. Then we have to use the standard formulae for the work done and the heat transferred for the respective processes.

**Complete step-by-step solution:**

For process $1 \to 2$:

This process is represented by a straight line in the given V-T diagram. We know that the slope of a straight line is constant.

We know from the ideal gas equation that

$PV = nRT$

$V = \dfrac{{nR}}{P}T$

Comparing with the equation of a line $y = mx$, we get the slope of the line representing the process $1 \to 2$ as

$m = \dfrac{{nR}}{P}$

Since the slope is constant, so $\dfrac{{nR}}{P}$ is also a constant. Now, the quantity of the gas doesn’t change during the cycle, so the number of moles $n$ remains constant. This means that in this process the pressure $P$ is constant.

Now, substituting the values of the state

So the process $1 \to 2$ is an isobaric process.

For process $2 \to 3$:

This process is represented by a horizontal line in the given T-V diagram. So this means that the volume is constant in this process.

So the process $2 \to 3$ is an isochoric process.

Similarly we can show that the processes $3 \to 4$ and $4 \to 1$ are isobaric and isochoric, respectively.

Thus, the above thermodynamic cycle exhibits only isobaric and isochoric processes.

Hence, the option C is incorrect.

Now, we know that the work done in an isobaric process is given by

$W = nR\Delta T$

According to the question, we have one mole of gas. Therefore we have

$W = R\Delta T$.....................(1)

So the work done in the process $1 \to 2$ is given by

${W_{1 \to 2}} = R\left( {{T_2} - {T_1}} \right)$

From the given diagram, ${T_2} = 2{T_0}$, and ${T_1} = {T_0}$. Substituting these above we get

$\Rightarrow$ \[{W_{1 \to 2}} = R{T_0}\]......................(2)

Also from (1) the work done in the process $3 \to 4$ is given by

${W_{3 \to 4}} = R\left( {{T_4} - {T_3}} \right)$

From the given diagram, ${T_3} = {T_0}$, and ${T_4} = {T_0}/2$. Substituting these above we get

$\Rightarrow$ \[{W_{1 \to 2}} = - \dfrac{1}{2}R{T_0}\].......................(3)

Now, we know that the work done in an isochoric process is equal to zero. Therefore we have

$\Rightarrow$ \[{W_{2 \to 3}} = 0\]..........................(4)

$\Rightarrow$ \[{W_{4 \to 1}} = 0\]...............................(5)

The net work done in the complete cycle is

$\Rightarrow$ $W = {W_{1 \to 2}} + {W_{2 \to 3}} + {W_{3 \to 4}} + {W_{4 \to 1}}$

Putting (2) (3) (4) and (5) above, we get

$\Rightarrow$ $W = R{T_0} + 0 - \dfrac{1}{2}R{T_0} + 0$

$ \Rightarrow W = \dfrac{1}{2}R{T_0}$

Thus, the work done in the complete cycle is $\left| W \right| = \dfrac{1}{2}R{T_0}$.

Hence, the option A is correct.

Now, we know that the heat transfer in an isobaric process is given by

$\Rightarrow$ $Q = n{C_P}\Delta T$

Since there in one mole of gas, so we have

$\Rightarrow$ $Q = {C_P}\Delta T$

So the heat transfer the process $1 \to 2$ is

$\Rightarrow$ ${Q_{1 \to 2}} = {C_P}\left( {{T_2} - {T_1}} \right)$

Substituting ${T_2} = 2{T_0}$, and ${T_1} = {T_0}$ we get

$\Rightarrow$ ${Q_{1 \to 2}} = {C_P}{T_0}$....................................(6)

Similarly the heat transfer in the process $3 \to 4$ is

${Q_{3 \to 4}} = - {C_P}\dfrac{{{T_0}}}{2}$..........................(7)

Also, we know that the work done in an isochoric process is given by

$Q = n{C_v}\Delta T$

Since there in one mole of gas, so we have

$Q = {C_v}\Delta T$

So the heat transfer the process $2 \to 3$ is

${Q_{2 \to 3}} = {C_v}\left( {{T_3} - {T_2}} \right)$

Substituting ${T_2} = 2{T_0}$, and ${T_3} = {T_0}$ we get

\[{Q_{2 \to 3}} = - {C_v}{T_0}\] ………………………..(8)

Similarly, the heat transfer in the process $4 \to 1$ is

${Q_{4 \to 1}} = \dfrac{1}{2}{C_v}{T_0}$ …….(9)

Dividing (6) by (8) we get

$\Rightarrow$ \[\dfrac{{{Q_{1 \to 2}}}}{{{Q_{2 \to 3}}}} = \dfrac{{{C_P}{T_0}}}{{ - {C_v}{T_0}}}\]

$\Rightarrow$ \[\dfrac{{{Q_{1 \to 2}}}}{{{Q_{2 \to 3}}}} = - \dfrac{{{C_P}}}{{{C_v}}}\]

We know that \[\dfrac{{{C_P}}}{{{C_v}}} = \gamma \], and for a monatomic gas, we have $\gamma = \dfrac{5}{3}$. Therefore we have

$\Rightarrow$ \[\dfrac{{{Q_{1 \to 2}}}}{{{Q_{2 \to 3}}}} = - \dfrac{5}{3}\]

Taking modulus both sides, we get

$\Rightarrow$ \[\left| {\dfrac{{{Q_{1 \to 2}}}}{{{Q_{2 \to 3}}}}} \right| = \dfrac{5}{3}\]

Therefore, the option B is correct.

Now, dividing (6) by (7) we have

$\Rightarrow$ \[\dfrac{{{Q_{1 \to 2}}}}{{{Q_{3 \to 4}}}} = \dfrac{{{C_P}{T_0}}}{{ - {C_p}{T_0}/2}}\]

\[\dfrac{{{Q_{1 \to 2}}}}{{{Q_{3 \to 4}}}} = - 2\]

Taking modulus both sides, we get

$\Rightarrow$ \[\left| {\dfrac{{{Q_{1 \to 2}}}}{{{Q_{3 \to 4}}}}} \right| = 2\]

Therefore, option D is incorrect.

**Hence, the correct options are A and B.**

**Note:**We could also attempt this question by obtaining the P-V diagram from the given T-S diagram. From there we could have directly got the value of the net work done in the cycle by calculating the area of the loop.

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