A cyclic process contains four steps AB,BC,CD and DA. Heat involved in different processes is given as `Q_(AB)=+200J,Q_(BC)=+600J,Q_(CD)=-300J,Q_(DA)=0`, then efficiency of process is

Options

A. `(5)/(8)`
B. `(3)/(8)`
C. `(1)/(4)`
D. `(1)/(2)`

Answer

To find the efficiency of the cyclic process, we will follow these steps:

### Step 1: Identify the heat inputs and outputs
The heat involved in the different processes is given as:
- \( Q_{AB} = +200 \, \text{J} \) (heat added)
- \( Q_{BC} = +600 \, \text{J} \) (heat added)
- \( Q_{CD} = -300 \, \text{J} \) (heat removed)
- \( Q_{DA} = 0 \, \text{J} \) (no heat transfer)

### Step 2: Calculate the total heat input
The total heat input \( Q_{in} \) is the sum of the heat added in processes AB and BC:
\[ Q_{in} = Q_{AB} + Q_{BC} = 200 \, \text{J} + 600 \, \text{J} = 800 \, \text{J} \]

### Step 3: Calculate the total heat output
The total heat output \( Q_{out} \) is the sum of the heat removed in process CD:
\[ Q_{out} = Q_{CD} = -300 \, \text{J} \]

### Step 4: Calculate the net heat interaction
The net heat interaction \( Q_{net} \) is the sum of all heat interactions:
\[ Q_{net} = Q_{AB} + Q_{BC} + Q_{CD} + Q_{DA} = 200 \, \text{J} + 600 \, \text{J} - 300 \, \text{J} + 0 \, \text{J} = 500 \, \text{J} \]

### Step 5: Calculate the net work done
According to the first law of thermodynamics for a cyclic process, the net work done \( W_{net} \) is equal to the net heat interaction:
\[ W_{net} = Q_{net} = 500 \, \text{J} \]

### Step 6: Calculate the efficiency
The efficiency \( \eta \) of the process is given by the formula:
\[ \eta = \frac{W_{net}}{Q_{in}} = \frac{500 \, \text{J}}{800 \, \text{J}} = \frac{5}{8} \]

### Conclusion
Thus, the efficiency of the process is \( \frac{5}{8} \).

---