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_(CA)=-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 can follow these steps:

### Step 1: Identify the heat transfers in each step of the cycle.
We are given:
- \( Q_{AB} = +200 \, \text{J} \)
- \( Q_{BC} = +600 \, \text{J} \)
- \( Q_{CD} = -300 \, \text{J} \)
- \( Q_{DA} = 0 \, \text{J} \)

### Step 2: Calculate the total heat absorbed by the system.
The total heat absorbed, \( Q_H \), is the sum of all positive heat transfers:
\[ Q_H = Q_{AB} + Q_{BC} = 200 \, \text{J} + 600 \, \text{J} = 800 \, \text{J} \]

### Step 3: Calculate the total work done by the system.
The total work done, \( W \), can be calculated using the first law of thermodynamics:
\[ W = Q_{in} - Q_{out} \]
Here, \( Q_{in} \) is the total heat absorbed, and \( Q_{out} \) is the heat released:
\[ W = Q_H - |Q_{CD}| = 800 \, \text{J} - 300 \, \text{J} = 500 \, \text{J} \]

### Step 4: Calculate the efficiency of the process.
Efficiency \( \eta \) is given by the formula:
\[ \eta = \frac{W}{Q_H} \]
Substituting the values we calculated:
\[ \eta = \frac{500 \, \text{J}}{800 \, \text{J}} = 0.625 \]
To express efficiency as a percentage:
\[ \eta = 0.625 \times 100 = 62.5\% \]

### Final Answer:
The efficiency of the process is \( 62.5\% \).

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