A system undergoes a cyclic process in which it absorbs Q, heat and gives out `Q_2.` heat. The efficiency of the process is n and the work done is W.

Options

A. `W=Q_(1)-Q_(2)`
B. `eta=(W)/(Q_(1))`
C. `eta=(Q_(2))/(Q_(1))`
D. `eta=1-(Q_(2))/(Q_(1))`

Answer

To solve the problem, we need to analyze the cyclic process in thermodynamics where a system absorbs heat \( Q_1 \), gives out heat \( Q_2 \), and does work \( W \). We are also given the efficiency \( \eta \) of the process.

### Step-by-Step Solution:

1. **Understanding the First Law of Thermodynamics**:
The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In a cyclic process, the internal energy returns to its initial state, so the change in internal energy is zero.
\[ \Delta U = Q - W = 0 \implies Q = W \]
Here, \( Q \) is the net heat added to the system.

2. **Relating Heat and Work**:
Since the system absorbs \( Q_1 \) and gives out \( Q_2 \), the work done \( W \) can be expressed as:
\[ W = Q_1 - Q_2 \]
This equation shows that the work done by the system is equal to the heat absorbed minus the heat rejected.

3. **Calculating Efficiency**:
The efficiency \( \eta \) of a cyclic process is defined as the ratio of the work done \( W \) to the heat input \( Q_1 \):
\[ \eta = \frac{W}{Q_1} \]
Substituting the expression for \( W \):
\[ \eta = \frac{Q_1 - Q_2}{Q_1} \]
This can be rearranged to:
\[ \eta = 1 - \frac{Q_2}{Q_1} \]

4. **Identifying Correct Relations**:
Now, we can summarize the relationships derived:
- Work done: \( W = Q_1 - Q_2 \)
- Efficiency: \( \eta = \frac{W}{Q_1} \) or \( \eta = 1 - \frac{Q_2}{Q_1} \)

### Conclusion:
From the analysis:
- The correct relations are:
- **A**: \( W = Q_1 - Q_2 \) (Correct)
- **B**: \( \eta = \frac{W}{Q_1} \) (Correct)
- **C**: \( \eta = \frac{Q_2}{Q_1} \) (Incorrect)
- **D**: \( \eta = 1 - \frac{Q_2}{Q_1} \) (Correct)

Thus, options A, B, and D are correct.