Two heat engines are operating in such a way that the heat rejected by the first engine is used as the heat input of the second. If both the engines are `40%` efficient, then the overall efficiency of the system is

Options

A. 0.64
B. 0.8
C. 0.5
D. 0.9

Answer

To solve the problem, we need to find the overall efficiency of two heat engines operating in series, where the heat rejected by the first engine is used as the heat input for the second engine. Both engines have an efficiency of 40%.

### Step-by-Step Solution:

1. **Understanding the Efficiency of the First Engine:**
- The efficiency (η) of a heat engine is defined as:
\[ \eta = 1 - \frac{Q_2}{Q_1} \]
- Given that the efficiency of the first engine is 40%, we can express this as:
\[ \frac{40}{100} = 1 - \frac{Q_2}{Q_1} \]
- Rearranging gives:
\[ \frac{Q_2}{Q_1} = 1 - 0.4 = 0.6 \]
- This means:
\[ Q_2 = 0.6 Q_1 \quad \text{(Equation 1)} \]

2. **Heat Input for the Second Engine:**
- The heat rejected by the first engine (Q2) becomes the heat input for the second engine (Q1').
- Therefore, we have:
\[ Q_1' = Q_2 = 0.6 Q_1 \]

3. **Understanding the Efficiency of the Second Engine:**
- The efficiency of the second engine is also 40%, so we can write:
\[ \eta' = 1 - \frac{Q_2'}{Q_1'} \]
- Substituting the efficiency:
\[ \frac{40}{100} = 1 - \frac{Q_2'}{Q_1'} \]
- Rearranging gives:
\[ \frac{Q_2'}{Q_1'} = 1 - 0.4 = 0.6 \]
- This means:
\[ Q_2' = 0.6 Q_1' \quad \text{(Equation 2)} \]

4. **Substituting for Q1':**
- Since \( Q_1' = 0.6 Q_1 \), we can substitute this into Equation 2:
\[ Q_2' = 0.6 (0.6 Q_1) = 0.36 Q_1 \]

5. **Finding Overall Efficiency:**
- The overall efficiency (η_total) of the system can be calculated as:
\[ \eta_{total} = 1 - \frac{Q_2'}{Q_1} \]
- Substituting \( Q_2' = 0.36 Q_1 \):
\[ \eta_{total} = 1 - \frac{0.36 Q_1}{Q_1} = 1 - 0.36 = 0.64 \]
- Converting to percentage:
\[ \eta_{total} = 0.64 \times 100 = 64\% \]

### Final Answer:
The overall efficiency of the system is **64%**.