A Carnot's engine is made to work between `200^(@)C` and `0^(@)C` first and then between `0^(@)C` and `-200^(@)C`. The ratio of efficiencies of the engine in the two cases is

Options

A. `1.73 : 1`
B. `1 : 1.73`
C. `1 : 1`
D. `1 : 2`

Explanation

To find the ratio of efficiencies of a Carnot engine working between two different temperature ranges, we can follow these steps:

### Step 1: Understand the formula for efficiency of a Carnot engine
The efficiency (\( \eta \)) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_C}{T_H} \]
where \( T_C \) is the absolute temperature of the cold reservoir and \( T_H \) is the absolute temperature of the hot reservoir.

### Step 2: Convert temperatures to Kelvin
For the first case, the temperatures are:
- Hot reservoir (\( T_H \)) = \( 200^\circ C = 200 + 273 = 473 \, K \)
- Cold reservoir (\( T_C \)) = \( 0^\circ C = 0 + 273 = 273 \, K \)

For the second case, the temperatures are:
- Hot reservoir (\( T_H \)) = \( 0^\circ C = 273 \, K \)
- Cold reservoir (\( T_C \)) = \( -200^\circ C = -200 + 273 = 73 \, K \)

### Step 3: Calculate efficiencies for both cases
**Efficiency for the first case (\( \eta_1 \))**:
\[ \eta_1 = 1 - \frac{T_C}{T_H} = 1 - \frac{273}{473} = 1 - 0.577 = 0.423 \]

**Efficiency for the second case (\( \eta_2 \))**:
\[ \eta_2 = 1 - \frac{T_C}{T_H} = 1 - \frac{73}{273} = 1 - 0.267 = 0.733 \]

### Step 4: Calculate the ratio of efficiencies
Now, we find the ratio of the efficiencies:
\[ \frac{\eta_1}{\eta_2} = \frac{0.423}{0.733} \]

Calculating this gives:
\[ \frac{\eta_1}{\eta_2} \approx 0.577 \]

### Step 5: Finalize the answer
The ratio of efficiencies of the engine in the two cases is approximately:
\[ \frac{\eta_1}{\eta_2} \approx 0.577 \]