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

Answer

To solve the problem of finding the ratio of efficiencies of a Carnot engine operating between two different temperature ranges, we will follow these steps:

### Step 1: Convert the temperatures from Celsius to Kelvin
- For the first case:
- Higher temperature (Th1) = 200°C = 200 + 273 = 473 K
- Lower temperature (Tl1) = 0°C = 0 + 273 = 273 K

- For the second case:
- Higher temperature (Th2) = 0°C = 0 + 273 = 273 K
- Lower temperature (Tl2) = -200°C = -200 + 273 = 73 K

### Step 2: Write the formula for the efficiency of a Carnot engine
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_l}{T_h} \]
where \(T_l\) is the lower temperature and \(T_h\) is the higher temperature.

### Step 3: Calculate the efficiency for the first case
Using the temperatures from Step 1:
\[ \eta_1 = 1 - \frac{T_{l1}}{T_{h1}} = 1 - \frac{273}{473} \]
Calculating this:
\[ \eta_1 = 1 - 0.577 = 0.423 \]

### Step 4: Calculate the efficiency for the second case
Using the temperatures from Step 1:
\[ \eta_2 = 1 - \frac{T_{l2}}{T_{h2}} = 1 - \frac{73}{273} \]
Calculating this:
\[ \eta_2 = 1 - 0.267 = 0.733 \]

### Step 5: Find the ratio of efficiencies
Now we can find the ratio of the efficiencies of the two cases:
\[ \text{Ratio} = \frac{\eta_1}{\eta_2} = \frac{0.423}{0.733} \]
Calculating this ratio:
\[ \text{Ratio} \approx 0.577 \]

### Step 6: Express the ratio in simplest form
To express the ratio in a more understandable form, we can multiply both the numerator and the denominator by a common factor to avoid decimals. However, since we are looking for a simple ratio, we can express it as:
\[ \text{Ratio} \approx 1 : 1.73 \]

### Final Answer
Thus, the ratio of efficiencies of the engine in the two cases is approximately:
\[ 1 : 1.73 \]