If we consider solar system consisting of the earth and sun only as one of the ideal thermodynamic system. The sun works as source of energy having temperature 6000 K and the earth as sink having temperature 300K, the efficiency of solar system would be on the basis of exchange of radiations

Options

A. 0.3
B. 0.65
C. 0.75
D. 0.95

Explanation

To find the efficiency of the solar system, we can use the formula for efficiency in a thermodynamic system, which is given by:

\[ \eta = 1 - \frac{T_2}{T_1} \]

where:
- \( T_1 \) is the temperature of the source (the Sun),
- \( T_2 \) is the temperature of the sink (the Earth).

### Step 1: Identify the temperatures
From the problem, we have:
- Temperature of the Sun (source), \( T_1 = 6000 \, K \)
- Temperature of the Earth (sink), \( T_2 = 300 \, K \)

### Step 2: Substitute the values into the efficiency formula
Now, we can substitute these values into the efficiency formula:

\[ \eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{6000} \]

### Step 3: Simplify the fraction
Next, we simplify the fraction:

\[ \frac{300}{6000} = \frac{1}{20} \]

### Step 4: Calculate the efficiency
Now we can substitute this back into the efficiency equation:

\[ \eta = 1 - \frac{1}{20} \]

Calculating this gives:

\[ \eta = 1 - 0.05 = 0.95 \]

### Step 5: Conclusion
Thus, the efficiency of the solar system is:

\[ \eta = 0.95 \]

### Final Answer
The efficiency of the solar system is \( 0.95 \).

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