Solar Pond Area Calculation

This solution calculates the necessary area for a solar pond based on its efficiency, power output, and solar insolation.

Key Parameters

• Efficiency ($ \eta $): 5% (0.05)
• Power Output ($ P_{out} $): 500 mW. Important Note: The provided correct answer ($10 \text{ km}^2$) implies the unit should be Megawatts (MW), not milliwatts (mW). Calculations herein assume $ P_{out} = 500 \text{ MW} = 500 \times 10^6 $ W.
• Solar Insolation ($ I $): $ 1000 \text{ W/m}^2 $

Calculation Steps

1. Find the total input solar power ($ P_{in} $) needed. The efficiency formula is $ \eta = P_{out} / P_{in} $. Rearranging gives $ P_{in} = P_{out} / \eta $. $ P_{in} = \frac{500 \times 10^6 \text{ W}}{0.05} = 10 \times 10^9 \text{ W} $
2. Calculate the required area (A) using the relationship $ P_{in} = I \times A $. Thus, $ A = P_{in} / I $. $ A = \frac{10 \times 10^9 \text{ W}}{1000 \text{ W/m}^2} = 10 \times 10^6 \text{ m}^2 $
3. Convert the area from square meters ($ \text{m}^2 $) to square kilometers ($ \text{km}^2 $). Use the conversion $ 1 \text{ km}^2 = 10^6 \text{ m}^2 $. $ A = \frac{10 \times 10^6 \text{ m}^2}{10^6 \text{ m}^2/\text{km}^2} = 10 \text{ km}^2 $

Result

The required area for the solar pond is determined to be $ 10 \text{ km}^2 $. This corresponds to Option B.