The sun delivers `10^3W//m^2` of electromagnetic flux to the earth's surface.The total power that is inclident on a roof of dimensions `8mxx20m`, will be

Options

A. `2.56xx10^4W`
B. `6.4xx10^5W`
C. `4.0xx10^5W`
D. `1.6xx10^5W`

Explanation

To find the total power incident on a roof of dimensions \(8 \, \text{m} \times 20 \, \text{m}\) when the sun delivers \(10^3 \, \text{W/m}^2\) of electromagnetic flux, we can follow these steps:

### Step 1: Calculate the Area of the Roof
The area \(A\) of the roof can be calculated using the formula:
\[ A = \text{length} \times \text{width} \]
Given the dimensions:
\[ \text{length} = 8 \, \text{m}, \quad \text{width} = 20 \, \text{m} \]
So,
\[ A = 8 \, \text{m} \times 20 \, \text{m} = 160 \, \text{m}^2 \]

### Step 2: Calculate the Total Power Incident on the Roof
The total power \(P\) incident on the roof can be calculated using the formula:
\[ P = I \times A \]
where \(I\) is the intensity of the electromagnetic flux. Given that:
\[ I = 10^3 \, \text{W/m}^2 \]
Now substituting the values:
\[ P = 10^3 \, \text{W/m}^2 \times 160 \, \text{m}^2 \]
Calculating this gives:
\[ P = 160000 \, \text{W} = 1.6 \times 10^5 \, \text{W} \]

### Final Answer
The total power that is incident on the roof is:
\[ P = 1.6 \times 10^5 \, \text{W} \]

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