The sun delivers `10^(3) W//m^(2)` of electromagnetic flux is incident on a roof of dimension `8m xx 20m` is ` 1.6 xx 10^(5)` W, the radiation force on the roof will be-

Options

A. `3.33 xx 10^(-5)` N
B. `5.33 xx 10^(-4)` N
C. `7.33 xx 10^(-3)` N
D. `9.33 xx 10^(-2)` N

Explanation

To find the radiation force on the roof due to the electromagnetic flux from the sun, we can follow these steps:

### Step 1: Calculate the Area of the Roof
The dimensions of the roof are given as \(8 \, m \times 20 \, m\).
\[ \text{Area} = 8 \, m \times 20 \, m = 160 \, m^2 \]

### Step 2: Calculate the Power Incident on the Roof
The intensity of the electromagnetic flux from the sun is given as \(10^3 \, W/m^2\). The power \(P\) incident on the roof can be calculated using the formula:
\[ P = \text{Intensity} \times \text{Area} \]
Substituting the values:
\[ P = 10^3 \, W/m^2 \times 160 \, m^2 = 1.6 \times 10^5 \, W \]

### Step 3: Relate Power to Force
The radiation force \(F\) can be calculated using the relationship between power and force. The formula relating power \(P\), force \(F\), and the speed of light \(c\) is:
\[ F = \frac{P}{c} \]
Where the speed of light \(c\) is approximately \(3 \times 10^8 \, m/s\).

### Step 4: Calculate the Radiation Force
Now, substituting the values we have:
\[ F = \frac{1.6 \times 10^5 \, W}{3 \times 10^8 \, m/s} \]
Calculating this gives:
\[ F = \frac{1.6 \times 10^5}{3 \times 10^8} = 5.33 \times 10^{-4} \, N \]

### Final Answer
Thus, the radiation force on the roof is approximately:
\[ F \approx 5.3 \times 10^{-4} \, N \]

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