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A "50 W m"^(-2) energy density of sunlig...

A `"50 W m"^(-2)` energy density of sunlight is incident normally on the surface of a solar panel. Some part of incident energy `(25%)` is reflected from the surface and the rest is absorbed. The force exerted on `1m^(2)` surface area will be close to
`(c=3xx10^(8)ms^(-1))`

A

`25xx10^(-8)N`

B

`20xx10^(-8)N`

C

`10xx10^(-8)N`

D

`35xx10^(-8)N`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Calculate the reflected intensity of sunlight. Given that the total energy density of sunlight incident on the solar panel is \( I = 50 \, \text{W/m}^2 \) and \( 25\% \) of it is reflected, we can calculate the reflected intensity \( I_R \). \[ I_R = 0.25 \times I = 0.25 \times 50 \, \text{W/m}^2 = 12.5 \, \text{W/m}^2 \] ### Step 2: Calculate the absorbed intensity of sunlight. The absorbed intensity \( I_A \) can be calculated by subtracting the reflected intensity from the total intensity. \[ I_A = I - I_R = 50 \, \text{W/m}^2 - 12.5 \, \text{W/m}^2 = 37.5 \, \text{W/m}^2 \] ### Step 3: Calculate the pressure due to absorbed radiation. The pressure \( P_A \) due to the absorbed radiation is given by the formula: \[ P_A = \frac{I_A}{c} \] Substituting the values: \[ P_A = \frac{37.5 \, \text{W/m}^2}{3 \times 10^8 \, \text{m/s}} = 1.25 \times 10^{-7} \, \text{N/m}^2 \] ### Step 4: Calculate the pressure due to reflected radiation. The pressure \( P_R \) due to the reflected radiation is given by the formula: \[ P_R = \frac{2I_R}{c} \] Substituting the values: \[ P_R = \frac{2 \times 12.5 \, \text{W/m}^2}{3 \times 10^8 \, \text{m/s}} = \frac{25 \, \text{W/m}^2}{3 \times 10^8 \, \text{m/s}} = \frac{25}{3} \times 10^{-8} \, \text{N/m}^2 \approx 8.33 \times 10^{-8} \, \text{N/m}^2 \] ### Step 5: Calculate the total pressure on the solar panel. The total pressure \( P \) on the solar panel is the sum of the pressures due to absorbed and reflected radiation. \[ P = P_A + P_R = 1.25 \times 10^{-7} \, \text{N/m}^2 + 8.33 \times 10^{-8} \, \text{N/m}^2 = 2.083 \times 10^{-7} \, \text{N/m}^2 \] ### Step 6: Calculate the force exerted on the solar panel. The force \( F \) exerted on a surface area \( A \) is given by: \[ F = P \times A \] Given that \( A = 1 \, \text{m}^2 \): \[ F = 2.083 \times 10^{-7} \, \text{N/m}^2 \times 1 \, \text{m}^2 = 2.083 \times 10^{-7} \, \text{N} \] ### Final Answer The force exerted on the \( 1 \, \text{m}^2 \) surface area of the solar panel is approximately \( 20.83 \times 10^{-8} \, \text{N} \). ---
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