Question

The pressure of solar radiation on earth is (given solar constant $ 8.14Jc{m^{ - 2}}{\min ^{ - 1}} $ ).
$ \left( A \right)3.29 \times {10^{ - 5}}N{m^{ - 2}} \\
  \left( B \right)4.53 \times {10^{ - 5}}N{m^{ - 2}} \\
  \left( C \right)3.29 \times {10^{ - 4}}N{m^{ - 2}} \\
  \left( D \right)4.53 \times {10^{ - 6}}N{m^{ - 2}} \\ $

Answer 1

Hint :In order to solve this question, we first have to convert the solar constant to the SI units so that the pressure can also be obtained in the SI units. Secondly, as we know that the pressure is the energy flux divided by the unit speed of light and energy flux is solar constant. Thus, pressure is obtained.
The formula used here says that the pressure is the energy flux divided by the unit speed of light and energy flux is solar constant.
 $ pressure = \dfrac{{{E_f}}}{c} $
Where $ {E_f} $ is energy flux and $ c $ is the speed of light.

Complete Step By Step Answer:
Let us take the value of solar constant
It is Given that the value of solar constant is $ 8.14Jc{m^{ - 2}}{\min ^{ - 1}} $
Let us first convert the units of the solar constant to SI unit
Therefore, it becomes
$ \dfrac{{8.14 \times {{\left( {100} \right)}^2}}}{{60}}J{s^{ - 1}}{m^{ - 2}} \\
  1356.66W{m^{ - 2}} \\ $
As we know that the pressure absorbed is given by
 $ {P_A} = \dfrac{{{E_f}}}{c} $
Where $ {E_f} $ is the energy flux and $ c $ is the speed of light.
This implies
$ {P_A} = \dfrac{{1356.66}}{{3 \times {{10}^8}}} = 452.22 \times {10^{ - 8}}Pa \\
   \Rightarrow {P_A} = 4.52 \times {10^{ - 6}}Pa \\ $
Therefore, the pressure of the solar radiation is $ {P_A} = 4.52 \times {10^{ - 6}}N{m^{ - 2}} $
Hence, option $ \left( D \right)4.53 \times {10^{ - 6}}N{m^{ - 2}} $ is the correct answer.

Note :
Alternatively, this question can also be solved in the following way. As we know that the power is given by
 $ power = force \times velocity $
And the pressure is force per unit area
Now, $ \dfrac{{power}}{{velocity}} = force $
This implies the force is calculated as , equating the solar constant to the power and taking the velocity as speed of light, we get

$ force = \dfrac{{1356.66}}{{3 \times {{10}^8}}} \\
   \Rightarrow force = 4.52 \times {10^{ - 6}}N \\ $
Now as
$ pressure = \dfrac{{force}}{{area}} \\
   \Rightarrow pressure = \dfrac{{4.52 \times {{10}^{ - 6}}N}}{{{m^2}}} = 4.52 \times {10^{ - 6}}N{m^{ - 2}} \\$
Hence, by this method also, we get the correct option $ \left( D \right)4.53 \times {10^{ - 6}}N{m^{ - 2}} $