Question:
A light of energy flux
\[
9\ \text{W cm}^{-2}
\]
is incident for \(20\) minutes on a black surface of area
\[
100\ \text{cm}^2.
\]
Then the maximum average force exerted on the surface is
\[
(c=3\times10^8\ \text{m s}^{-1})
\]
A light of energy flux
\[
9\ \text{W cm}^{-2}
\]
is incident for \(20\) minutes on a black surface of area
\[
100\ \text{cm}^2.
\]
Then the maximum average force exerted on the surface is
\[
(c=3\times10^8\ \text{m s}^{-1})
\]
Show Hint
For a perfectly absorbing surface:
\[
P_{\text{rad}}=\frac{I}{c}.
\]
For a perfectly reflecting surface:
\[
P_{\text{rad}}=\frac{2I}{c}.
\]
Force is obtained from
\[
F=P_{\text{rad}}A.
\]
Updated On: Jul 29, 2026
- \[ 3\times10^{-3}\ \text{N} \]
- \[ 3\times10^{-6}\ \text{N} \]
- \[ 3\times10^{-8}\ \text{N} \]
- \[ 10\ \text{N} \]
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
For a perfectly absorbing (black) surface,
\[
\text{Radiation Pressure}
=
\frac{I}{c},
\]
where \(I\) is the energy flux (intensity).
The force exerted on area \(A\) is
\[
F=PA=\frac{IA}{c}.
\]
Step 1: Convert the intensity into SI units. Given, \[ I=9\ \text{W cm}^{-2}. \] Since \[ 1\ \text{cm}^2=10^{-4}\ \text{m}^2, \] \[ I = 9\times10^{4}\ \text{W m}^{-2}. \]
Step 2: Convert the area into SI units. \[ A=100\ \text{cm}^2. \] \[ A=100\times10^{-4} =10^{-2}\ \text{m}^2. \]
Step 3: Calculate the force. \[ F = \frac{IA}{c}. \] \[ = \frac{(9\times10^{4})(10^{-2})} {3\times10^{8}}. \] \[ = \frac{9\times10^{2}} {3\times10^{8}}. \] \[ = 3\times10^{-6}\ \text{N}. \] Thus, \[ \boxed{F=3\times10^{-6}\ \text{N}} \]
Note: The given time of \(20\) minutes is irrelevant because the force depends on power flux, not on the duration of exposure. Therefore, \[ \boxed{\text{Answer = (B)}} \]
Step 1: Convert the intensity into SI units. Given, \[ I=9\ \text{W cm}^{-2}. \] Since \[ 1\ \text{cm}^2=10^{-4}\ \text{m}^2, \] \[ I = 9\times10^{4}\ \text{W m}^{-2}. \]
Step 2: Convert the area into SI units. \[ A=100\ \text{cm}^2. \] \[ A=100\times10^{-4} =10^{-2}\ \text{m}^2. \]
Step 3: Calculate the force. \[ F = \frac{IA}{c}. \] \[ = \frac{(9\times10^{4})(10^{-2})} {3\times10^{8}}. \] \[ = \frac{9\times10^{2}} {3\times10^{8}}. \] \[ = 3\times10^{-6}\ \text{N}. \] Thus, \[ \boxed{F=3\times10^{-6}\ \text{N}} \]
Note: The given time of \(20\) minutes is irrelevant because the force depends on power flux, not on the duration of exposure. Therefore, \[ \boxed{\text{Answer = (B)}} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Electromagnetic waves Questions
- The RMS value of the electric field of an electromagnetic wave emitted by a source is \( 660 \) N/C. The average energy density of the electromagnetic wave is:
- The average value of electric energy density in an electromagnetic wave is: [where \( E_0 \) is the peak value]
- If the electric field of an electromagnetic wave is increased by 100%, then the energy density of the wave increases by
- Electromagnetic waves of energy flux $ 75 \times 10^4 \, \text{W/m}^2 $ incidents normally on a surface of area $ 40 \, \text{cm}^2 $. If the surface absorbs the flux completely, the total momentum delivered to the surface in one second is: Options:
- If $E_o$ and $B_o$ are the magnitudes of the electric and magnetic fields respectively of an electromagnetic wave in vacuum, then among the following the correct relation is ($\mu_o$ -- permeability of free spa, $\varepsilon_o$ -- permittivity of free spa)
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions