Q75 / 90 · JEE Main 2024 (08 Apr Shift 1) · Physics

Average force exerted on a non-reflecting surface at normal incidence is $2.4 \times 10^{-4} \mathrm{~N}$. If $360 \mathrm{~W} / \mathrm{cm}^2$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:

  • A
    $0.2 \mathrm{~m}^2$
  • B
    $20 \mathrm{~m}^2$
  • C
    $0.1 \mathrm{~m}^2$
  • Correct answer
    D
    $0.02 \mathrm{~m}^2$

Solution

$\begin{aligned} & \text { Pressure }=\frac{I}{C}=\frac{F}{A} \\ & \Rightarrow \frac{360}{10^{-4} \times 3 \times 10^8}=\frac{2.4 \times 10^{-4}}{A} \\ & \Rightarrow \mathrm{A}=2 \times 10^{-2} \mathrm{~m}^2=0.02 \mathrm{~m}^2\end{aligned}$

Hence, the correct answer is option: 'D'.

Relevant Questions