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Dual Nature of Matter

JEE Main level, single correct, medium

111 of 200

Question 111 of 200

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:

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Answer: (D) $0.02 \mathrm{~m}^2$

\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}

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Dual Nature of Matter (Photoelectric Effect), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

Average force exerted on a non-reflecting surface at normal incidence is 2.4×10−4 N2.4 \times 10^{-4} \mathrm{~N}2.4×10−4 N. If 360 W/cm2360 \mathrm{~W} / \mathrm{cm}^2360 W/cm2 is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:

JEE Main question paper PDF

More Physics practice

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: