Physics Q40 – JEE-MAIN 2025

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2m away from it, is

Clue 1 of 3: Radiation Pressure Concept

Light carries momentum, and when it interacts with a surface, it exerts a force, leading to pressure.

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Ninja Strategy•Order of Magnitude Estimation

Estimate the intensity ( $I \approx 9 {W/m}^{2}$ ) and then the pressure ( $p_{r} = 2 I / c \approx 2 \times 9 / (3 \times 10^{8}) = 6 \times 10^{- 8} Pascals$ ) to quickly identify the correct option.

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Step 1: Calculate the intensity of light✦ Active

The intensity $I$ of light from an isotropic source of power $P$ at a distance $r$ is given by the formula:

I = \frac{P}{4 π r^{2}}

Given $P = 450$ W and $r = 2$ m, substitute these values:

I = \frac{450 W}{4 π (2 m)^{2}} = \frac{450}{16 π} = \frac{225}{8 π} {W/m}^{2}

Step 2: Calculate the radiation pressure for a perfectly reflecting surface○ Expand

For a perfectly reflecting surface, the radiation pressure $p_{r}$ is given by the formula:

p_{r} = \frac{2 I}{c}

Where $c$ is the speed of light, $c = 3 \times 10^{8}$ m/s. Substitute the calculated intensity $I$ :

p_{r} = \frac{2 \times \frac{225}{8 π}}{3 \times 10^{8}} = \frac{225}{4 π \times 3 \times 10^{8}} = \frac{75}{4 π \times 10^{8}} Pascals

Now, substitute the value of $π \approx 3.14159$ :

p_{r} = \frac{75}{4 \times 3.14159} \times 10^{- 8} \approx \frac{75}{12.56636} \times 10^{- 8} \approx 5.968 \times 10^{- 8} Pascals

Step 3: Compare with options○ Expand

The calculated value $5.968 \times 10^{- 8}$ Pascals is approximately $6 \times 10^{- 8}$ Pascals. This matches option 2.

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Physics Question 40 – JEE-MAIN 2025

Clue 1 of 3: Radiation Pressure Concept

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