Average force exerted on a non-reflecting surface at normal incidence is 2.4 × 10⁻⁴ N. If 360 W/cm² is the light energy flux during span of 1 hour 30 minutes, then the area of the surface is: (a) 20 m² (b) 0.2 m² (c) 0.1 m² (d) 0.02 m²

Answer: (d). Non-reflecting (absorbing) surface: F = IA/c, independent of the exposure time I = 360 W/cm² = 3.6 × 10⁶ W/m² A = Fc/I = 2.4 × 10⁻⁴ × 3 × 10⁸3.6 × 10⁶ = 0.02 m² So the answer is (d).

Average force exerted on a non-reflecting surface at normal incidence… | JEE Main 2024 (8 Apr, Shift 1)

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Momentum & Radiation Pressure

Average force exerted on a non-reflecting surface at normal incidence is $2.4 \times 10^{-4}\text{ N}$ . If $360\text{ W/cm}^2$ 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)

Non-reflecting (absorbing) surface: $F = \dfrac{IA}{c}$ , independent of the exposure time
$I = 360\ \text{W/cm}^2 = 3.6 \times 10^6\ \text{W/m}^2$
$A = \dfrac{Fc}{I} = \dfrac{2.4 \times 10^{-4} \times 3 \times 10^8}{3.6 \times 10^6} = 0.02\ \text{m}^2$
So the answer is (d).

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