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Dual Nature of Radiation question

2025 · 3 Apr · Shift 1 · Q69
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Dual Nature of Radiation question

2025 · 3 Apr · Shift 1 · Q69

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is
  1. A
    3×10−83 \times 10^{-8}3×10−8 Pascals
  2. B
    0
  3. C
    1.5×10−81.5 \times 10^{-8}1.5×10−8 Pascals
  4. D
    6×10−86 \times 10^{-8}6×10−8 Pascals
View written solutionFree

Correct answer: D

  1. Intensity at distance r=2 mr = 2\,\text{m}r=2m

For a source radiating uniformly in all directions, the intensity at distance rrr is

I=P4πr2I = \frac{P}{4\pi r^2}I=4πr2P​

Given:

  • Power, P=450 WP = 450\,\text{W}P=450W
  • Distance, r=2 mr = 2\,\text{m}r=2m

So,

I=4504π(2)2=45016πI = \frac{450}{4\pi (2)^2} = \frac{450}{16\pi}I=4π(2)2450​=16π450​

Using π≈3.14\pi \approx 3.14π≈3.14,

I≈45050.24≈8.96 W/m2I \approx \frac{450}{50.24} \approx 8.96\,\text{W/m}^2I≈50.24450​≈8.96W/m2
  1. Radiation pressure on a perfectly reflecting surface

For a perfectly reflecting surface, radiation pressure is

p=2Icp = \frac{2I}{c}p=c2I​

where c=3×108 m/sc = 3 \times 10^8\,\text{m/s}c=3×108m/s.

Thus,

p=2×8.963×108p = \frac{2 \times 8.96}{3 \times 10^8}p=3×1082×8.96​ p=17.923×108≈5.97×10−8 Pap = \frac{17.92}{3 \times 10^8} \approx 5.97 \times 10^{-8}\,\text{Pa}p=3×10817.92​≈5.97×10−8Pa
  1. Final answer
p≈6×10−8 Pap \approx 6 \times 10^{-8}\,\text{Pa}p≈6×10−8Pa

So the correct option is D.

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