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MHT CET 2026 (13 May, Shift 1) · Thermal Radiation: Black Body, Stefan & Wien

Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2026 (13 May, Shift 1)

Q19MHT CET 2026 · 13 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien

The total radiant energy per unit area normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature 'T' K is (σ\sigma = Stefan's constant)

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Answer: (a)

  1. Total power of the star P=4πr2σT4P = 4\pi r^2 \sigma T^4.
  2. At distance RR this spreads over a sphere of area 4πR24\pi R^2.
  3. Energy per unit area per unit time =4πr2σT44πR2=σr2T4R2= \dfrac{4\pi r^2 \sigma T^4}{4\pi R^2} = \dfrac{\sigma r^2 T^4}{R^2}
  4. So the answer is (a).
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