Physics Circle home Student Login

Home›Chapters›Heat Transfer and Radiation PYQs›Thermal Radiation: Black Body, Stefan & Wien›Question 48

MHT CET 2025 (5 May, Shift 2) · Thermal Radiation: Black Body, Stefan & Wien

Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2025 (5 May, Shift 2)

Q48MHT CET 2025 · 5 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien

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

Show answer and solution

Answer: (a)

  1. Total power of the star P=4πr2σT4P = 4\pi r^2 \sigma T^4.
  2. At distance RR it spreads over 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).
Report a mistake

More questions on Thermal Radiation: Black Body, Stefan & Wien