Physics Circle home Student Login

Home›Chapters›Heat Transfer and Radiation PYQs›Thermal Radiation: Black Body, Stefan & Wien›Question 102

JEE Main 2006 (AIEEE) · Thermal Radiation: Black Body, Stefan & Wien

Thermal Radiation: Black Body, Stefan & Wien question from JEE Main 2006 (AIEEE)

Q102JEE Main 2006 · AIEEEMedium

Thermal Radiation: Black Body, Stefan & Wien

Assuming the Sun to be a spherical body of radius RR at a temperature of TKTK, evaluate the total radiant power incident on Earth at a distance rr from the Sun

Where r0 is the radius of the Earth and σ\sigma is Stefan's constant.

Show answer and solution

Answer: (b)

  1. Power of the sun: P=4πR2σT4P = 4\pi R^2 \sigma T^4.
  2. Intensity at distance rr: I=P4πr2=R2σT4r2I = \frac{P}{4\pi r^2} = \frac{R^2 \sigma T^4}{r^2}.
  3. Earth intercepts this over its cross-section πr02\pi r_0^2.
  4. PE=πr02R2σT4r2P_E = \pi r_0^2 R^2 \sigma \frac{T^4}{r^2}, so the answer is (b).
Report a mistake

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