Physics Circle home Student Login

Home›Chapters›Heat Transfer and Radiation PYQs›Thermal Radiation: Black Body, Stefan & Wien›Question 53

JEE Main 2025 (22 Jan, Shift 1) · Thermal Radiation: Black Body, Stefan & Wien

Thermal Radiation: Black Body, Stefan & Wien question from JEE Main 2025 (22 Jan, Shift 1)

Q53JEE Main 2025 · 22 JAN · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien

Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K. If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding temperature to be negligible),

Show answer and solution

Answer: (c)

  1. P=σe(4πr2)T4⇒P∝r2T4P = \sigma e (4\pi r^2) T^4 \Rightarrow P \propto r^2 T^4 (same material, same ee).
  2. PbigPsmall=(0.80.2)2(400800)4=16×116=1\frac{P_{big}}{P_{small}} = \left(\frac{0.8}{0.2}\right)^2 \left(\frac{400}{800}\right)^4 = 16 \times \frac{1}{16} = 1
  3. So the answer is (c) EE.
Report a mistake

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