Using Stefan-Boltzmann’s law: P = e . A 2 σ T 4 P = e.A_{2}\sigma T^{4} P 2 P 1 = e . A 2 σ T 2 4 e A 1 σ T 1 4 \frac{P_{2}}{P_{1}} = \frac{e.A_{2}\sigma{\text{\ }T}_{2}^{4}}{eA_{1}\sigma{\text{\ }T}_{1}^{4}} P 2 E = ( 0.8 0.2 ) 2 ( 400 800 ) 4 = ( 4 1 ) 2 ( 1 4 ) 4 = 1 \frac{P_{2}}{E} = \left( \frac{0.8}{0.2} \right)^{2}\left( \frac{400}{800} \right)^{4} = \left( \frac{4}{1} \right)^{2}\left( \frac{1}{4} \right)^{4} = 1 T h u s P 2 = E Thus\ P_{2} = E

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Question 1:

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).

256E

16E

64E

Answer:

Solution:

Using Stefan-Boltzmann’s law:

$P = e.A_{2}\sigma T^{4}$

$\frac{P_{2}}{P_{1}} = \frac{e.A_{2}\sigma{\text{\ }T}_{2}^{4}}{eA_{1}\sigma{\text{\ }T}_{1}^{4}}$

$\frac{P_{2}}{E} = \left( \frac{0.8}{0.2} \right)^{2}\left( \frac{400}{800} \right)^{4} = \left( \frac{4}{1} \right)^{2}\left( \frac{1}{4} \right)^{4} = 1$

$Thus\ P_{2} = E$

Prev Exam(s):

JEE MAIN

- 2025

, 22 Jan Shift-1

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HEAT TRANSFER

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PHYSICS