JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 4)
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),
64 E
16 E
E
256 E
Explanation
We know Stefan-Boltzmann law states that energy radiated by a black body per unit
time is,
$$P = \sigma eA{T^4}$$
$$ \Rightarrow P \propto A{T^4}$$
$$ \Rightarrow P \propto {R^2}{T^4}$$ (As $$A = 4\pi {R^2}$$)
so, $${{{P_1}} \over {{P_2}}} = {\left( {{{{R_1}} \over {{R_2}}}} \right)^2}{\left( {{{{T_1}} \over {{T_2}}}} \right)^4}$$
$$ \Rightarrow {E \over {E'}} = {\left( {{{0.2} \over {0.8}}} \right)^2}{\left( {{{800} \over {400}}} \right)^4}$$
$$ \Rightarrow {{\rm E} \over {E'}} = {1 \over {16}} \times 16 \Rightarrow E' = E$$
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