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),

Heat and Thermodynamics PYQ — JEE Main 2025 · 22 Jan · Shift 1 · Q53 with Solution

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Correct answer: C

Use Stefan–Boltzmann law

For a body radiating energy per unit time, $P = e\sigma A T^4$ where:

$e$ = emissivity,
$\sigma$ = Stefan constant,
$A$ = surface area,
$T$ = absolute temperature.

Since both spherical bodies are made of the same material and placed in the same atmosphere, we take emissivity same for both. Also, surrounding temperature is negligible, so net radiation is proportional to $AT^4$ .

Surface area of a sphere

$A = 4\pi r^2$

Hence, $P \propto r^2 T^4$

Given data

Smaller sphere: $r_1 = 0.2\,\text{m},\; T_1 = 800\,\text{K}$
Bigger sphere: $r_2 = 0.8\,\text{m},\; T_2 = 400\,\text{K}$

If the smaller body radiates energy $E$ , then $\frac{P_2}{P_1} = \frac{r_2^2 T_2^4}{r_1^2 T_1^4}$

Substitute ratios

$\frac{P_2}{P_1} = \left(\frac{0.8}{0.2}\right)^2 \left(\frac{400}{800}\right)^4$

$= 4^2 \left(\frac{1}{2}\right)^4$

$= 16 \cdot \frac{1}{16} = 1$

So, $P_2 = P_1 = E$

Conclusion

The energy radiated from the bigger body is $\boxed{E}$

So the correct option is C.

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