Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The ratio of the energy radiated per second by the first sphere to that by the second is (a) 1: 1 (b) 16: 1 (c) 4: 1 (d) 1: 9

Answer: (a). P ∝ r² T⁴ for the same material. (P₁)/(P₂) = (1/4)² ((4000)/(2000))⁴ = 1/16 × 16 = 1 So the answer is (a) 1: 1.

Two spheres of the same material have radii 1 m and 4 m and… | JEE Main 2002 (AIEEE)

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Thermal Radiation: Black Body, Stefan & Wien

Two spheres of the same material have radii $1\text{ m}$ and $4\text{ m}$ and temperatures $4000\text{ K}$ and $2000\text{ K}$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

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Answer: (a)

$P \propto r^2 T^4$ for the same material.
$\frac{P_1}{P_2} = \left(\frac{1}{4}\right)^2 \left(\frac{4000}{2000}\right)^4 = \frac{1}{16} \times 16 = 1$
So the answer is (a) $1 : 1$ .

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