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If the radius of star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy protection is Q?
Options
(a) Q / 4 πR²σ
(b) (Q / 4 πR²σ)⁻¹/²
(c) (4 πR²Q / σ)¹/⁴
(d) (Q / 4 πR²σ)¹/⁴
Correct Answer:
(Q / 4 πR²σ)¹/⁴
Explanation:
Stefan’s law for black body radiation Q = σe AT⁴
T = [Q / σ(4πR²) ]¹/⁴ Here e= 1
A = 4πR²
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Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
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