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A Black metal foil receives radiation of power P from a hot sphere at absolute temperature T, kept at a distance d. If the temperature is doubled and distance is halved, then power will be

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64 P

16 P

4 P

8 P

answer is A.

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Detailed Solution

T₁ = T, T₂ = 2T ; d₁ = d, d₂ = d/2, p₁ = p
$P \propto \frac{T^{4}}{d^{2}}; \frac{P_{1}}{P_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} {(\frac{d_{2}}{d_{1}})}^{2} = \frac{1}{16} (\frac{1}{4})$
P₂ = 64P

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