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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64P

16P

answer is A.

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Detailed Solution

Power radiated from the surface of a hot spherical surface is P = eσAT⁴,

Intensity: $I = \frac{P}{4 π d^{2}}$

$∴ I \propto T^{4} / d^{2}$

$|T_{1} = T, T_{2} = 2 T; d_{1} = d, d_{2} = d / 2, P_{1} = P|$

$\Rightarrow P_{rcvd} = (\frac{P}{4 {πd}^{2}}) \times {area}_{foil} ∵ P \propto \frac{T^{4}}{d^{2}}; \Rightarrow \frac{P_{1}}{P_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} {(\frac{d_{2}}{d_{1}})}^{2} = \frac{1}{16} (\frac{1}{4})$

$\Rightarrow P_{2} = 64 P$

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