A highly conducting solid sphere of radius R, density $ρ$ and specific heat s is kept in an evacuated chamber. A parallel beam of electromagnetic radiation having uniform intensity I is incident on its surface. Assume that the surface of the sphere is perfectly black. Calculate the rate of increase of temperature.

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$\frac{3 (I - 4 σ T^{4})}{4 R ρ s}$

$\frac{3 (I - 2 σ T^{4})}{2 R ρ s}$

$\frac{3 (I + 4 σ T^{4})}{4 R ρ s}$

$\frac{4 (I - 4 σ T^{4})}{4 R ρ s}$

answer is B.

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Detailed Solution

As the radiation is incident as parallel beams.

The effective surface area of the sphere which receives the radiation is ${πR}^{2}$ .

And when the temperature of the sphere is T, the sphere will emit radiation obeying stefan's law

$m s (\frac{d T}{d t}) = [I π R^{2} - 4 π R^{2} σ T^{4}]$

$\frac{4}{3} π R^{3} ρ S (d T)$ $= [I π R^{2} - 4 π R^{2} σ T^{4}] d t$

$\frac{d T}{d t} = \frac{3}{4 R ρ S} [I - 4 σ T^{4}]$

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