A black sphere has radius $R$ whose rate of radiation is E at temperature T . If radius is made half and temperature 4 T , the rate of radiation will be

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Aicharya · Accepted Answer

$\mathrm{E}=\mathrm{e} \mathrm{~A} \sigma \mathrm{~T}^4$
But for sphere, $A=4 \pi R^2$ $\begin{array}{ll} \therefore & \quad \mathrm{E}=\mathrm{e}\left(4 \pi \mathrm{R}^2\right) \sigma \mathrm{T}^4 \\ \therefore & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{R}_1^2 \mathrm{~T}_1^4}{\mathrm{R}_2^2 \mathrm{~T}_2^4} \\ \therefore & \frac{\mathrm{E}}{\mathrm{E}_2}=\frac{\mathrm{R}^2 \mathrm{~T}^4}{\left(\frac{\mathrm{R}}{2}\right)^2(4 \mathrm{~T})^4} \\ \therefore & \mathrm{E}_2=64 \mathrm{E} \end{array}$

A black sphere has radius $R$ whose rate of radiation is E at temperature T . If radius is made half and…

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