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A sphere and a cube of same material and same volume are heated upto same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiations emitted will be
$1 :1$
$\frac{{4\pi }}{3}:1$
${\left( {\frac{\pi }{6}} \right)^{\frac{1}{3}}}:1$
$\frac{1}{2}{\left( {\frac{{4\pi }}{3}} \right)^{\frac{2}{3}}}:1$
Solution
$\Delta \mathrm{E}=\mathrm{eA} \sigma\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right)$
$\frac{(\Delta \mathrm{E})_{\text {sphere }}}{(\Delta \mathrm{E})_{\text {cube}}}=\frac{\text { Surface area of sphere }}{\text { Surface area of cube }}=\frac{4 \pi \mathrm{R}^{2}}{6 \mathrm{a}^{2}}$
where $R$ is the radius of sphere and a is the side of the cube.
Given $\quad \frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{a}^{3} \Rightarrow \mathrm{a}=\left(\frac{4}{3} \pi\right)^{1 / 3} \cdot \mathrm{R}$
$\therefore \quad \frac{(\Delta E)_{\text {sphere }}}{(\Delta E)_{\text {cube }}}=\frac{4 \pi R^{2}}{6\left\{\left(\frac{4}{3} \pi\right)^{1 / 3} \cdot R\right\}^{2}}=\left(\frac{\pi}{6}\right)^{1 / 3}$
