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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
$1:1$
$\frac{{4\pi }}{3}\,\,:\,\,1$
${\left( {\frac{\pi }{6}} \right)^{1/3}}:\,\,1$
$\frac{1}{2}\,{\left( {\frac{{4\pi }}{3}} \right)^{2/3}}:\,\,1$
Solution
(c) $Q$ = $\sigma$ $A$ t ($T_4$ -$T_0^4$)
If $T, T_0, \sigma $ and t are same for both bodies then $\frac{{{Q_{sphere}}}}{{{Q_{cube}}}} = \frac{{{A_{sphere}}}}{{{A_{cube}}}} = \frac{{4\pi {r^2}}}{{6{a^2}}}$ …..$(i)$
But according to problem, volume of sphere = Volume of cube
==> $\frac{4}{3}\pi {r^3} = {a^3}$
==> $a = {\left( {\frac{4}{3}\pi } \right)^{1/3}}r$
Substituting the value of a in equation $(i)$ we get
$\frac{{{Q_{sphere}}}}{{{Q_{cube}}}} = \frac{{4\pi {r^2}}}{{6{a^2}}} = \frac{{4\pi {r^2}}}{{6{{\left\{ {{{\left( {\frac{4}{3}\pi } \right)}^{1/3}}r} \right\}}^2}}}$
$ = \frac{{4\pi {r^2}}}{{6\,{{\left( {\frac{4}{3}\pi } \right)}^{2/3}}{r^2}}} = {\left( {\frac{\pi }{6}} \right)^{1/3}}:1$
