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
Diffcult
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(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$
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