A sphere and a cube of same material and same volume are heated u… - Vidyadip

MCQ

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

A
$1:1$
B
$\frac{{4\pi }}{3}\,\,:\,\,1$
✓
${\left( {\frac{\pi }{6}} \right)^{1/3}}:\,\,1$
D
$\frac{1}{2}\,{\left( {\frac{{4\pi }}{3}} \right)^{2/3}}:\,\,1$

✓

Answer

Correct option: C.

${\left( {\frac{\pi }{6}} \right)^{1/3}}:\,\,1$

c
(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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