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Transmission of Heat — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsTransmission of Heat1 Mark
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.
$\left(\frac{\pi}{6}\right)^{1 / 3}: 1$
$Q=\sigma A t(T-T)$If $T, T, \sigma$ and $t$ are same for both bodies then
$\frac{Q_{\text {sphere }}}{Q_{\text {cube }}}=\frac{A_{\text {sphere }}}{A_{\text {cube }}}=\frac{4 \pi r^2}{6 a^2}$
But according to problem, volume of sphere $=$ Volume of cube
$\Rightarrow \frac{4}{3} \pi r^3=a^3$
$ \Rightarrow a=\left(\frac{4}{3} \pi\right)^{1 / 3} r$
Substituting the value of a in equation (i) we get
$\frac{Q_{\text {sphere }}}{Q_{\text {cube }}}$
$=\frac{4 \pi r^2}{6 a^2}$
$=\frac{4 \pi r^2}{6\left\{\left(\frac{4}{3} \pi\right)^{1 / 3} r\right\}^2} $
$=\frac{4 \pi r^2}{6\left(\frac{4}{3} \pi\right)^{2 / 3} r^2}$
$=\left(\frac{\pi}{6}\right)^{1/31}$

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