Question

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.
  
• C.
• D.

Answer 1

The correct option is C

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