Question

If a sphere and a cube of same material having same volume are heated upto same temperature and allowed to cool in the same surrounding then ratio of amounts of radiations emitted by them will be

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

Answer 1

The correct option is D ${\left(\frac{\pi }{6}\right)}^{1/3}:1$

The radiation emitted,
$Q=\sigma AT(T^4-T_0^4)$
Here, for sphere and cube, $\sigma ,t,T$ and $T_0$ are same.
$\therefore \frac{Q_{sphere}}{Q_{cube}}=\frac{A_{sphere}}{A_{cube}}=\frac{4\pi r^2}{6a^2}.....\left(i\right)$
Given, Volume of sphere $=$ Volume of cube
$\frac{4}{3}\pi r^3=a^3$
$a={\left(\frac{4}{3}\pi \right)}^{1/3}\cdot r....\left(ii\right)$
Substituting value of Eqs. (ii) in Eq. (i), we get
$\frac{Q_{sphere}}{Q_{cube}}=\frac{4\pi r^2}{6{\{{\left(\frac{4}{3}\pi \right)}^{1/3}\cdot r\}}^2}={\left(\frac{\pi }{6}\right)}^{1/3}:1$.