Question

A sphere and a cube of same material and same volume are heated up to same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiations emitted in equal time1 intervals will be

• A.
• 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

$\frac{Q_{sphere}}{Q_{cube}}$ = $\frac{A_{sphere}}{A_{cube}}$ = $\frac{4\pi r^2}{6a^2}$

But according to problem, volume of sphere = volume of cube

$\Rightarrow$ $\frac{4}{3}\pi r^3$ = $a^3$

$\Rightarrow$ $a$ = $(\frac{4}{3}\pi {)}^{\frac{1}{3}}r$

Substituting the value of $a$ in Eq. 9(i), we get

$\frac{Q_{sphere}}{Q_{cube}}$ = $\frac{4\pi r^2}{6a^2}$

= $\frac{4\pi r^2}{6\left(\right(\frac{4}{3}\pi {)}^{\frac{1}{3}}r{)}^2}$ = $\frac{4\pi r^2}{6(\frac{4}{3}\pi {)}^{\frac{1}{3}}{r}^2}$

= $(\frac{\pi }{6}{)}^{\frac{1}{3}}$ : 1