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

Answer 1

Answer: (C)

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

Answer 2

Q = σ A t (T⁴ – T₀⁴)

If T, T₀, σ 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 ⇒ $\frac{4}{3}\pi r^{3} = a^{3}$ ⇒ $a = \left( \frac{4}{3}\pi \right)^{1/3}r$

Substituting the value of a in equation (i) we get

$\frac{Q_{sphere}}{Q_{cube}} = \frac{4\pi r^{2}}{6a^{2}} = \frac{4\pi r^{2}}{6\left\{ \left( \frac{4}{3}\pi \right)^{1/3}r \right\}^{2}} = \frac{4\pi r^{2}}{6\left( \frac{4}{3}\pi \right)^{2/3}r^{2}} = \left( \frac{\pi}{6} \right)^{1/3}:1$