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. 1:01
• 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

According to Stefen Boltzmann law, $Q = \sigma \,A\,t\left(T^{4}-T_{0}^{4}\right)$ If T, $T_{0}, \sigma$ and t are same for both bodies. Then, $\Rightarrow\, \frac{Q_{sphere}}{Q_{cube}}=\frac{A_{sphere}}{A_{cube}}=\frac{4\pi r^{2}}{6a^{2}}\quad\quad\quad...\left(i\right)$ Given, Volume of sphere = Volume of cube $\Rightarrow\quad \frac{4}{3}\pi r^{3}=a^{3}$ $\Rightarrow\,a = r\left(\frac{4}{3}\pi\right)^{1/3}$ 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}} $=\frac{4\pi r^{2}}{6\left(\frac{4}{3}\pi\right)^{1/3} r^{2}}=\left(\frac{\pi}{6}\right)^{1/3} :1$