Question

Two spheres of the same material have radii $ 1 \,m $ and $ 4\, m $ and temperatures $ 4000\, K $ and $ 2000\, K $ respectively., The ratio of energy radiated per second by the first sphere to the second is

• A. $ 1 : 1 $
• B. $ 16 : 1 $
• C. $ 4 : 1 $
• D. $ 1 : 9 $

Answer 1

Answer: (A)

Energy radiated per second by a body which has surface area $A$ at temperature $T$ is given by Stefan's law,
$E = \sigma AT^4$
Therefore
$\frac{E_{1}}{E_{2}} = \left(\frac{r_{1}}{r_{2}}\right)^{2}\left(\frac{T_{1}}{T_{2}}\right)^{4}$
$ = \left(\frac{1}{4}\right)^{2}\left(\frac{4000}{2000}\right)^{4}$
$ \frac{E_{1}}{E_{2}} = \frac{16}{16} = \frac{1}{1}$