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

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1 : 1

16 : 1

4 : 1

1 : 9

answer is A.

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Detailed Solution

$E = A T^{4}$

$\frac{E_{1}}{E _{2}} = \frac{π r_{1}^{2} T_{1}^{4}}{π r_{2}^{2} T_{2}^{4}} = \frac{1}{16} \times \frac{{(4 \times 10^{3})}^{4}}{{(2 \times 10^{3})}^{4}}$

$= \frac{1}{16} \times \frac{4 \times 4 \times 4 \times 4}{2 \times 2 \times 2 \times 2} = \frac{16}{16} = \frac{1}{1}$

$E_{1} : E_{2} = 1 : 1$

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