Two spheres of same material have radius 1 m and 4 m and temperature 4000 K and 2000 K respectively. The energy radiated per second by the first sphere is

Options

A. greater than that by the second
B. less than that by the second
C. equal in both cases
D. the information is incomplete to draw any conclusion

Answer

C

Explanation

The energy radiated per second E, by a body of surface area A, at temperature T K, is given by
`E = esigma AT^(4)`
Hence, `E_(1) = esigma4pi(1)^(2) xx (4000)^(4) = esigmapi xx 1024 xx 10^(12) J s^(-1)`
and `E_(2) = esigma4pi(4)^(2)(2000)^(4) = esigmapi xx 1024 xx 10^(12) J s^(-1)`
So, `E_(1) = E_(2)`