Two bodies A and B having equal surface areas are maintained at temperature 10°C and 20°C. The thermal radiation emitted in a given time by A and B are in the ratio

(a) 1 : 1.15
(b) 1 : 2
(c) 1 : 4
(d) 1 : 16

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Solution

(a) 1 : 1.15

From Stefan-Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by
$u = σ A T^{4}$
Here, $σ$ is Stefan-Boltzmann constant.

The thermal radiation emitted in a given time by A and B will be in the ratio
$\frac{u_{A}}{u_{B}} = \frac{{T_{A}}^{4}}{{T_{B}}^{4}} \frac{u_{A}}{u_{B}} = \frac{(273 + 10)^{4}}{(273 + 20)^{4}} \frac{u_{A}}{u_{B}} = \frac{1}{1.15}$

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