A) \[{{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{2}}\]
B) \[{{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}}\]
C) \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}\]
D) \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}\]
Correct Answer: A
Solution :
Radius of 1st body\[={{r}_{1}},\]radius of IInd body \[={{r}_{2}}\] Temperature of \[\text{Ist}\]body \[={{T}_{1}}\] Temperature of \[\text{IInd}\]body \[={{T}_{2}}\] The emissivity of a black body or radiated power is given by \[E=A\sigma {{T}^{4}}\times t\propto A{{T}^{4}}\] or\[A\propto \frac{E}{{{T}^{4}}}\](where A is the surface area of the spherical black, body) As in the condition of question the power radiated by first and second body is same. Hence, \[A\propto \frac{1}{{{T}^{4}}}\] so \[\frac{{{A}_{1}}}{{{A}_{2}}}={{\left( \frac{{{T}_{2}}}{T} \right)}^{4}}\] \[\frac{4\pi r_{1}^{2}}{4\pi r_{2}^{2}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}}\] Thus \[\frac{{{r}_{1}}}{{{r}_{2}}}={{\left( \frac{{{T}_{2}}}{T} \right)}^{2}}\]
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