A) \[\sqrt{3}:1\]
B) \[\sqrt{2}:1\]
C) \[1:\sqrt{5}\]
D) \[1:\sqrt{8}\]
Correct Answer: B
Solution :
Radiating power of a body of area \[{{A}_{1}},\]emissivity \[{{e}_{1}}\] and surface temperature \[{{T}_{1}}\] is \[{{P}_{1}}=\sigma {{e}_{1}}T_{1}^{4}{{A}_{1}}\] ?(i) Similarly, radiating power of a body of area \[{{A}_{2}},\] emissivity \[{{e}_{2}}\] and surface temperature \[{{T}_{2}}\] is \[{{P}_{2}}=\sigma {{e}_{2}}T_{2}^{4}{{A}_{2}}\] ?(ii) Given, \[{{P}_{1}}={{P}_{2}}\] and \[{{A}_{1}}={{A}_{2}}\] \[\sigma {{e}_{1}}T_{1}^{4}{{A}_{1}}=\sigma {{e}_{2}}T_{2}^{4}{{A}_{2}}\] \[\Rightarrow \] \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}=\left( \frac{{{e}_{2}}}{{{e}_{1}}} \right)\,\left( \frac{{{A}_{2}}}{{{A}_{1}}} \right)\] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{e}_{2}}}{{{e}_{1}}} \right)}^{1/4}}\] \[(\because \,{{A}_{1}}={{A}_{2}})\] Hence, \[{{e}_{1}}=0.2\] and \[{{e}_{2}}=0.8\] Hence, \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{0.8}{0.2} \right)}^{1/4}}={{\left( \frac{4}{1} \right)}^{1/4}}={{\left( \frac{{{2}^{2}}}{{{1}^{1}}} \right)}^{1/4}}\] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{2}{1} \right)}^{1/2}}\] \[=\sqrt{\frac{2}{1}}\] \[{{T}_{1}}:{{T}_{2}}=\sqrt{2}:1\]
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