question_answer

Two identical metal balls at temperature \[{{200}^{o}}C\]and \[{{400}^{o}}C\] kept in air at \[{{27}^{o}}C\]. The ratio of net heat loss by these bodies is                                                           [CPMT 2002]

A)            1/4                                           

B)            1/2

C)            1/16                                         

D)                    If temperature of surrounding is considered then net loss of energy of a body by radiation \[Q=A\varepsilon \sigma ({{T}^{4}}-T_{0}^{4})t\]Þ\[Q\propto ({{T}^{4}}-T_{0}^{4})\] Þ \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{T_{1}^{4}-T_{0}^{4}}{T_{2}^{4}-T_{0}^{4}}\] \[=\frac{{{(273+200)}^{4}}-{{(273+27)}^{4}}}{{{(273+400)}^{4}}-{{(273+27)}^{4}}}\]\[=\frac{{{(473)}^{4}}-{{(300)}^{4}}}{{{(673)}^{4}}-{{(300)}^{4}}}\]