AIIMS AIIMS Solved Paper-2001

  • question_answer
    A black body is heated from 27°C to \[{{127}^{o}}C\]. The ratio of their energies of radiation emitted will be :

    A)  9 : 16    

    B)                         27 : 64 

    C)         81 : 256               

    D)         3 : 4

    Correct Answer: C

    Solution :

                    From Stefahs law, the total radiant energy emitted per second per unit surface area of a black body is proportional to the fourth power of the absolute temperature of the body. That is  \[E=\sigma {{T}^{4}}\] where ct is Stefans constant. Given, initial temperature \[{{T}_{1}}=273+27=300K\] final temperature \[{{T}_{2}}=127+273=400K\] Hence, we have \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{T_{1}^{4}}{T_{2}^{4}}=\frac{{{(300)}^{4}}}{{{(400)}^{4}}}=\frac{81}{256}\] Therefore,   \[{{E}_{1}}:{{E}_{2}}=81:256\]


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