BHU PMT BHU PMT Solved Paper-2003

  • question_answer
    The temperature of the black body increases from T to 2T. The factor by which the rate of emission will increase is:                                             [BHU M-2003]

    A)  4                                            

    B)  2

    C)  16                                         

    D)  8

    Correct Answer: C

    Solution :

                     From Stefan?s 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 \[\left( T \right)\] of the body,   that is                                    \[E=\sigma {{T}^{4}}\] Where \[\sigma \] is Stefan?s constant. Given, \[{{T}_{1}}=T,\,\,{{T}_{2}}=2T\]                 \[\therefore \]                  \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{T_{1}^{4}}{T_{2}^{4}}={{\left( \frac{T}{2T} \right)}^{4}}=\frac{1}{16}\]                 \[\Rightarrow \]                               \[{{E}_{2}}=16E\] Note: As the temperature of surface rises more and more radiant energy is emitted by it.           


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