JEE Main & Advanced Sample Paper JEE Main Sample Paper-24

  • question_answer
    A black body at high temperature T radiates energy at the rate of U (in \[W{{m}^{-2}}\]). When the temperature T falls to half (i.e. \[\frac{T}{2}\]), The radiated energy (in \[W{{m}^{-2}}\]) will be:

    A)  \[\frac{U}{8}\]                         

    B)  \[\frac{U}{16}\]

    C)  \[\frac{U}{4}\]                                     

    D)  \[\frac{U}{2}\]

    Correct Answer: B

    Solution :

    According to Stefan's law of radiation \[U\propto \,{{T}^{4}}\,(\therefore \,\,U\] is the energy) \[\Rightarrow \,\,\frac{{{U}_{1}}}{{{U}_{2}}}\,={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}({{T}_{1}}=T)\] \[\Rightarrow \,\frac{{{U}_{1}}}{{{U}_{2}}}={{\left( \frac{T}{T/2} \right)}^{4}}(\therefore \,\,{{T}_{2}}=\frac{T}{2})\] Or \[\frac{{{U}_{1}}}{{{U}_{2}}}={{\left( \frac{2}{1} \right)}^{4}}\,\,\,or\,\,\frac{{{U}_{1}}}{{{U}_{2}}}\,=\left( \frac{16}{1} \right)\] Or \[{{U}_{2}}=\frac{{{U}_{1}}}{16}\,\Rightarrow \,{{U}_{2}}=\frac{U}{16}\]


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