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JEE Main & Advanced Sample Paper JEE Main Sample Paper-17

  • question_answer
    The energy emitted per seconds by a black body at \[27{{\,}^{o}}C\]is\[\text{ }\!\!~\!\!\text{ 10 J}\text{.}\] If the temperature of the black body is increased to \[327{{\,}^{o}}C,\] the energy emitted per second will be

    A)  80 J                                       

    B)  160 J

    C)  \[2.15\times {{10}^{5}}\,J\]                       

    D)  \[120\,J\]

    Correct Answer: B

    Solution :

     For a black body, rate at which energy is  emitted by it is given by \[P=\sigma A{{T}^{4}},\]where T is kelvin, \[{{P}_{t}}=10\,J=\sigma A\times {{(300)}^{4}}\] \[{{P}_{f}}=\sigma A{{(600)}^{4}}=16\times 10=160\,J\]


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