BHU PMT BHU PMT (Mains) Solved Paper-2008

  • question_answer
    The power of a black body at temperature 200 K is 544 W. Its surface area is \[(\sigma =5.67\times {{10}^{-8}}W{{m}^{-2}}{{K}^{-4}})\]

    A)  \[6\times {{10}^{-2}}{{m}^{2}}\]              

    B)  \[6{{m}^{2}}\]

    C)  \[6\times {{10}^{-6}}{{m}^{2}}\]              

    D)  \[6\times {{10}^{2}}{{m}^{2}}\]

    Correct Answer: B

    Solution :

                     According to Stefan's law \[\frac{Q}{At}=\sigma {{T}^{4}}\] Here, \[\frac{Q}{t}=544,\] \[T=200K,\]                 \[\sigma =5.67\times {{10}^{-8}}W{{m}^{-2}}{{K}^{-4}}\] \[\therefore \]\[A=\frac{Q}{\sigma t{{T}^{4}}}=\frac{544}{5.67\times {{10}^{-8}}\times {{(200)}^{4}}}=6\,{{m}^{2}}\]


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