Punjab Medical Punjab - MET Solved Paper-2008

  • question_answer
    The power of a black body at temperature \[200K\] is\[544W\]. 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 Stefans 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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