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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2002

  • question_answer
    The energy spectrum of a black body exhibits a maximum around a wavelength\[{{\lambda }_{0}}\]. The temperature of the black body is now changed such that the energy is maximum around a wavelength 3\[{{\lambda }_{0}}\]/4. The power radiated by the black body will now increase by a factor of:

    A)  64/27

    B)  256/81

    C)  4/3

    D)  16/9

    Correct Answer: B

    Solution :

     \[\lambda T=\] constant \[{{\lambda }_{0}}T=\frac{3{{\lambda }_{0}}}{4}{{T}_{1}}\] \[{{T}_{1}}=\frac{4}{3}T\] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{T_{1}^{4}}{T_{2}^{4}}=\frac{{{T}^{4}}}{{{\left( \frac{4}{3}T \right)}^{4}}}\] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{81}{256}\] \[{{E}_{2}}=\frac{256}{81}{{E}_{1}}\]


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