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CET Karnataka Engineering CET - Karnataka Engineering 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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