NEET Physics Transmission of Heat NEET PYQ-Transmission Of Heat

  • question_answer
    The power radiated by a black body is P and it radiates maximum energy at wavelength, \[{{\lambda }_{0}}\]. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength \[\frac{3}{4}{{\lambda }_{0}}\], the power radiated by it becomes \[\text{nP}\]. The value of n is                                                                                                                                         [NEET - 2018]

    A)  \[\frac{256}{81}\]     

    B)       \[\frac{4}{3}\]

    C)  \[\frac{3}{4}\]                       

    D)       \[\frac{81}{256}\]

    Correct Answer: A

    Solution :

    [a] We know,
    \[{{\lambda }_{\text{max}}}\text{T=}\]constant (Wien's law)
    So, \[{{\lambda }_{\text{ma}{{\text{x}}_{\text{1}}}}}{{\text{T}}_{\text{1}}}\text{=}{{\lambda }_{\text{ma}{{\text{x}}_{\text{2}}}}}{{\text{T}}_{\text{2}}}\]
    \[\Rightarrow {{\lambda }_{0}}T=\frac{3{{\lambda }_{0}}}{4}T'\]
    \[\Rightarrow T'=\frac{4}{3}T\]
    So, \[\frac{{{P}_{2}}}{{{P}_{1}}}={{\left( \frac{T'}{T} \right)}^{4}}={{\left( \frac{4}{3} \right)}^{4}}=\frac{256}{81}\]


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