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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2002

  • question_answer
    The temperature of star is 6060 K and maximum radiation emitted at 4653 A. For another star, the maximum radiation emitted at 4545 A. The temperature of another star will be

    A)  300 K           

    B)  6204 K

    C)  3000 K          

    D)  120 K

    Correct Answer: B

    Solution :

     From Wien's displacement law, \[{{\lambda }_{m}}T=\]constant \[\therefore \] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] Or \[\frac{4653}{4545}=\frac{{{T}_{2}}}{6060}\] Or         \[{{T}_{2}}=6204K\]


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