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JIPMER Jipmer Medical Solved Paper-1999

  • question_answer
    The critical angle for glass-water boundary is C. If \[{{\mu }_{g}}=1.5\] 1.5 and \[{{\mu }_{\omega }}=\frac{4}{3}\] then the value of C is:

    A)  \[{{\sin }^{-1}}\left( \frac{3}{2} \right)\]               

    B)         \[{{\sin }^{-1}}\left( \frac{4}{3} \right)\]               

    C)  \[{{\sin }^{-1}}\left( \frac{8}{9} \right)\]         

    D)         \[{{\sin }^{-1}}\left( \frac{1}{2} \right)\]

    Correct Answer: C

    Solution :

    From the relation \[\sin C=\frac{{{\mu }_{w}}}{{{\mu }_{g}}}=\frac{4/3}{3/2}=\frac{8}{9}\] \[C={{\sin }^{-1}}\frac{8}{9}\]


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