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EAMCET Medical EAMCET Medical Solved Paper-2003

  • question_answer
    Two circular loops A and B or radii rA and rB respectively are made from a uniform wire. The ratio of their moments of inertia about axes passing through their centres and perpendicular to their planes is \[\frac{{{I}_{B}}}{{{I}_{A}}}=8,\] then \[\left( \frac{{{r}_{B}}}{{{r}_{A}}} \right)\] equal to:

    A)  2                                            

    B)  4

    C)  6                                            

    D)  8

    Correct Answer: A

    Solution :

                     Moment of inertia of a loop about axis passing through their centre     and perpendicular to their plane which is \[\frac{{{I}_{B}}}{{{I}_{A}}}=8\]                 Hence,                  \[\frac{\frac{1}{2}mr_{B}^{2}}{\frac{1}{2}mr_{A}^{2}}=8\]                 or            \[\frac{r_{B}^{2}}{r_{A}^{2}}=\frac{8}{2}=4\]or \[\frac{{{r}_{B}}}{{{r}_{A}}}=2\] So,          \[\frac{{{r}_{B}}}{{{r}_{A}}}=\frac{2}{1}\]


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