CMC Medical CMC-Medical Ludhiana Solved Paper-2008

  • question_answer
    Two discs have same mass and same thickness. Their materials are of densities \[{{\rho }_{1}}\] and\[{{\rho }_{2}}\]. The ratio of their moments of inertia about central axis will be

    A)  \[{{\rho }_{1}}{{\rho }_{2}}:1\]                 

    B)  \[1:{{\rho }_{1}}{{\rho }_{2}}\]

    C)  \[{{\rho }_{1}}:{{\rho }_{2}}\]                   

    D)  \[{{\rho }_{2}}:{{\rho }_{1}}\]

    Correct Answer: D

    Solution :

                    From the formula \[{{I}_{1}}=\frac{1}{2}mr_{1}^{2}\]                 and        \[{{I}_{2}}=\frac{1}{2}mr_{2}^{2}\]                          Then,                \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{\frac{1}{2}mr_{1}^{2}}{\frac{1}{2}mr_{2}^{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}\] It t is thickness of each plate, then \[\pi r_{1}^{2}t{{\rho }_{1}}=\pi r_{2}^{2}t{{\rho }_{2}}\] or            \[\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{{{\rho }_{2}}}{{{\rho }_{1}}}\] \[\therefore \]  \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{{{\rho }_{2}}}{{{\rho }_{1}}}\]


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