BCECE Engineering BCECE Engineering Solved Paper-2011

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

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

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

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

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

    Correct Answer: D

    Solution :

    Moment of inertia of disc \[I=\frac{1}{2}M{{R}^{2}}=\frac{1}{2}M\left( \frac{M}{\pi \rho t} \right)\] \[\therefore \]  \[I=\frac{1}{2}\frac{{{M}^{2}}}{\pi \rho t}\] \[\left( \text{As}\,\rho =\frac{\text{Mass}}{\text{Volume}}\text{=}\frac{M}{\rho {{R}^{2}}t},\text{therefore,}{{\text{R}}^{2}}=\frac{M}{\pi \rho t} \right)\] \[\therefore \]  \[I\propto \frac{1}{\rho }\](If M and t are constants) \[\Rightarrow \]               \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{\rho }^{2}}}{{{\rho }_{1}}}\]


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