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BCECE Engineering BCECE Engineering Solved Paper-2013

  • question_answer
    Three particles each of mass m gram, are situated at the vertices of an equilateral triangle ABC of the side 1 cm. The moment of inertia of the system (as shown in figure) about a line AX perpendicular to AB and in the plane of ABC in gram- cm2 units will be

    A)  \[\frac{3}{2}m{{l}^{2}}\]                              

    B)         \[\frac{3}{4}m{{l}^{2}}\]                              

    C)         \[2\,m{{l}^{2}}\]                             

    D)         \[\frac{5}{4}m{{l}^{2}}\]

    Correct Answer: D

    Solution :

    Moment of inertia of the system \[l={{l}_{1}}+{{l}_{2}}+{{l}_{3}}\] \[=0+m{{\left( \frac{l}{2} \right)}^{2}}+m{{l}^{2}}\] \[=\frac{m{{l}^{2}}}{4}+m{{l}^{2}}\] \[=\frac{5}{4}m{{l}^{2}}\]


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