2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre and perpendicular to its plane is I0 as shown in the figure. A cavithy DEF is cut out from the lamina, where D,E,F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is [JEE Online 08-04-2017]

    A) \[\frac{15}{16}{{I}_{0}}\]                                              

    B) \[\frac{3{{I}_{0}}}{4}\]

    C) \[\frac{7}{8}{{I}_{0}}\]                                  

    D) \[\frac{31{{I}_{0}}}{32}\]

    Correct Answer: A

    Solution :

                    \[{{I}_{0}}=km{{l}^{2}}\]               \[BC=l\] \[{{I}_{DEF}}=K\frac{m}{4}{{\left( \frac{l}{2} \right)}^{2}}\] \[=\frac{k}{16}m{{l}^{2}}\] \[{{I}_{DEF}}=\frac{{{I}_{0}}}{16}\] \[{{I}_{remain}}={{I}_{0}}=\frac{{{I}_{0}}}{16}\] \[=\frac{15{{I}_{0}}}{16}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner