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QJEEM20Ph8d82: JEE Main 2020 Physics - Class 11 JEEM20Ph8d82 Solution: ABC is a plane lamina of the shape of an equilateral triangle.…

ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid points of AB, AC and G is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is I0 . If part ADE is removed, the moment of inertia of the remaining part about the same axis is NI016{{N{I_0}} \over {16}} where N is an integer. Value of N is ______. JEE Main 2020 (Online) 4th September Morning Slot Physics - Rotational Motion Question 127 English

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JEE Main 2020 (Online) 4th September Morning Slot Physics - Rotational Motion Question 127 English Explanation

Let  I0=kml2\,{I_0} = km{l^2}

IDEF=I′=K(m4)(l2)2{I_{DEF}} = I' = K\left( {{m \over 4}} \right){\left( {{l \over 2}} \right)^2}

= ml216{{m{l^2}} \over {16}} =(I016)= \left( {{{{I_0}} \over {16}}} \right)

and IADE=IBDE=IEFC=I{I_{ADE}} = {I_{BDE}} = {I_{EFC}} = I

∴\therefore 3I + I' = I0

⇒\Rightarrow 3I=I0−I016=15I0163I = {I_0} - {{{I_0}} \over {16}} = {{15{I_0}} \over {16}}

⇒I=5I016\Rightarrow I = {{5{I_0}} \over {16}}

Iremaining=2I+I016=11I016{I_{remaining}} = 2I + {{{I_0}} \over {16}} = {{11{I_0}} \over {16}}

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