Rotational Motion

JEE Main level, single correct, medium

87 of 280

Question 87 of 280

Moment of inertia of an equilateral triangular lamina ABCABC, about the axis passing through its centre OO and perpendicular to its plane is I0{I}_{0} as shown in the figure. A cavity DEFDEF is cut out from the lamina, where D,E,FD,E,F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is:

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Answer: (B) 1516I0\dfrac{15}{16}{I}_{0}

Given,
AB=BC=AC=lAB=BC=AC=l

Moment of inertia of a triangular lamina ABCABC

I0=kml2{I}_{0}=km{l}^{2}

DE=EF=DF=12DE=EF=DF=\dfrac{1}{2},
AB=l2AB=\dfrac{l}{2}

∴∴ Moment of inertia of ∆DEF∆DEF

IDEF=km4(l2)2{I}_{DEF}=k\dfrac{m}{4}{\left(\dfrac{l}{2}\right)}^{2}

IDEF=k16ml2{I}_{DEF}=\dfrac{k}{16}m{l}^{2}.

IDEF=I016{I}_{DEF}=\dfrac{{I}_{0}}{16}

Moment of inertia of the remaining part

Iremain=I0−I016{I}_{remain}={I}_{0}-\dfrac{{I}_{0}}{16}

=15I016=\dfrac{15{I}_{0}}{16}

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