JEE Main 2017 · 8 Apr · Shift 1 · Physics Q6

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PhysicsSingle correct+4 / -1Question 6 of 90

Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is $I_o$ as shown in the figure. A cavity 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 :

Equilateral triangle ABC with centre O; D, E and F are the midpoints of AB, BC and CA, and the inner triangle DEF is marked.

\dfrac78I_o

\dfrac{15}{16}I_o

\dfrac{3I_o}4

\dfrac{31I_o}{32}

Answer and worked solution

Official answer: B

Triangle DEF has $\frac14$ of the mass and half the side, and its centroid is also O. Moment of inertia scales as mass × side², so
$I_{DEF}=\frac14\cdot\frac14I_o=\frac{I_o}{16},\qquad I=I_o-\frac{I_o}{16}=\frac{15}{16}I_o.$

Topics:Moment of inertia

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