An equilateral triangle ABC is cut from a thin solid sheet of wood. (See figure) D , E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I 0 . If the smaller triangle DEF is removed from ABC , the moment of inertia of the remaining figure about the same axis is I . Then:

Question

An equilateral triangle ABC is cut from a thin solid sheet of wood. (See figure) D , E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I 0 . If the smaller triangle DEF is removed from ABC , the moment of inertia of the remaining figure about the same axis is I . Then:, I = 3 4 I 0, I = 15 16 I 0, I = 9 16 I 0, I = I 0 4

MARKS App · Accepted Answer

Dimension analysis I o = K M a 2 Now for small lamina I ′ = K M 4 a 2 2 = k m a 2 16 I ′ = I o 16 So moment of Inertia of remaining part I L = I o - I ′ = I o - I o 16 I L = 15 I o 16

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