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 :
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 :
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DEF has one quarter of the mass and half the linear size of ABC, and its centroid coincides with G. Its moment of inertia about G is \(\frac{1}{4}\times\frac{1}{4} = \frac{1}{16}\) of \(I_0\).
Updated On: Nov 5, 2026
- \(I = \frac{I_0}{4}\)
- \(I = \frac{15}{16}I_0\)
- \(I = \frac{9}{16}I_0\)
- \(I = \frac{3}{4}I_0\)
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The Correct Option is B
Solution and Explanation
For a plane lamina of mass \(M\) and linear size \(a\), \(I_0 \propto Ma^2\).
Triangle DEF has side \(\frac{a}{2}\); being cut from the same uniform sheet, its mass is \(\frac{M}{4}\).
Its centroid is also G, so its moment of inertia about the same axis is \(I_{DEF} \propto \frac{M}{4}\cdot\frac{a^2}{4} = \frac{1}{16}\,Ma^2\), i.e. \(I_{DEF} = \frac{I_0}{16}\).
Remaining figure: \(I = I_0 - \frac{I_0}{16} = \frac{15}{16}I_0\).
Answer: (2) \(I = \frac{15}{16}I_0\)
Triangle DEF has side \(\frac{a}{2}\); being cut from the same uniform sheet, its mass is \(\frac{M}{4}\).
Its centroid is also G, so its moment of inertia about the same axis is \(I_{DEF} \propto \frac{M}{4}\cdot\frac{a^2}{4} = \frac{1}{16}\,Ma^2\), i.e. \(I_{DEF} = \frac{I_0}{16}\).
Remaining figure: \(I = I_0 - \frac{I_0}{16} = \frac{15}{16}I_0\).
Answer: (2) \(I = \frac{15}{16}I_0\)
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