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  4. 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 and perpendicular to the plane of the triangle is I0. It 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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Q. 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 and perpendicular to the plane of the triangle is $I_0$. It the smaller triangle $DEF$ is removed from $ABC$, the moment of inertia of the remaining figure about the same axis is $I$. Then:Physics Question Image

JEE MainJEE Main 2019System of Particles and Rotational Motion

Solution:

Suppose $M$ is mass and a is side of larger triangle, then $\frac{M}{4}$ and $\frac{a}{2}$ will be mass and side length of smaller triangle.
$\frac{I_{\text{removed}}}{I_{\text{original}}} = \frac{\frac{M}{4}}{M} . \frac{\left(\frac{a}{2}\right)^{2}}{\left(a\right)^{2}} $
$I_{\text{removed}} = \frac{I_{0}}{16} $
So, $I = I_{0} - \frac{I_{0}}{16} = \frac{15I_{0}}{16} $