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 $I_0$. It the smaller triangle $DEF$ is removed from $ABC$, the moment of inertia of the remaining figure about the same axis is $I$. Then:

• A. $I=\frac{9}{16}{I}_0$
• B. $I=\frac{3}{4}{I}_0$
• C. $I=\frac{I_0}{4}$
• D. $I=\frac{15}{16}{I}_0$

Answer 1

Answer: (D)

$I=\frac{15}{16}{I}_0$

Suppose $M$ is mass and a is the side of larger triangle, then $\frac{M}{4}$ and $\frac{a}{2}$ will be mass and side length of the smaller triangle

$\frac{I_{removed}}{I_{original}}=\frac{\frac{M}{4}{\left(\frac{a}{2}\right)}^2}{(a{)}^2}$

$I_{removed=\frac{I_0}{16}}$

So,$I=I_0-\frac{I_0}{16}=\frac{15I_0}{16}$