Question

An equilateral triangle $ABC$ is cut from a thin solid sheet of wood. (See figure) $D,E\text{and}F$ 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:

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

Answer 1

The correct option is A $I=\frac{15}{16}{I}_0$

$\text{Let mass of the larger triangle}=M$
$\text{Side of larger triangle}=l$

Moment of inertia about an axis passing through $G$ and perpendicular to the plane of the larger triangle is given by,

$I_0=\frac{M{l}^2}{12}$

$\text{Length of smaller triangle},l^{\prime }=\frac{l}{2}$

As the mass of the thin triangle is proportional to the area,

$m\propto l^2$

$\Rightarrow \text{Mass of smaller triangle},M^{\prime }=\frac{M}{4}$

Therefore, the Moment of inertia of the removed triangle about an axis passing through $G$ and perpendicular to the plane of the triangle is given by,

$I_{\text{removed}}=\frac{1}{12}\frac{M}{4}{\left(\frac{l}{2}\right)}^2$

$\therefore \frac{I_{\text{removed}}}{I_0}=\frac{\frac{M}{4}}{M}.\frac{{\left(\frac{l}{2}\right)}^2}{(l{)}^2}$

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

The moment of the inertia of the remaining figure about an axis passing through $G$ and perpendicular to the plane of the triangle is given by,

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

Hence, option $\left(A\right)$ is correct.