Question

A thin square plate $ABCD$ of uniform thickness is shown in the figure. $I_1,I_2,I_3$ and $I_4$ are respectively the moments of inertia about axis $1,2,3$ and $4$ which are in the plane of the square plate. Then, the moment of inertia $I_0$ of the plate about an axis passing through its centre $\text{O}$ and perpendicular to the plane of the plate will be equal to : (Choose the correct option(s))

• A. $I_1+I_2$
• B. $I_3+I_4$
• C. $I_1+I_3$
• D. $I_1+I_2+I_3+I_4$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $I_1+I_2$
B $I_3+I_4$
C $I_1+I_3$

Let $I_0$ be the moment of inertia about an axis perpendicular to the square plate and passing through the centre $\text{O}$.
The axis $1$ and $2$ are perpendicular and their point of intersection is $\text{O}$, hence by perpendicular axis theorem,
$I_0=I_1+I_2...\left(i\right)$
The axis $3$ and $4$ are perpendicular and their point of intersection is $\text{O}$, hence by perpendicular axis theorem,
$I_0=I_3+I_4...\left(ii\right)$

Adding Eq $\left(i\right)$ and $\left(ii\right)$ gives
$I_0+I_0=(I_1+I_2)+(I_3+I_4)$
$\because$ square plate is symmetrical with respect to axis $1,2$ and $3,4$ hence $I_1=I_2,\&I_3=I_4$

$\Rightarrow 2I_0=2\left(I_1\right)+2\left(I_3\right)$
$\therefore I_0=I_1+I_3$

$\Rightarrow$ Options $\left(a\right),\left(b\right),\left(c\right)$ are correct.

For option $\left(d\right)$, from eq $\left(i\right)\&\left(ii\right)$
$(I_1+I_2)+(I_3+I_4)=I_0+I_0=2I_0$
Option (d) is wrong.