Question

ABCD is a square plate with centre O. The moments of inertia of the plate about the perpendicular axis through O is $I_0$ and about the axes 1, 2, 3 and 4 are $I_1,I_2,I_3$ & $I_4$ respectively. It follows that :

• A. $I_2=I_3$
• B. $I_0=I_1+I_4$
• C. $I_0=I_2+I_4$
• D. $I_1=I_3$

Answer 1

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

The correct options are
A $I_0=I_1+I_4$
B $I_1=I_3$
C $I_0=I_2+I_4$
D $I_2=I_3$

According to Theorem of Perpendicular Axes, Moment of Inertia of the plate about an axis passing through the center O and perpendicular to the plane is,

$I_0=I_1+I_2=I_3+I_4$ ................(1) (Since, diagonals are mutually perpendicular to each other)

But, by symmetry rule,

$I_1=I_2=I$ (say) and $I_3=I_4=I^{\prime }$(say)

Hence, $I_0=2I=2I^{\prime }$ ...................(2)

Now, for square plate,

$I=\frac{M{L}^2}{12}$

Therefore,

$I_0=2\left(\frac{M{L}^2}{12}\right)$

$I_0=\frac{M{L}^2}{6}$

Hence, from (1),

$I_1+I_2=I_3+I_4=\frac{M{L}^2}{6}$

$I=I^{\prime }=\frac{M{L}^2}{12}$ ...............from(2)

$\Rightarrow I_1=I_2=I_3=I_4=\frac{M{L}^2}{12}$

$\Rightarrow I_0=I_1+I_4$

$\Rightarrow I_0=I_2+I_4$