Question

The moment of inertia of a thin square plate $A B C D$, of uniform thickness about an axis passing through the centre $O$ and perpendicular to the plane of the plate is where $I_{1}, I_{2}, I_{3}$ and $I_{4}$ are respectively moments of inertia about axes $1,2,3$ and $4$ which are in the plane of the plate

• 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: (B)

$I_3+I_4$

Answer 2

Since, it is a square lamina
$I_{3}=I_{4}$
and $I_{1}=I_{2}$ (by symmetry)
From perpendicular axes theorem.
Moment of inertia about an axis perpendicular to square plate and passing from $O$ is
$I_{0}=I_{1}+I_{2}=I_{3}+I_{4}$
or $I_{0}=2 I_{2}=2 I_{3}$
Hence, $I_{2}=I_{3}$
Rather we can say $I_{1}=I_{2}=I_{3}=I_{4}$
Therefore, $I_{0}$ can be obtained by adding any two i.e.
$I_{0}=I_{1}+I_{2}=I_{1}+I_{3}$
$=I_{1}+I_{4}=I_{2}+I_{3}$
$=I_{2}+I_{4}=I_{3}+I_{4}$