The moment of inertia of thin square plate PQRS of uniform thickness, about an axis passing through centre '…

The moment of inertia of thin square plate PQRS of uniform thickness, about an axis passing through centre ' O ' and perpendicular to the plane of the plate is $\left(\mathrm{I}_1, \mathrm{I}_2, \mathrm{I}_3, \mathrm{I}_4\right.$ are respectively the moments of inertia about axis $1,2,3,4$ which are in the plane of the plate as shown in figure)
  1. $\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3$
  2. $\mathrm{I}_1+\mathrm{I}_3+\mathrm{I}_4$
  3. $\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3+\mathrm{I}_4$
  4. $\quad \mathrm{I}_1+\mathrm{I}_3$

Solution

Axis of $I_1$ and $I_2$ and that of $I_3$ and $I_4$ are pependicular to each other. By theorem of perpendicular axis, $\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2 \quad \text { or } \mathrm{I}=\mathrm{I}_3+\mathrm{I}_4...(i)$
As it is a square, $\mathrm{I}_1=\mathrm{I}_2 \quad \text { and } \mathrm{I}_3=\mathrm{I}_4$
From (i), $\begin{aligned} & \mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{I}}{2} \\ & \mathrm{I}_3=\mathrm{I}_4=\frac{\mathrm{I}}{2} \end{aligned}$ $\therefore \quad \mathrm{I}_3=\mathrm{I}_1$ $\therefore \quad$ Momentof inertia of the plate $=\mathrm{I}=\mathrm{I}_1+\mathrm{I}_3$

Asked in: MHT CET 2024 (09 May Shift 1)

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