MHT CET · Physics · Rotational Motion

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)

A \(\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3\)
B \(\mathrm{I}_1+\mathrm{I}_3+\mathrm{I}_4\)
C \(\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3+\mathrm{I}_4\)
D \(\quad \mathrm{I}_1+\mathrm{I}_3\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\quad \mathrm{I}_1+\mathrm{I}_3\)

Step-by-step Solution

Detailed explanation

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 \(=\mathrm{I}=\mathrm{I}_1+\mathrm{I}_3\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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