Animated Solution for Physics - Rotational Motion: Let $I$ be the moment of inertia of a uniform square plate about an axis $AB$ that passes through its centre and is parallel to two of its sides. $CD$ is a line in the plane of the plate that passes through the centre of the plate and makes an angle $\theta$ with $AB$. The moment of inertia of the plate about the axis $CD$ is then equal to

The moment of inertia is often perceived as a rigid, purely mathematical property, but it is deeply intertwined with the physical symmetry of an object. When we encounter a highly symmetric body like a uniform square plate, we can often bypass tedious integration and arrive at elegant solutions using pure logic and fundamental theorems.

Imagine you are looking straight down at a uniform square plate resting on a table. We are given an axis, $AB$, that passes directly through the center of the square and is parallel to its edges. The moment of inertia about this axis is given as $I$.

Now, let's draw a second axis, $A^{\prime }{B}^{\prime }$, which also passes through the center but is exactly perpendicular to $AB$. If you were to rotate the square plate by $90^{\circ }$, it would look completely unchanged. This perfect rotational symmetry implies that the mass distribution of the plate around the axis $A^{\prime }{B}^{\prime }$ is physically indistinguishable from the mass distribution around $AB$.

Because the mass is distributed identically relative to both axes, their moments of inertia must be exactly the same. Therefore, we can confidently state:

This simple observation is the crucial first step in unlocking the problem.

With two perpendicular axes lying in the plane of the plate, the stage is perfectly set for the Perpendicular Axes Theorem. This powerful theorem states that for any planar lamina, the moment of inertia about an axis perpendicular to the plane (let's call it the $Z$-axis) is equal to the sum of the moments of inertia about any two mutually perpendicular axes lying in the plane and intersecting the $Z$-axis.

Applying this theorem to our axes $AB$ and $A^{\prime }{B}^{\prime }$, we get:

Substituting the values we know, we find the moment of inertia about the central perpendicular axis:

This value, $2I$, is an intrinsic property of the square plate and remains constant regardless of how we draw our in-plane axes, as long as they intersect at the center.

The problem introduces a new axis, $CD$, which passes through the center but is tilted at an arbitrary angle $\theta$ with respect to $AB$. We need to find the moment of inertia about this axis, $I_{CD}$.

To solve this, we employ the exact same strategy. We construct a new axis, $C^{\prime }{D}^{\prime }$, that is perfectly perpendicular to $CD$ and lies in the plane of the plate.

Once again, the symmetry of the square comes to our rescue. No matter how you orient a pair of perpendicular crosshairs through the center of a square, the mass distribution around one crosshair is identical to the mass distribution around the other. Thus, by symmetry:

We now have a new pair of mutually perpendicular in-plane axes, $CD$ and $C^{\prime }{D}^{\prime }$. We can apply the Perpendicular Axes Theorem to this pair as well. The sum of their moments of inertia must equal the moment of inertia about the $Z$-axis, which we already calculated.

Since $I_{C^{\prime }{D}^{\prime }}=I_{CD}$, we can substitute this into the equation:

Now, we simply equate our two different expressions for $I_Z$. We know that $I_Z=2I$, so:

Canceling the factor of 2 from both sides yields our final, elegant result:

This reveals a profound property of the square plate: its moment of inertia about any axis passing through its center in its plane is exactly the same, completely independent of the angle $\theta$.

In competitive exams like JEE, time is of the essence. We can verify this result almost instantly using boundary conditions. Consider what happens when $\theta =0^{\circ }$. The axis $CD$ perfectly aligns with $AB$, so $I_{CD}$ must equal $I_{AB}$, which is $I$.

Now consider when $\theta =90^{\circ }$. The axis $CD$ aligns with $A^{\prime }{B}^{\prime }$, and we already established that $I_{A^{\prime }{B}^{\prime }}=I$.

Looking at the given options, only option (a), which is the constant $I$, satisfies both of these extreme cases. The other options involving sine and cosine would yield different values at these angles. This logical check confirms our derived result with absolute certainty.