Let l be the moment of inertia of an 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 θ with AB . The moment of inertia of the plate about the axis CD is then equal to

The correct option is l.

We can solve this problem using the perpendicular axis theorem and the rotational symmetry of a square plate.

Step 1: Understand the given axes
Let the uniform square plate lie in the $xy$-plane, with its centre at the origin.
The axis $AB$ passes through the centre and is parallel to two of its sides. Let us align the $xt$-axis along $AB$.
Therefore, the moment of inertia about the $xt$-axis is:
$I_x=l$

Step 2: Use symmetry of the square plate
Let the $y$-axis be perpendicular to the $xt$-axis in the plane of the plate, passing through its centre. By symmetry, the moment of inertia of the square plate about the $y$-axis must be equal to that about the $xt$-axis:
$I_y=l$

Step 3: Apply the Perpendicular Axis Theorem
According to the perpendicular axis theorem, the moment of inertia about the $zt$-axis (which is perpendicular to the plane of the plate and passes through its centre) is given by:
$I_z=I_x+I_y$
Substituting the values:
$I_z=l+l=2l$

Step 4: Analyze the axis CD
The axis $CD$ lies in the plane of the plate, passes through the centre, and makes an angle $\theta$ with $AB$.
Let us define another axis, say $EF$, in the plane of the plate that passes through the centre and is perpendicular to $CD$.
Since $CD$ and $EF$ are two mutually perpendicular axes in the plane of the plate passing through its centre, we can apply the perpendicular axis theorem to them as well:
$I_z=I_{CD}+I_{EF}$

Due to the rotational symmetry of a square plate in its plane, the moment of inertia about any axis passing through its centre in the plane of the plate is identical. Therefore:
$I_{CD}=I_{EF}$

Substituting this back into the relation:
$I_z=2I_{CD}$
Since we already found $I_z=2l$:
$2l=2I_{CD}$
$I_{CD}=l$

Thus, the moment of inertia of the plate about the axis $CD$ is equal to $l$, independent of the angle $\theta$.