Final answer:
The moment of inertia (I) of a rectangular lamina with mass m for an axis through its center of mass and perpendicular to its diagonal is calculated by considering it as two rods with distributed mass. The simplified formula does not match the provided options, suggesting a potential typo in the question or the need for additional context.
Step-by-step explanation:
The moment of inertia (I) of an object is critical in understanding how the object will behave when it rotates around a fixed axis. For a rectangular lamina with mass m, length l, and width b, finding the moment of inertia about an axis passing through its center of mass and perpendicular to its diagonal involves a more complex calculation than for a point mass. According to the parallel-axis theorem, the moment of inertia for a rod about its end is mL², changing to mL²/12 if the rotation axis passes through its center. For the lamina, since the mass is evenly distributed and the axis is in the same plane as the lamina, we compute the moment of inertia for a compound shape.
To solve the question about the moment of inertia of a rectangular lamina, we can consider it a combination of two rods, one with length l and the other with width b. The mass distribution would be half of m for each rod since we are considering the axis through the center. The total moment of inertia would be the sum for each of these 'rods', giving us I = (1/2)ml²/12 + (1/2)mb²/12. After simplification, we obtain I = m(l² + b²)/24. This does not match any of the given options, suggesting a possible typo in the question or the need for additional context to understand the axis's orientation correctly.