Four equal masses, m each are placed at the comers of a square of length (l) as shown in the figure. The moment of inertia of the system about an axis passing through A and parallel to DB would be :

Solution

Physics sir · 4 steps

The axis through A runs parallel to DB, so every distance is a piece of the diagonal AC.

1. Find the distances. The diagonals of a square are perpendicular, so distances from the axis are measured along AC, which has length √2 l. A is on the axis. D and B lie on the line DB, which is half a diagonal away, l/√2. C is the full diagonal away, √2 l.

   I = m(0)² + m(l/√2)² + m(l/√2)² + m(√2 l)²

2. Add the terms. This gives I = 0 + ml²/2 + ml²/2 + 2ml² = 3ml².
3. Quick check. About DB itself, A and C are each l/√2 away, so I_(DB) = ml². The parallel axis theorem with total mass 4m and shift l/√2 gives ml² + 4m × l²/2 = 3ml², the same result. This works because DB passes through the centre of mass.
4. So. The moment of inertia about the axis through A parallel to DB is 3ml², which is option (d).

The idea to keep. For point masses, use I = Σ mr² with r as the perpendicular distance to the axis. For a square, the diagonals are perpendicular, so distances along one diagonal settle it.

Answer from the NTA final answer key. Solution by PrepWiser.