Four equal masses, m each are placed at the corners of a square of length l as shown in the figure. The…

Four equal masses, m each are placed at the corners 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 :

  1. ml2
  2. 2 ml2
  3. 3 ml2
  4. 3 ml2

Solution

Moment of inertia of point mass = mass × Perpendicular distance from axis2

Moment of Inertia 

=m02+ml22+ml22+ml22

=3 ml2

Asked in: JEE Main 2021 (16 Mar Shift 1)

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