Question:
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 :
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 :
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Find each mass's perpendicular distance from the axis through A parallel to diagonal DB: A is at 0, B and D at \(l/\sqrt{2}\), C at \(\sqrt{2}\,l\).
Updated On: Nov 4, 2026
- \(\sqrt{3}\ ml^2\)
- \(2\ ml^2\)
- \(ml^2\)
- \(3\ ml^2\)
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The Correct Option is D
Solution and Explanation
Take A bottom-left, B bottom-right, C top-right, D top-left, side \(l\). The axis passes through A and is parallel to the diagonal DB.
The distance from A to the line DB is \(l/\sqrt{2}\), so the axis is at distance \(l/\sqrt{2}\) from DB.
Distances of the masses from the axis: A: 0; B and D (both lie on DB): \(l/\sqrt{2}\) each; C lies on the other side of DB at \(l/\sqrt{2}\) from it, so its distance is \(2 \times l/\sqrt{2} = \sqrt{2}\,l\).
\(I = m(0)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m(\sqrt{2}\,l)^2 = \frac{ml^2}{2} + \frac{ml^2}{2} + 2ml^2 = 3ml^2\).
Answer: (4) \(3\ ml^2\)
The distance from A to the line DB is \(l/\sqrt{2}\), so the axis is at distance \(l/\sqrt{2}\) from DB.
Distances of the masses from the axis: A: 0; B and D (both lie on DB): \(l/\sqrt{2}\) each; C lies on the other side of DB at \(l/\sqrt{2}\) from it, so its distance is \(2 \times l/\sqrt{2} = \sqrt{2}\,l\).
\(I = m(0)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m\left(\frac{l}{\sqrt{2}}\right)^2 + m(\sqrt{2}\,l)^2 = \frac{ml^2}{2} + \frac{ml^2}{2} + 2ml^2 = 3ml^2\).
Answer: (4) \(3\ ml^2\)
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