Question

Four particles each of mass \(M\) are placed at the corners of a square of side \(L\). The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

• A. \(\text{L}\)
• B. \(\frac{\text{L}}{2}\)
• C. \(\frac{\text{L}}{4}\)
• D. \(\frac{\text{L}}{\sqrt{2}}\)

Hint:

For square, centre to corner distance = \(L/\sqrt{2}\).

Answer 1

The Correct Option is D

Concept:
Radius of gyration: \[ I = Mk^2 \] where \(I\) = moment of inertia, \(k\) = radius of gyration. Step 1: Find distance of each mass from centre. For a square of side \(L\), distance of corner from centre: \[ r = \frac{L}{\sqrt{2}} \]
Step 2: Calculate moment of inertia. Each mass contributes: \[ I = M r^2 = M \left(\frac{L}{\sqrt{2}}\right)^2 = \frac{ML^2}{2} \] For 4 masses: \[ I_{\text{total}} = 4 \times \frac{ML^2}{2} = 2ML^2 \]
Step 3: Find radius of gyration. Total mass = \(4M\) \[ I = (4M)k^2 \] \[ 2ML^2 = 4M k^2 \] \[ k^2 = \frac{L^2}{2} \]
Step 4: Conclusion. \[ k = \frac{L}{\sqrt{2}} \]