MHT CET · Physics · Rotational Motion
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 \(\frac{L}{2}\)
- B \(\frac{L}{\sqrt{2}}\)
- C 2L
- D \(\frac{L}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{L}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(r = \frac{\sqrt{L^2+L^2}}{2} = \frac{L\sqrt{2}}{2} = \frac{L}{\sqrt{2}}\) \(I = \sum m r^2 = 4 M \left(\frac{L}{\sqrt{2}}\right)^2 = 4 M \frac{L^2}{2} = 2 M L^2\)
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