Four particles each of mass m are placed at the corners of a square of side length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is
A
l√2
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B
l2
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C
l
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D
√2l
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Solution
The correct option is Al√2
For this system, Itotal=IA+IB+IC+ID
or Itotal=4×IA
calculation of moment of inertia for point A, IA=M(√2l2)2IA=Ml22
So, Itotal=4×Ml22=2Ml2 ....(1)
When converts into a closed ring, where, Iring=4Mk2 ....(2)