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Question

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
l2
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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 A l2

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)

Now, (1)=(2)
2ml2=4Mk2k=l2

hence, option A is correct.



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