Four point size bodies each of mass m are fixed at four corners of a light square frame of side length L. The radius of gyration of these four bodies about an axis perpendicular to the plane of frame passing through its centre is:
A
√2L
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B
2L
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C
L√2
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D
L2
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Solution
The correct option is CL√2 I=4⎛⎝m(L√2)2⎞⎠ =2mL2 So we have radius of gyration equivalence equation as, 4mk2=2mL2 k=L√2