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A thin rod of length L spins about a perpendicular axis through its centre. Its radius of gyration is
AL/12
BL/√3
CL/√12
DL/2
Answer & Solution
Correct answer: C. L/√12
With I = ML^2/12 we get k^2 = L^2/12, so k equals L divided by the square root of 12.
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