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The radius of gyration of a uniform rod of length l, about an axis passing through a point away from the center of the rod, and perpendicular to it is
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a
748L
b
38L
c
18L
d
14L
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Correct option is A
Using Parallel Axis Theorem, IC+Md2 =IP ⇒ML212+ML42=MK2 here K = Radius of Gyration⇒L212+L216=K2 ⇒ K=748LSimilar Questions
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