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Question

# The radius of gyration of rod of length â€²Lâ€² and mass â€²Mâ€² about an axis perpendicular to its length and passing through a point at a distance L/3 from one of its ends is

A
76L
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B
L29
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C
L3
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D
52L
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Solution

## The correct option is C L3Moment of inertia of a rod about its centre of mass is Icm=ML212 To calculate the moment of inertia about a point at a distance of L3 from the end we need to find the value of x which is the distance between the COM and this point. As the COM for a rod lies at the centre of the rod at L2 we have, We have from parallel axis theorem, I=Icm+Mx2 I=ML212+M(L6)2 I=ML2[112+136] I=ML2[436] If K is the radius of gyration, MK2=I=ML2[436] K2=L2[436] ⇒K=L×26=L3

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