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Question

# The radius of gyration of a plane lamina of mass M, length L and breadth B about an axis passing through its center of gravity and perpendicular to its plane will be

A
(L2+B2)12
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B
(L2+B2)8
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C
(L2+B2)2
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D
(L3+B3)12
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Solution

## The correct option is B √(L2+B2)12The moment of inertia about a rectangular lamina with area density ρ is ∫∫ρr2dA. Here r is the distance from each point to the axis of rotation and dA is the differential of area.Take the center of the rectangle to be at (0,0) so the vertices are (L/2,B/2), etc. Then the distance from (x,y) to the axis of rotation (passing through (0,0) perpendicular to the plate) is x2+y2 and the moment of inertia is∫L/2−L/2∫B/2B/2ρ(x2+y2)dxdyor∫L/2−L/2ρ(Bx2+112B3)or112ρ(L3B+LB3)Now as M=ρLB, thus we get moment of inertia as M12(L2+B2)so radius of gyration is K=√IM⟹K=√112(L2+B2)

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