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(C) The moment of inertia of a rectangular sheet of mass $M$,length $L = 80 	ext{ cm}$,and breadth $B = 60 	ext{ cm}$ about an axis passing through

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$1/4$

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(C) The moment of inertia of a rectangular sheet of mass $M$,length $L = 80 	ext{ cm}$,and breadth $B = 60 	ext{ cm}$ about an axis passing through its center of mass $O$ and perpendicular to the sheet is given by:$I_O = \frac{M}{12} [L^2 + B^2] = \frac{M}{12} [80^2 + 60^2] = \frac{M}{12} [6400 + 3600] = \frac{10000M}{12} = \frac{2500M}{3}$.Using the parallel axis theorem,the moment of inertia about an axis passing through the corner point $O'$ and perpendicular to the sheet is:$I_{O'} = I_O + Md^2$,where $d$ is the distance between $O$ and $O'$.The distance $d = \sqrt{(L/2)^2 + (B/2)^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = 50 	ext{ cm}$.$I_{O'} = \frac{2500M}{3} + M(50)^2 = \frac{2500M}{3} + 2500M = \frac{2500M + 7500M}{3} = \frac{10000M}{3}$.The ratio of the moments of inertia is:$\frac{I_O}{I_{O'}} = \frac{2500M/3}{10000M/3} = \frac{2500}{10000} = \frac{1}{4}$.

For a uniform rectangular sheet shown in the figure,the ratio of moments of inertia about the axes perpendicular to the sheet and passing through $O$ (the centre of mass) and $O'$ (corner point) is

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