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A uniform square sheet has a side of length $2R$. A circular sheet of maximum possible area is removed from one of the quadrants of the square sheet. The distance of the center of mass of the remaining portion from the center of the original sheet is

\[\begin {array} {1 1} (a)\;\frac{\pi R}{\sqrt {2}[16-\pi]} & \quad (b)\;\frac{R}{[16-\pi]} \\ (c)\;\frac{R}{\pi[16-\pi]} & \quad  (d)\;\frac{R\pi}{16-\pi} \end {array}\]

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Consider 2 as particles system with COM at $C_1$
If man of square $=M$
man of circle $=M_1$
Then $(M-M_1)d_1=M_1.d_2$
answered Nov 21, 2013 by meena.p

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