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A uniform circular sheet has a radius r. A square sheet whose diagonal length is equal to radius of the plate is removed . The maximum distance of the center of mass of the remaining plate from the center of this original sheet is

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$\frac{r}{2 (2 π - 1)}$

$\frac{r}{(π - 2)}$

$\frac{2 r}{π - 3}$

$\frac{r}{2 (π - 1)}$

answer is A.

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Detailed Solution

$σ = \frac{M}{π R^{2}}$

$\sqrt{2} a = R$

$a = \frac{R}{\sqrt{2}}$

$m_{1} = {(\frac{R}{\sqrt{2}})}^{2} \times \frac{M}{π R^{2}} = \frac{M}{2 π}$

$m_{2} = M - m_{1} = M - \frac{M}{2 π}$ $= \frac{M}{2 π} (2 π - 1)$

$x_{cm} = \frac{mass removed \times distance from CM}{Mass remains}$

$= \frac{\frac{M}{2 π} \times (\frac{R}{2})}{\frac{M}{2 π} (2 π - 1)} = \frac{R}{2 (2 π - 1)}$

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