\[\begin {array} {1 1} (a)\;\frac{r}{2(2 \pi-1)} & \quad (b)\;\frac{r}{(\pi-2)} \\ (c)\;\frac{2r}{\pi-3} & \quad (d)\;\frac{r}{2(\pi-1)}\end {array}\]

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Let $m_1$ be te mass of the square that is chopped off.

$m_1 = \large\frac{M}{\pi r^2}$$ \times \large\frac{r^2}{2}$$ = \large\frac{M}{ 2 \pi}$

$x_{m_1} = \large\frac{M\times0 - m_1\times \large\frac{r}{2} }{M - \large\frac{M}{2 \pi }}$

$x_{m_1} = \large\frac{-\large\frac{M}{2 \pi } \large\frac{r}{2}}{M - \large\frac{M}{2 \pi }}$

$x_{m_1} = \large\frac{-r}{2(2 \pi - 1)}$

Thus, the centre of mass of the remaining part lies at a distance $x_{m_1} = \large\frac{-r}{2(2 \pi - 1)}$ towards right of the origin, or the initial centre of mass of the disc.

Therefore the correct option is (A) $\large\frac{r}{2(2\pi -1)}$

Please note: The new centre of mass does not lie at the origin any more.

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