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(C) The original square plate is symmetric about both the $X$ and $Y$ axes,so its center of mass is at the origin $(0, 0)$.When four identical squares

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In the $III$ quadrant

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(C) The original square plate is symmetric about both the $X$ and $Y$ axes,so its center of mass is at the origin $(0, 0)$.When four identical squares are removed from the corners,the remaining plate is still symmetric about both axes,so the center of mass remains at $(0, 0)$.If we remove only square $1$ (located in the $I$ quadrant),the mass distribution shifts away from the $I$ quadrant.The center of mass of the removed square $1$ is at $(x_1, y_1)$,where $x_1 > 0$ and $y_1 > 0$.The center of mass of the remaining plate $(X_{cm}, Y_{cm})$ can be found using the formula: $X_{cm} = \frac{M X_{total} - m x_1}{M - m}$ and $Y_{cm} = \frac{M Y_{total} - m y_1}{M - m}$.Since $X_{total} = 0$ and $Y_{total} = 0$,we get $X_{cm} = \frac{-m x_1}{M - m}$ and $Y_{cm} = \frac{-m y_1}{M - m}$.Since $x_1 > 0$ and $y_1 > 0$,both $X_{cm}$ and $Y_{cm}$ are negative.$A$ point with both negative coordinates lies in the $III$ quadrant.

The figure shows a uniform square plate from which four identical small squares have been removed from the corners. If square $1$ is removed,where will the center of mass $(C.M.)$ be located?

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