Question

From a uniform square plate of side 2L a circular portion is removed from one of the quadrant then shift im centre of mass is.

Answer 1

Let square plate is placed in Cartesian system in such a way that centre of plate lies on origin.
we know, centre of mass of regular square plate lies on geometrical centre of plate.
see figure, centre of mass of square plate is (0,0)

Let a circular portion is removed from 1st quadrant as shown in figure. we also know centre of mass of circular plate lies on geometrical centre of circle. e.g., centre of mass of circular portion is (-L/2, L/2)

mass of circular portion = {mass of square plate/area of square plate} × area of circular plate
= m/(2L)² × π(L/2)²
= mπ/16

now, use basic formula of centre of mass,
$x=\frac{m_1x_1-m_2x_2}{m_1-m_2}$

$x=\frac{m\times0-\frac{m\pi}{16}\times\frac{L}{2}}{m-\frac{m\pi}{16}}$

$x=-\frac{\frac{m\pi L}{32}}{m\frac{(16-\pi)}{16}}$

$x=-\frac{\pi L}{2(16-\pi)}$

similarly, you can y by using formula,
$y=\frac{m_1y_1-m_2y_2}{m_1-m_2}$

so, $y=-\frac{\pi L}{2(16-\pi)}$