JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Conservation of Momentum

Treat the shape as the full sheet minus the piece cut away. The full sheet of mass $M$ has its centre of mass at $\left(\frac{a}{2},\frac{b}{2}\right)$. The removed corner piece is a quarter of the sheet, so its mass is $\frac{M}{4}$ and, being the corner quarter of size $\frac{a}{2}\times \frac{b}{2}$, its own centre sits at $\left(\frac{3a}{4},\frac{3b}{4}\right)$. The remaining mass is $\frac{3M}{4}$. Balancing the x-coordinate, $M\cdot \frac{a}{2}=\frac{3M}{4}{x}_{cm}+\frac{M}{4}\cdot \frac{3a}{4}$, which solves to $x_{cm}=\frac{5a}{12}$. The geometry is symmetric in $x$ and $y$, so the same working gives $y_{cm}=\frac{5b}{12}$. The centre of mass of what is left is $\left(\frac{5a}{12},\frac{5b}{12}\right)$.