Q 11-06-216JEE MainJEE Main 2019 (8 Apr, Shift 2)Easy
A uniform rectangular thin sheet $ABCD$ of mass $M$ has length $a$ and breadth $b$, as shown in the figure. If the shaded portion $HBGO$ is cut-off, the coordinates of the centre of mass of the remaining portion will be
Answer: (A) $\left(\dfrac{5a}{12}, \dfrac{5b}{12}\right)$
The removed quarter $HBGO$ has mass $M/4$ and centre $\left(\frac{3a}{4}, \frac{3b}{4}\right)$. The full sheet has centre $\left(\frac a2, \frac b2\right)$. Treating the cut-off part as negative mass:
$$x_{cm} = \frac{M\cdot\frac a2 - \frac M4\cdot\frac{3a}{4}}{M - \frac M4} = \frac{\frac a2 - \frac{3a}{16}}{\frac34} = \frac{5a}{16}\cdot\frac43 = \frac{5a}{12}$$
By symmetry of the calculation, $y_{cm} = \dfrac{5b}{12}$.
Solution by Sreeraj P, M.Sc Physics