Question

A uniform square sheet has a side of length $2R$, where $R=12\text{cm}$. A circular sheet of maximum possible area is removed from one of the quadrants of the square sheet. The distance of the center of mass of the remaining portion from the center of the original sheet is:

• A. $5.1\text{cm}$
• B. $2.07\text{cm}$
• C. $3.8\text{cm}$
• D. $4.2\text{cm}$

Answer 1

The correct option is B $2.07\text{cm}$


Let $A_1$ be the area of the circular sheet of radius $R$ and $A_2$ be the area of the remaining portion of the square sheet after removing the circular sheet.
Then, $A_1=\frac{\pi R^2}{4}$ and $A_2=4R^2-\frac{\pi R^2}{4}$

Considering the centre of the square sheet to be the origin.
Coordinates of COM of circular sheet
$x_1=\frac{R}{2}$, $y_1=\frac{R}{2}$
Let the coordinates of COM of remaining portion be
$x_2=x$ and $y_2=y$

Then, $y_{com}=\frac{A_1{y}_1+A_2{y}_2}{A_1+A_2}$
i.e $0=\frac{\frac{\pi R^2}{4}\left(\frac{R}{2}\right)+(4R^2-\frac{\pi R^2}{4})y}{A_1+A_2}$
$\Rightarrow y=-\frac{\pi R}{2(16-\pi )}$
Similarly $x=-\frac{\pi R}{2(16-\pi )}$

Therefore, centre of mass of $A_2$ is at $(-\frac{\pi R}{2(16-\pi )},-\frac{\pi R}{2(16-\pi )})$
Distance of COM from origin $=\frac{\pi R}{\sqrt{2}\times (16-\pi )}$
$=\frac{\pi \times 12}{\sqrt{2}(16-\pi )}=2.07\text{cm}$
(Putting $R=12\text{cm}$ (given))