Question

For a square sheet of side $1\text{m}$ having uniform surface density, the position of $\text{COM}$ is at $x_1$. On the other hand, if surface density is varying as $\sigma =2x{\text{kg/m}}^2$, $\text{COM}$ is observed to be at position $x_2$. The distance between $x_2$ and $x_1$ is :

• A. $\frac{1}{3}\text{m}$
• B. $\frac{1}{6}\text{m}$
• C. $\frac{1}{2}\text{m}$
• D. $\frac{2}{3}\text{m}$

Answer 1

The correct option is B $\frac{1}{6}\text{m}$

For a square of side $1\text{m}$ with uniform surface density, its $\text{COM}$ will lie at the geometric centre.

i.e $\text{COM}:(x_1,y_1)=(\frac{1}{2},\frac{1}{2})$

For surface density $\sigma =2x{\text{kg/m}}^2$:

Since density is varying with $x$ only, $y_{\text{COM}}$ will remain the same
Let the area of a small element shown in figure be $dA$.
$dA=1dx$


Mass per unit area $\left(\sigma \right)=\frac{dm}{dA}$
$\Rightarrow 2x=\frac{dm}{1dx}$
or $dm=2xdx$

Let $x_{\text{CM}}$ be the position of $COM$ along $x$ direction. Applying the basic formula:
$x_{\text{CM}}=\frac{\int xdm}{\int dm}$
The limits for integration are $x=0$ to $x=1\text{m}$

$\Rightarrow x_{\text{CM}}=\frac{{\int }_0^{1}x\left(2xdx\right)}{{\int }_0^{1}2xdx}=\frac{2{\int }_0^1{x}^{2}dx}{2{\int }_0^{1}xdx}$
$\Rightarrow x_{\text{CM}}=\frac{{\left[\frac{x^3}{3}\right]}_0^1}{{\left[\frac{x^2}{2}\right]}_0^1}=\frac{\left(\frac{1}{3}\right)}{\left(\frac{1}{2}\right)}=\frac{2}{3}\text{m}$
$\therefore x_2=\frac{2}{3}\text{m}$

$\therefore$ Distance between $x_1$ and $x_2$
$=|x_2-x_1|=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\text{m}$