Question

If linear density of a rod of length 3 m varies as $\lambda$ =2+ $x$ , then the position of the centre of gravity of the rod is

• A. $\frac{7}{3}m$
• B. $\frac{12}{7}m$
• C. $\frac{10}{7}m$
• D. $\frac{9}{7}m$

Answer 1

Answer: (B)

Let rod is placed along $x$ -axis. Mass of element $PQ$ of length $dx$ situated at $x=x$ is


$dm=\lambda dx=\left(2+x\right)dx$
The CM of the element has coordinates ( $x$ , 0, 0).
Therefore, $x$ -coordinates of CM of the rod will be
$x_{CM}=\frac{{\int }_0^{3}xdm}{{\int }_0^{3}dm}$
$=\frac{{\int }_0^{3}x\left(2+x\right)dx}{{\int }_0^3\left(2+x\right)dx}$
$=\frac{{\int }_0^3\left(2x+x^2\right)dx}{{\int }_0^3\left(2+x\right)dx}$
$=\frac{{\left[\frac{2x^2}{2}+\frac{x^3}{3}\right]}_0^3}{{\left[2x+\frac{x^2}{2}\right]}_0^3}$
$=\frac{\left[{\left(3\right)}^2+\frac{{\left(3\right)}^3}{3}\right]}{\left[2\times 3+\frac{{\left(3\right)}^2}{2}\right]}=\frac{9+9}{6+9/2}$
$=\frac{18\times 2}{21}=\frac{12}{7}$ m