Question

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

• A. $\frac{7}{12}m$
• B. $\frac{5}{13}m$
• C. $\frac{12}{7}m$
• D. $\frac{14}{4}m$

Answer 1

The correct option is C $\frac{12}{7}m$

Linear density of rod varies with distance
$\frac{dm}{dx}=\lambda$
$x_{cm}=\frac{\int dm\times x}{\int dm}$
$=\frac{{\int }_0^3\left(\lambda dx\right)x}{{\int }_0^3\lambda dx}=\frac{{\int }_0^3(2+x)xdx)}{{\int }_0^3(2+x)dx}$
$=\frac{{[x^2+\frac{x^3}{3}]}_0^3}{{[2x+\frac{x^3}{3}]}_0^3}$
$=\frac{9+9}{6+\frac{9}{2}}=\frac{36}{21}=\frac{12}{7}m$