Question

The linear density of a thin rod of length $1.0$m varies as $\lambda =2$kg/m$+\left(\frac{2kg}{m^2}\right)x$, where x is the distance from its one end. The distance of its centre of mass from its end is?

• A. $\frac{2}{3}$m
• B. $\frac{5}{9}$m
• C. $\frac{4}{3}$m
• D. $\frac{1}{2}$m

Answer 1

The correct option is B $\frac{5}{9}$m

REF.Image

$\lambda =2\frac{kg}{meter}+\frac{2kg}{mete{r}^2}.x$

$\lambda =(2+2x)kg/m$ ; x is in meter

suppose there is an element of length dx

& at a distance x from left must end.

mass of this element

is $dm=\lambda dx=(2+2x)dx$

$\Rightarrow$ mass of rod = sum of marvels of these elements

$M={\int }_{x=0}^{1}dm={\int }_0^1(2+2x)dx=[2x+\frac{2x}{2}{]}_0^1$

$m=3kg$

$x_{c.m}=\frac{{\int }_0^{1}xdm}{\int dm}=\frac{{\int }_0^1(2+2x)xdx}{{\int }_0^1(2+2x)dx}$

$x_{cm}=\frac{{\int }_0^1(2x+2x^2)}{3}dx=\frac{[\frac{2x^2}{2}+\frac{2x^3}{3}{]}_0^1}{3}$

$x_{cm}=\frac{1+\frac{2}{3}}{3}$ meter = $\frac{5}{9}$ meter

from are end (having less density)