Question

Particles of masses $m,2m,3m,...,nm$ are placed on the same line at distances $L,2L,3L,...,nL$ from $O$. The distance of center of mass from $O$ is

• A. $\frac{(2n+1)L}{4}$
• B. $\frac{L}{(2n+1)}$
• C. $\frac{n(n^2+1)L}{2}$
• D. $\frac{(2n+1)L}{3}$

Answer 1

The correct option is D $\frac{(2n+1)L}{3}$

$\because x_{CM}=\frac{mL+2m\times 2L+...+nm\times nL}{m+2m+...+nm}$
$=\frac{L(1^2+2^2+...+n^2)}{(1+2+...+n)}$
$=L\frac{\sum n^2}{\sum n}=L\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}$
$=\frac{(2n+1)L}{3}$