Question

Particles of masses $m,2m,3m,....nm$ $gram$ are placed on the same line at distance $l,2l,3l,....nl$ $cm$ from a fixed point. The distance of centre of mass of the particles from the fixed point in centimeter is

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

Answer 1

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

$COM=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+............+m_n{x}_n}{m_1+m_2+m_3+............+m_n}$

Here,

$COM=\frac{1^{2}ml+2^{2}ml+3^{2}ml+..............+n^{2}m{l}^2}{1m+2m+3m+..............+nm}$

$COM=l\left[\frac{1^2+2^2+3^2+..........+n^2}{1+2+3+........+n}\right]$

$COM=l\left[\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}\right]$

$COM=\frac{l\cdot (2n+1)}{3}$