Question

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

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

Answer 1

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

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

$=\frac{\left(m\right)\left(l\right)+\left(2m\right)\left(2l\right)+\left(3m\right)\left(3l\right)+-------+\left(nm\right)\left(nl\right)}{m+2m+3m+-------+nm}$

$=\frac{ml(1^2+2^2+3^2+-------+n^2)}{m(1+2+3+--------+n)}$

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

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