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 centimeters in:

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

Answer 1

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

Center of mass $X_{com}=\frac{m_1{x}_1+m_2{x}_2+.......m_n{x}_n}{m_1+m_2+.......+m_n}$

$X_{com}=\frac{ml+\left(2m\right)\left(2l\right)+.......\left(nm\right)\left(nl\right)}{m+2m+.......+nm}$

Or $X_{com}=\frac{1+2^2+.......n^2}{1+2+3+.....n}l$

Or $X_{com}=\frac{\frac{\left(n\right)(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}l=\frac{(2n+1)}{3}l$