Question

Particles of masses $m,2m,3m,...\left(n\right)m$ grams are placed on the same line at distances $l,2l,3l,...\left(n\right)l$ cm from a fixed point. The distance of the centre of mass of the particles from the fixed point in centimetres 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 $\frac{(2n+1)l}{3}$

Distance of center of mass$=\frac{m1x1+m2x2+......}{m1+m2+....}$\\

$=\frac{ml(1+4+9+....}{m(1+2+\mathrm{3...})}$

$=\frac{2nl(n+1)(2n+1)}{6.n(n+1)}$

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