Q. Particles of masses $m,\ 2m,\ 3m,\ \ldots\ldots,\ nm$ grams are placed on the same line at distances $l,\ 2l,\ 3l,\ldots\ldots.,\ nl\ cm$ from a fixed point. The distance of centre of mass of the particles from the fixed point in centimeters is
A) $\frac{(2n + 1)l}{3}$
B) $\frac{l}{n + 1}$
C) $\frac{n\left( n^{2} + 1 \right)l}{2}$
D) $\frac{2l}{n(n^{2} + 1)}$
Explanation
(a) $x_{CM} = \frac{m_{1}x_{1} + m_{2} + x_{2} + \ldots}{m_{1} + m_{2} + \ldots}$
$= \frac{ml + 2m.2l + 3m.3l + \ldots}{m + 2m + 3m + \ldots}$
$= \frac{ml(1 + 4 + 9 + \ldots)}{m(1 + 2 + 3 + \ldots)} = \frac{\frac{l\ n(n + 1)(2n + 1)}{6}}{\frac{n(n + 1)}{2}} = \frac{l(2n + 1)}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.