Question

Particles of masses m, 2m, 3m..........nm grams are placed on the same line at distances $l$, 2$l$,3$l$,.......n$l$ 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. $\frac{n(n^2+1)l}{2}$
• D. $\frac{2l}{n(n^2+1)}$

Answer 1

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

$l_{cm}=\frac{\sum _{i=1}^n{m}_i{l}_i}{\sum _{i=1}^n{m}_i}=\frac{ml\sum _{i=1}^n{n}^2}{m\sum n}$

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