Particles of masses $m, 2 m, 3 m, \ldots n m$ grams are placed on the same line at distance $l, 2 l, 3 l…
Particles of masses $m, 2 m, 3 m, \ldots n m$ grams are placed on the same line at distance $l, 2 l, 3 l \ldots n l$ $\mathrm{cm}$ from a fixed point. The distance of centre of mass of the particles from the fixed point in centimetre is
$\frac{(2 n+1) l}{3}$
$\frac{l}{n+1}$
$\frac{n\left(n^2+1\right) l}{2}$
$\frac{2 l}{n\left(n^2+1\right)}$
Solution
Distance of centre of mass
$
\begin{aligned}
x_{c m} & =\frac{m_1 x_1+m_2 x_2+\ldots m_n x_n}{m_1+m_2+\ldots m_n} \\
& =\frac{m l+2 m 2 l+\ldots n m \times n l}{m+2 m+3 m+\ldots n m} \\
& =\frac{m l\left(1+4+9+\ldots n^2\right)}{m(1+2+3+\ldots n)} \\
1+4+9 & +\ldots n^2=\Sigma n^2 \\
& =\frac{n(n+1)(2 n+1)}{6} \\
1+2+3+ & \ldots n=\Sigma n=\frac{n(n+1)}{2} \\
\therefore \quad x_{c m} & =\frac{\ln (n+1)(2 n+1)}{6}=\frac{l(2 n+1)}{3}
\end{aligned}
$