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Particles of masses $m, 2 m, 3 m, \ldots, n m$ are placed on the same line at distances $L, 2 L, 3 L, \ldots, n L$ from $O$. The distance of centre of mass from $O$ is
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The correct answer is:
$\frac{(2 n+1) L}{3}$
$\because \quad x_{C M}=\frac{m L+2 m \times 2 L+\ldots+n m \times n L}{m+2 m+\ldots+n m}$
$\begin{aligned} & =\frac{L\left(1^2+2^2+\ldots+n^2\right)}{(1+2+\ldots+n)} \\ & =L \frac{\Sigma n^2}{\Sigma n}=L \frac{\frac{n(n+1)(2 n+1)}{6}}{\frac{n(n+1)}{2}} \\ & =\frac{(2 n+1) L}{3}\end{aligned}$
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