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If n particles of mass 1, 2, 3, …, n kg are placed at points 1, 4, 9, …, $n^{2}$ meter respectively along +x-axis. The distance of centre of mass from origin in meter is

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$\sqrt{\frac{n^{2} {(n + 1)}^{2}}{4}}$

$\frac{n (n + 1) (2 n + 1)}{12}$

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

$\frac{2 n + 1}{3}$

answer is B.

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Detailed Solution

$1^{2} + 2^{2} + 3^{2} + ..... + n^{2} = {\frac{n (n + 1)}{2}}^{2} C_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2} + ..... + m_{n} x_{n}}{m_{1} + m_{2} + ... + m_{n}}$

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