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Particles of masses $m, 2 m, 3 m, ..... n m$ 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 centimeter is

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

$\frac{l}{n + 1}$

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

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

answer is A.

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Detailed Solution

Distance of centre of mass from the fixed point is

$= \frac{(n (n + 1)) (2 n + 1) / 6}{\frac{n (n + 1)}{2}} = \frac{(2 n + 1) l}{3}$

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