Particles of masses m, 2m, 3m ... nm gram are placed on the same line at distance l, 2l, 3l, ..., nl cm from a fixed point. The distance of centre of mass of the particles from the fixed point in cm is

Options

A. `((2n+1)1)/(3)`
B. `(1)/(n+1)`
C. `(n(n^(2)+1)I)/(2)`
D. `(2I)/(n(n^(2)+I)`

Answer

(a)

Explanation

Centres of mass , `x_(cm)=(summ_(i)x_(i))/(summ_(i))`
`=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3)+…+m_(n)x_(n))/(m_(1)+m_(2)+m_(3)+…+m_(n))`
`=([(m)xx.l]+(2mxx2l)+(3mxx3l)+....+(nmxxnl))/(m+2m+3m+......+nm)`
`(ml(1+4+9+....+n^(2)))/(m(1+2+3+....+n))`
`=l[(n(n+1)(2n+1))/((6)/((n(n+1))/(2)))]`
`l[(2n+1)/(3)]`