Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$x_{cm}$ = $\frac{m\left(l\right)+2m\left(2l\right)+3m\left(3l\right)+.........+nm\left(nl\right)}{m+2m+3m+............+nm}$

= $\frac{ml(1^1+2^2+3^3+............+n^2)}{m[1+2+3+.........+n]}$

= $\frac{l\left[\frac{n(n+1)(2n+1)}{6}\right]}{\frac{n(n+1)}{2}}$ [Hence, sum of squares of 'n' terms = $\frac{n(n+1)(2n+1)}{6}$ and

sum of 'n' terms = $\frac{n(n+1)}{2}$]

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