Question

10. Particles of masses m, 2m, 3m, nm grams
are placed on the same line at distances l,
21, 31, ... nl cm from a fixed point. The
distance of center of mass of the particles
from the fixed point in centimetres is:
(a) (2n+1)
(b) ?
n
+1
(c) n(n2+1)
(d) 21
n(n2+1)​

Answer 1

Answer:

The answer will be  l (2n+1)/3 cm

Explanation:

I think there is some typing mistakes that's why the value of the distances will be 2l and 3l not 21 and 31 respectively.

According to the problem the distance of center of mass from the fixed point  is =

   ml +2m x 2l + 3m x 3l+ ..... nm x nl/ m+ 21+31... nl

= ml(1+2^2+3^2 + ..+n^2)/m[ 1+2+....+n]

= ml [ n (n+1)/6 (2n+1)] / m[ n (n+1)/2] = l (2n+1)/3 cm

The distance will be  l (2n+1)/3 cm