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 is

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 is (A) ((2n+1)l/3) (B) (l/n + 1) (C) (n(n2 +1)l/2) (D) (2l/n(n2 +1))

Tardigrade Admin · Accepted Answer

XCM =(m1x1 +m2x2+.../m1+m2+...) = (ml+2m.2l+3m.3l+.../m+2m+3m+....) =(ml(1+4+9+...)/m(1+2+3+....)) =(l (n(n+1)(2n+1)/6)/(n(n+1))2)=(l(2n+1)/3)

Q. Particles of masses $m, 2 m, 3 m, ..., nm$ 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 centimetres is

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

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

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

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

$X_{CM} = \frac{m _{1} x _{1} + m _{2} x _{2} + ...}{m _{1} + m _{2} + ...}$
$= \frac{m l + 2 m .2 l + 3 m .3 l + ...}{m + 2 m + 3 m + ....}$
$= \frac{m l ( 1 + 4 + 9 + ... )}{m ( 1 + 2 + 3 + .... )}$
$= \frac{l \frac{n ( n + 1 ) ( 2 n + 1 )}{6}}{\frac{n ( n + 1 )}{2}} = \frac{l ( 2 n + 1 )}{3}$

Q. 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 is

3(2n+1)l​

n+1l​

2n(n2+1)l​

n(n2+1)2l​