Particles of masses m, 2m, 3m, ldots ., nm are placed on the same line at distance L, 2L, 3L,..nL from O. The distance of the centre of mass from O is

Question

Particles of masses m, 2m, 3m, ldots ., nm are placed on the same line at distance L, 2L, 3L,..nL from O. The distance of the centre of mass from O is (A) ((2 n + 1) L/4) (B) (L/(2 n + 1)) (C) (n (n2 + 1) L/2) (D) ((2 n + 1) L/3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-2rt7ohvotp9e.png" /> L ∵ xC M=(m L + 2 m × 2 L + ldots + n m × n L/m + 2 m + ldots + n m) =(L (12 + 22 + ldots + n2)/(1 + 2 + ldots + n)) =L( displaystyle ∑ n2/ displaystyle ∑ n)=L((n (n + 1) (2 n + 1)/6)/(n (n + 1))2) =((2 n + 1) L/3)

Q. Particles of masses $m, 2 m, 3 m, \dots ., nm$ are placed on the same line at distance $L, 2 L, 3 L, .. n L$ from $O .$ The distance of the centre of mass from $O$ is

$\frac{( 2 n + 1 ) L}{4}$

$\frac{L}{( 2 n + 1 )}$

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

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

L
$∵ x_{CM} = \frac{m L + 2 m \times 2 L + \dots + nm \times n L}{m + 2 m + \dots + nm}$
$= \frac{L ( 1 ^{2} + 2 ^{2} + \dots + n ^{2} )}{( 1 + 2 + \dots + n )}$
$= L \frac{\sum n ^{2}}{\sum n} = L \frac{\frac{n ( n + 1 ) ( 2 n + 1 )}{6}}{\frac{n ( n + 1 )}{2}}$
$= \frac{( 2 n + 1 ) L}{3}$

Q. Particles of masses m,2m,3m,….,nm are placed on the same line at distance L,2L,3L,..nL from O. The distance of the centre of mass from O is

4(2n+1)L​

(2n+1)L​

2n(n2+1)L​

3(2n+1)L​