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Physics Question on Centre of mass

Particles of masses m,2m,3m,...,nmm, 2m, 3m, ... , nm grams are placed on the same line at distances l,2l,3l,...,nll, 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)l3\frac{\left(2n+1\right)l}{3}

B

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

C

n(n2+1)l2\frac{n\left(n^2 +1\right)l}{2}

D

2ln(n2+1)\frac{2l}{n\left(n^2 +1\right)}

Answer

(2n+1)l3\frac{\left(2n+1\right)l}{3}

Explanation

Solution

XCM=m1x1+m2x2+...m1+m2+...X_{CM} =\frac{m_{1}x_{1} +m_{2}x_{2}+...}{m_{1}+m_{2}+...} =ml+2m.2l+3m.3l+...m+2m+3m+....= \frac{ml+2m.2l+3m.3l+...}{m+2m+3m+....} =ml(1+4+9+...)m(1+2+3+....) =\frac{ml\left(1+4+9+...\right)}{m\left(1+2+3+....\right)} =ln(n+1)(2n+1)6n(n+1)2=l(2n+1)3=\frac{l \frac{n\left(n+1\right)\left(2n+1\right)}{6}}{\frac{n\left(n+1\right)}{2}}=\frac{l\left(2n+1\right)}{3}