question_answer

Particles of masses m, 2m, 3m ... nm grams are placed on the same line at distance \[l,2l,3l...nlcm\] from a fixed point. The distance of centre of mass of the particles from the fixed point in centimetre is

A) \[\frac{2n+1)l}{3}\]

B) \[\frac{l}{n+1}\]

C) \[\frac{n({{n}^{2}}+1)l}{2}\]  

D) \[\frac{2l}{n({{n}^{2}}+1)}\]

Correct Answer: A

Solution :

[a] The distance of centre of mass of given configuration of the particles from the fixed point O is \[{{X}_{CM}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}+...+{{m}_{n}}{{x}_{n}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+...+{{m}_{n}}}\] \[{{X}_{CM}}=\frac{(m)(l)+(2m)(2l)+(3m)(3l)+...+(nm)(nl)}{m+2m+3m+...+nm}\] \[=\frac{ml[{{1}^{2}}{{2}^{2}}+{{3}^{2}}+...+{{n}^{2}}]}{m[1+2+3+...+n]}\] \[=\frac{\frac{(l)(n)(n+1)(2n+1)}{6}}{\frac{(n)(n+1)}{2}}=\frac{(2n+1)l}{3}cm\]