Q: Particles of masses m , 2m , 3m , …, nm grams are placed on the same line at distance l , 2l , 3l , …., nl cm from a fixed point. The distance of centre of mass of the particles from the fixed point in centimeter is
(a) $\displaystyle \frac{(2n + 1)l}{3}$
(b) $\displaystyle \frac{l}{n+3}$
(c) $\displaystyle \frac{n (n^2 + 1)l}{2}$
(d) $\displaystyle \frac{2 l}{n (n^2 + 1)}$
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Ans: (a)
Sol: $\displaystyle x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3 + …. + m_n x_n}{m_1 + m_2 + m_3 + …..+ m_n}$
$\displaystyle x_{cm} = \frac{m \times l + 2 m \times 2l + 3 m \times 3l + …. + nm \times n l}{m + 2m + 3m + …..+ n m}$
$\displaystyle x_{cm} = \frac{m l (1 + 2^2 + 3^2 + ……+ n^2) }{m (1 + 2 + 3 + …..+ n }$
$\displaystyle x_{cm} = l( \frac{n(n+1)(2n+1)/6}{n(n+1)/2} )$
$\displaystyle x_{cm} = \frac{(2n + 1) l}{3} $