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Particles of masses $100\;g$ and $300\;g$ have position vectors $(2 \hat {i}+5 \hat {j}+13 \hat {k})$ and $(-6 \hat i +4 \hat j+2 \hat k).$ Position vector of their center of Mass is

\[\begin {array} {1 1} (a)\;\frac{-16}{4}\hat{i}+\frac{17}{4} \hat{j}+\frac{7}{4} \hat {k} \\ (b)\;\frac{20}{4} \hat{i}+\frac{17}{4} \hat {j}+\frac{7}{4} \hat{k} \\ (c)\;\frac{-16}{4} \hat{i}+\frac{17}{4} \hat {j}+\frac{19}{4} \hat{k} \\ (d)\;\frac{-16}{4} \hat{i}+\frac{13}{4} \hat {j}+\frac{19}{4} \hat{k} \end {array}\]

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$\overrightarrow{r}_{cm}=\large\frac{m_1\overrightarrow{r_1}+m_2 \overrightarrow {r_2}}{(m_1+m_2)}$
So the correct answer is $\large\frac{-16}{4} \hat{i}+\frac{17}{4} \hat {j}+\frac{19}{4} \hat{k}$
answered Nov 20, 2013 by meena.p
edited Jun 16, 2014 by lmohan717
 

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