Three identical spheres,each of mass 2 M,are | vedclass

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(B) Let the positions of the three spheres be at $(0, 0)$,$(4, 0)$,and $(0, 4)$ in the $XY$-plane,each with mass $m = 2M$.The position of the centre o

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$3$

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(B) Let the positions of the three spheres be at $(0, 0)$,$(4, 0)$,and $(0, 4)$ in the $XY$-plane,each with mass $m = 2M$.The position of the centre of mass $\overrightarrow{r}_{	ext{COM}}$ is given by:$\overrightarrow{r}_{	ext{COM}} = \frac{m_1 \overrightarrow{r}_1 + m_2 \overrightarrow{r}_2 + m_3 \overrightarrow{r}_3}{m_1 + m_2 + m_3}$$\overrightarrow{r}_{	ext{COM}} = \frac{2M(0\hat{i} + 0\hat{j}) + 2M(4\hat{i} + 0\hat{j}) + 2M(0\hat{i} + 4\hat{j})}{2M + 2M + 2M}$$\overrightarrow{r}_{	ext{COM}} = \frac{8M\hat{i} + 8M\hat{j}}{6M} = \frac{4}{3}\hat{i} + \frac{4}{3}\hat{j}$The magnitude of the position vector is:$|\overrightarrow{r}_{	ext{COM}}| = \sqrt{(\frac{4}{3})^2 + (\frac{4}{3})^2} = \sqrt{\frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{32}{9}} = \frac{4\sqrt{2}}{3}$Comparing this with $\frac{4\sqrt{2}}{x}$,we get $x = 3$.

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