Five masses are placed in a plane as shown in | vedclass

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(C) From the figure,the coordinates $(x, y)$ and masses $m$ are:$m_1 = 1 	ext{ kg}$ at $(0, 0)$$m_2 = 2 	ext{ kg}$ at $(2, 0)$$m_3 = 3 	ext{ kg}$ a

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$(1.1, 1.3)$

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(C) From the figure,the coordinates $(x, y)$ and masses $m$ are:$m_1 = 1 	ext{ kg}$ at $(0, 0)$$m_2 = 2 	ext{ kg}$ at $(2, 0)$$m_3 = 3 	ext{ kg}$ at $(0, 2)$$m_4 = 4 	ext{ kg}$ at $(2, 2)$$m_5 = 5 	ext{ kg}$ at $(1, 1)$The total mass $M = 1 + 2 + 3 + 4 + 5 = 15 	ext{ kg}$.The $x$-coordinate of the centre of mass is:$x_{cm} = \frac{\sum m_i x_i}{M} = \frac{(1 	imes 0) + (2 	imes 2) + (3 	imes 0) + (4 	imes 2) + (5 	imes 1)}{15} = \frac{0 + 4 + 0 + 8 + 5}{15} = \frac{17}{15} \approx 1.13$The $y$-coordinate of the centre of mass is:$y_{cm} = \frac{\sum m_i y_i}{M} = \frac{(1 	imes 0) + (2 	imes 0) + (3 	imes 2) + (4 	imes 2) + (5 	imes 1)}{15} = \frac{0 + 0 + 6 + 8 + 5}{15} = \frac{19}{15} \approx 1.27$Thus,the coordinates are approximately $(1.1, 1.3)$. The correct option is $C$.

Column-$I$	Column-$II$
$(1)$ $\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$	$(a)$ Reduced mass of a two-particle system
$(2)$ $\frac{{{r_1} + {r_2}}}{2}$	$(b)$ Position vector of the center of mass for a system of two equal masses

Five masses are placed in a plane as shown in the figure. The coordinates of the centre of mass are nearest to:

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Match Column-$I$ with Column-$II$.
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