In a system,two particles of masses m_1 = 3 t | vedclass

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(D) The displacement of the center of mass is given by the formula: $\Delta X_{	ext{COM}} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2}$.Since

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$3$

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(D) The displacement of the center of mass is given by the formula: $\Delta X_{	ext{COM}} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2}$.Since the center of mass must remain at its original position,the net displacement of the center of mass is zero,i.e.,$\Delta X_{	ext{COM}} = 0$.Let the displacement of $m_1$ be $\Delta x_1 = +2 	ext{ cm}$ (towards the center of mass) and the displacement of $m_2$ be $\Delta x_2 = -x$ (towards the center of mass).Substituting the values into the equation:$0 = \frac{3 	imes 2 + 2 	imes (-x)}{3 + 2}$$0 = \frac{6 - 2x}{5}$$6 - 2x = 0$$2x = 6$$x = 3 	ext{ cm}$.Therefore,the particle of mass $m_2$ must move by $3 	ext{ cm}$ towards the center of mass.

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