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(C) The position of the centre of mass $R_{cm}$ is given by $R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$.For the centre of mass to remain at the sam

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$\left(\frac{m_1}{m_2}\right) d$

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(C) The position of the centre of mass $R_{cm}$ is given by $R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$.For the centre of mass to remain at the same position,the change in the position of the centre of mass must be zero,i.e.,$\Delta R_{cm} = 0$.Therefore,$m_1 \Delta r_1 + m_2 \Delta r_2 = 0$.Here,the particle of mass $m_1$ is moved by a distance $d$ towards the centre of mass,so $\Delta r_1 = -d$ (assuming the direction towards the centre of mass is negative).Substituting this into the equation: $m_1 (-d) + m_2 \Delta r_2 = 0$.$m_2 \Delta r_2 = m_1 d$.$\Delta r_2 = \left(\frac{m_1}{m_2}\right) d$.Thus,the particle of mass $m_2$ must be moved by a distance of $\left(\frac{m_1}{m_2}\right) d$ to keep the centre of mass at the original position.

$A$ system of two particles has masses $m_1$ and $m_2$. If the particle of mass $m_1$ is pushed towards the centre of mass of the particles through a distance $d$,by what distance should the particle of mass $m_2$ be moved to keep the centre of mass of the particles at the original position?

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