In a system of two particles of masses ' \(\mathrm{m}_{1}\) ' and ' \(\mathrm{m}_{2}\) ', the second particle is moved
by a distance 'd' towards the centre of mass. To keep the centre of mass
unchanged, the first particle will have to be moved by a distance

Correct option is E)
given masses are \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\)
let \(x\) and \(y\) be the distance of \(m_{1}\) and \(m_{2}\) from the centre of mass respectively, then
\(\mathrm{m}_{1} \mathrm{x}=\mathrm{m}_{2} \mathrm{y}\)
If the mass \(m_{1}\) is moved by a distance \(d_{1}\), and mass \(m_{2}\) be moved by a distance \(\mathrm{d}_{2}\), then
\(\mathrm{m}_{1}\left(\mathrm{x}-\mathrm{d}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{y}-\mathrm{d}_{2}\right)\)
\(\mathrm{m}_{1} \mathrm{x}-\mathrm{m}_{1} \mathrm{~d}_{1}=\mathrm{m}_{2} \mathrm{y}-\mathrm{m}_{2} \mathrm{~d}_{2}\)
\(\mathrm{m}_{1} \mathrm{~d}_{1}=\mathrm{m}_{2} \mathrm{~d}_{2}\)
\(\mathrm{d}_{2}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d}_{1} \quad(\because\) constants \()\)
\(\mathrm{d}_{2}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d} \quad\left(\because \mathrm{d}_{1}=\mathrm{d}\right)\)