AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

The expression for the centre of the mass of the spheres is, $x_{CM} = \;\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3}$ The centre of the mass of the spheres along the $x$ axis is calculated as, $x_{CM} \;\; = \;\frac{m}{3m}(0+2\sqrt{3}+\sqrt{3})$ $\;\; = \;\frac{3\sqrt{3}}{3}$ $\;\; = \sqrt{3}$ The centre of the mass of the spheres along the $y$ axis is calculated as, $y_{CM} \;\; = \;\frac{m}{3m}(0+0+3)$ $\;\; = 1$ The coordinates of the centre of the mass of the sphere is $C_{CM}=(\sqrt{3}, 1)$ According to the given condition, the centre of sphere when one sphere is removed is, $x_{CM}^{'} \;\; = \;\frac{0+2\sqrt{3}}{2}$ $\;\; = \sqrt{3}$ The below diagram is a representation of spheres after the exclusion of one sphere. The new coordinates of the centre of mass of the spheres is $C_{CM}^{'}=(\sqrt{3}, 0),$ which implies that the centre of the mass is shifted by $1\ \mathrm{m}$ in $y$ direction.