Three identical spheres each of diameter $2 \sqrt{3} \mathrm{~m}$ are kept on a horizontal surface such that…

1. 12 m
2. 1 m
3. 2 m
4. $\frac{3}{2} \mathrm{~m}$

Solution

The centre of mass of the spheres is given by $ x_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3} $
As the spheres are identical, $ \begin{aligned} \Rightarrow \quad x_{\mathrm{CM}} & =\frac{m}{3 m}(0+2 \sqrt{3}+\sqrt{3}) \\ & =\frac{3 \sqrt{3}}{3}=\sqrt{3} \end{aligned} $ Similarly, $ \begin{aligned} y_{\mathrm{CM}} & =\frac{m}{3 m}\left(y_1+y_2+y_3\right) \\ & =\frac{(0+0+3)}{3}=1 \end{aligned} $ So, the centre of mass, $C_{\mathrm{CM}}=(\sqrt{3}, 1)$ If one sphere is removed (say $C$ ), then $ \begin{aligned} & x_{\mathrm{CM}}^{\prime}=\frac{0+2 \sqrt{3}}{2}=\sqrt{3} \\ & y_{\mathrm{CM}}^{\prime}=0 \end{aligned} $
So, $\quad C_{\mathrm{CM}}^{\prime}=(\sqrt{3}, 0)$ Hence, the centre of mass shifted by $1 \mathrm{~m}$ in $-y$ direction. The correct option is (b)

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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