Three identical spheres each of diameter 2sqr | vedclass

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(B) The diameter of each sphere is $D = 2\sqrt{3} 	ext{ m}$,so the radius is $R = \sqrt{3} 	ext{ m}$.The centres of the three spheres form an equila

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$1 	ext{ m}$

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(B) The diameter of each sphere is $D = 2\sqrt{3} 	ext{ m}$,so the radius is $R = \sqrt{3} 	ext{ m}$.The centres of the three spheres form an equilateral triangle of side length $a = 2R = 2\sqrt{3} 	ext{ m}$.Let the coordinates of the centres be $A(0, 0)$,$B(2\sqrt{3}, 0)$,and $C(\sqrt{3}, 3)$.The centre of mass $(X_{CM}, Y_{CM})$ of the three identical spheres is:$X_{CM} = \frac{m(0) + m(2\sqrt{3}) + m(\sqrt{3})}{3m} = \frac{3\sqrt{3}}{3} = \sqrt{3} 	ext{ m}$$Y_{CM} = \frac{m(0) + m(0) + m(3)}{3m} = \frac{3}{3} = 1 	ext{ m}$So,the initial centre of mass is $C_{CM} = (\sqrt{3}, 1)$.If sphere $C$ is removed,the new centre of mass $(X'_{CM}, Y'_{CM})$ of the remaining two spheres $A$ and $B$ is:$X'_{CM} = \frac{m(0) + m(2\sqrt{3})}{2m} = \sqrt{3} 	ext{ m}$$Y'_{CM} = \frac{m(0) + m(0)}{2m} = 0 	ext{ m}$So,the new centre of mass is $C'_{CM} = (\sqrt{3}, 0)$.The shift in the centre of mass is $\Delta = \sqrt{(\sqrt{3} - \sqrt{3})^2 + (1 - 0)^2} = 1 	ext{ m}$.The correct option is $B$.

Three identical spheres each of diameter $2\sqrt{3} \text{ m}$ are kept on a horizontal surface such that each sphere touches the other two spheres. If one of the spheres is removed,then the shift in the position of the centre of mass of the system is

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