Four identical spheres each of radius \(10 \mathrm{~cm}\) and equal mass \(1 \mathrm{~kg}\) each are placed…

Four identical spheres each of radius \(10 \mathrm{~cm}\) and equal mass \(1 \mathrm{~kg}\) each are placed on horizontal surface touching each other, so that their centre are located at the vertices of a square of side \(20 \mathrm{~cm}\). What is the distance of their centre of mass from the centre of either sphere?

1. \(20 \sqrt{2}\)
2. \(30 \sqrt{2}\)
3. \(10 \sqrt{2}\)
4. \(40 \sqrt{2}\)

Solution

The given situation is shown in the figure,
\(A B=B C=C D=D A=20 \mathrm{~cm}\) Since, all spheres are identical with mass \(1 \mathrm{~kg}\) and radius \(10 \mathrm{~cm}\) each, hence according to figure, it is clear that all spheres are symmetrically arranged, hence centre of mass will be at intersection point of diagonals of square \(A B C D\). \(\begin{aligned} & \therefore \quad A C=\sqrt{20^2+20^2}=20 \sqrt{2} \mathrm{~cm} \\ & \therefore \quad A O=B O=C O=D O=\frac{20 \sqrt{2}}{2}=10 \sqrt{2} \mathrm{~cm} \end{aligned}\)

Asked in: AP EAMCET 2020 (21 Sep Shift 2)

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