Q145 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Physics
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?
- A\(20 \sqrt{2}\)
- B\(30 \sqrt{2}\)
- Correct answerC\(10 \sqrt{2}\)
- D\(40 \sqrt{2}\)
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 \(A B C D\). \(\begin{aligned} & \therefore \quad A C=\sqrt{20^2+20^2}=20 \sqrt{2} \mathrm{~cm} \\ & \therefore \quad A O=B O=C O=D O=\frac{20 \sqrt{2}}{2}=10 \sqrt{2} \mathrm{~cm} \end{aligned}\)
\(A B=B C=C D=D A=20 \mathrm{~cm}\) 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 \(A B C D\). \(\begin{aligned} & \therefore \quad A C=\sqrt{20^2+20^2}=20 \sqrt{2} \mathrm{~cm} \\ & \therefore \quad A O=B O=C O=D O=\frac{20 \sqrt{2}}{2}=10 \sqrt{2} \mathrm{~cm} \end{aligned}\)
Hence, the correct answer is option: 'C'.