Four identical spheres, each of radius 10 ,cm | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given situation is shown in the figure where $A, B, C, D$ are the centers of the four spheres forming a square of side $a = 20 \,cm$.Since all

Vedclass · Accepted Answer

$10 \sqrt{2} \,cm$

Vedclass · Answer

(C) The given situation is shown in the figure where $A, B, C, D$ are the centers of the four spheres forming a square of side $a = 20 \,cm$.Since all four spheres are identical and have equal masses, the center of mass of the system will coincide with the geometric center of the square formed by their centers, denoted by point $O$.The distance of the center of mass from the center of any sphere (e.g., center $A$) is the distance $AO$.In a square of side $a = 20 \,cm$, the length of the diagonal $AC$ is given by $AC = \sqrt{a^2 + a^2} = a\sqrt{2} = 20\sqrt{2} \,cm$.The distance from the center of the square to any vertex is half the length of the diagonal:$AO = \frac{AC}{2} = \frac{20\sqrt{2}}{2} = 10\sqrt{2} \,cm$.

Four identical spheres, each of radius $10 \,cm$ and equal mass $1 \,kg$, are placed on a horizontal surface touching each other such that their centers are located at the vertices of a square of side $20 \,cm$. What is the distance of their center of mass from the center of any sphere?

Similar Questions

The mass per unit length of a rod of length $l$ is given by $\lambda = \frac{M_0 x}{l}$,where $M_0$ is a constant and $x$ is the distance from one end of the rod. The position of the center of mass of the rod is:

Three bodies each of mass $1 \,kg$ are situated at the vertices of an equilateral triangle of side $1 \,m$. The $x y$-coordinates of the centre of mass of the system are:

The coordinates of the center of mass of a uniform flag-shaped lamina (thin flat plate) of mass $4 \; kg$ (the coordinates of the same are shown in the figure) are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module