Question

Three identical spheres each of mass $M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $3 m$ each Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $\sqrt{ x } m$ The value of $x$ is

Answer 1

Correct Answer: 2

To solve this problem, we need to calculate the position vector of the center of mass for a system of three identical spheres placed at the corners of a right-angled triangle with sides of 3 m each. Let's denote the vertices A, B, and C, where A is the origin (0,0), B is (3,0), and C is (0,3). The position of the center of mass (CM) for these spheres is given by the average of their position vectors, weighted by their masses. Since all spheres have the same mass M, the masses cancel out in our calculation.
The formula for the center of mass of point masses in two dimensions is:

\( \text{CM}_x = \frac{M \cdot 0 + M \cdot 3 + M \cdot 0}{3M} = \frac{3}{3} = 1 \)

\( \text{CM}_y = \frac{M \cdot 0 + M \cdot 0 + M \cdot 3}{3M} = \frac{3}{3} = 1 \)

Thus, the coordinates of the center of mass are (1,1). To find the magnitude of the position vector from the origin to this point, we use the distance formula:

\(\text{Magnitude} = \sqrt{(1-0)^2 + (1-0)^2} = \sqrt{1+1} = \sqrt{2}\)

Therefore, the value of \(x\) in the expression \(\sqrt{x}\) is 2. This value falls within the specified range of 2,2, confirming the correctness of our solution.