Three identical spheres each of mass $2 \mathrm{M}$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $4 \mathrm{~m}$ each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is $\frac{4 \sqrt{2}}{x}$, where the value of $x$ is ___________ .

<strong>1. Center of Mass Coordinates:</strong> <p>The center of mass (COM) of a system of particles is calculated as:</p> <p>$X_{COM} = \frac{\sum_{i} m_i x_i}{\sum_{i} m_i}$</p> <p>$Y_{COM} = \frac{\sum_{i} m_i y_i}{\sum_{i} m_i} $$$</p> <p>where $m_i$ is the mass of the $i$-th particle, and $(x_i, y_i)$ are its coordinates.</p> <strong>2. Coordinate Setup:</strong> <p>Let's place the origin at the right angle of the triangle and align the sides along the x and y axes:</p> <p><ul> <li>Sphere 1: $(4, 0)$</li><br> <li>Sphere 2: $(0, 4)$</li><br> <li>Sphere 3: $(0, 0)$</li> </ul></p> <strong>3. Calculations:</strong> <p>Since all spheres have mass $2M$, we can simplify the COM calculations:</p> <p>$$X_{COM} = \frac{2M \cdot 4 + 2M \cdot 0 + 2M \cdot 0} {2M + 2M + 2M} = \frac{4}{3}$$</p> <p>$$Y_{COM} = \frac{2M \cdot 0 + 2M \cdot 4 + 2M \cdot 0} {2M + 2M + 2M} = \frac{4}{3}$$</p> <strong>4. Magnitude of Position Vector:</strong> <p>The position vector of the COM is $\left(\frac{4}{3}, \frac{4}{3}\right)$. Its magnitude is:</p> <p>$$|\vec{r}_{COM}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \frac{4\sqrt{2}}{3}$$</p> <strong>5. Finding x:</strong> <p>We are given that the magnitude of the position vector is of the form $\frac{4\sqrt{2}}{x}$. Comparing this to our result, we find that $x = 3$.</p> <strong>Answer:</strong> <p>The value of $x$ is 3.</p> <p></p>