The identical spheres each of mass $$2M$$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $$4$$ 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 ________.

We have three identical spheres each of mass $$2M$$ at points O(0,0), A(4,0) and B(0,4).

Formula for centre of mass coordinates: $$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}\quad -(1)$$ and $$y_{cm} = \frac{\sum m_i y_i}{\sum m_i}\quad -(2)$$.

Total mass = $$2M + 2M + 2M = 6M$$.

From $$(1)$$:
$$x_{cm} = \frac{2M\cdot0 + 2M\cdot4 + 2M\cdot0}{6M} = \frac{8M}{6M} = \frac{4}{3}$$.

From $$(2)$$:
$$y_{cm} = \frac{2M\cdot0 + 2M\cdot0 + 2M\cdot4}{6M} = \frac{8M}{6M} = \frac{4}{3}$$.

Position vector magnitude of centre of mass:
Formula: $$r = \sqrt{x_{cm}^2 + y_{cm}^2}\quad -(3)$$.

Substituting:
$$r = \sqrt{\Bigl(\frac{4}{3}\Bigr)^2 + \Bigl(\frac{4}{3}\Bigr)^2} = \frac{4}{3}\sqrt{2} = \frac{4\sqrt{2}}{3}$$.

Comparing with $$\frac{4\sqrt{2}}{x}$$ gives $$x = 3$$.

Answer: 3.