Question 39 of 130

The identical spheres each of mass $2M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $4m$ 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 $\dfrac{4\sqrt{2}}{x}$, where the value of $x$ is ________.

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Answer: 3

Position vector ${\vec{r}}_{COM}=\dfrac{{m}_{1}{\vec{r}}_{1}+{m}_{2}{\vec{r}}_{2}+{m}_{3}{\vec{r}}_{3}}{{m}_{1}+{m}_{2}+{m}_{3}}$
${\vec{r}}_{\text{COM}}=\dfrac{2M\times 0+2M\times 4\hat{i}+2M\times 4\hat{j}}{6M}$
$\Rightarrow \vec{r}=\dfrac{4}{3}\hat{i}+\dfrac{4}{3}\hat{j}$
$|\vec{r}|=\sqrt{{\left(\dfrac{4}{3}\right)}^{2}+{\left(\dfrac{4}{3}\right)}^{2}}=\dfrac{4\sqrt{2}}{3}$
Therefore, $x=3$.