Question 42 of 130

Three identical spheres each of mass $M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $3m$ 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

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Answer: 2

The diagram represents the locations of the masses as mentioned in the question.
For point $1$, $\vec{{r}_{1}}=0\hat{i}+0\hat{j}$ for point $2$, $\vec{{r}_{2}}=3\hat{i}+0\hat{j}$ and for point $3$, $\vec{{r}_{3}}=0\hat{i}+3\hat{j}$.
The formula for centre of mass is
${\vec{r}}_{com}=\dfrac{{m}_{1}\vec{{r}_{1}}+{m}_{2}\vec{{r}_{2}}+{m}_{3}\vec{{r}_{3}}}{{m}_{1}+{m}_{2}+{m}_{3}}$
$\Rightarrow {\vec{r}}_{com}=\dfrac{M\left(0\hat{i}+0\hat{j}\right)+M\left(3\hat{i}\right)+M\left(3\hat{j}\right)}{3M}$
$\Rightarrow {\vec{r}}_{com}=\hat{i}+\hat{j}$
$\left|{\vec{r}}_{com}\right|=\sqrt{{1}^{2}+{1}^{2}}=\sqrt{2}=\sqrt{x}$
$\Rightarrow x=2$