Question

3 particles of masses $2\text{kg}$ each, are placed such that $1^{st}$ one lies on negative $x-$ axis, $2^{nd}$ one lies on negative $y-$ axis and the third one lies on positive $z-$ axis at distances of $2\text{m},3\text{m}$ and $1\text{m}$ respectively from the origin. Then the square of the distance of centre of mass of the system from the origin is

• A. $1.55{\text{m}}^2$
• B. $\sqrt{1.55}{\text{m}}^2$
• C. $1.55\text{m}$
• D. $1.25\text{m}$

Answer 1

The correct option is A $1.55{\text{m}}^2$

Given that;
$m_1=2\text{kg};\overrightarrow{r_1}=-2\hat{i},m_2=2\text{kg};\overrightarrow{r_2}=-3\hat{j}$
$m_3=2\text{kg};\overrightarrow{r_3}=\hat{k}$
$\therefore {\overrightarrow{r}}_{CM}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}+m_3\overrightarrow{r_3}}{m_1+m_2+m_3}$
$=\frac{2(-2\hat{i})+2(-3\hat{j})+2\left(\text{k}\right)}{6}=\frac{-2\hat{i}-3\hat{j}+\hat{k}}{3}$
$\therefore |{\overrightarrow{r}}_{CM}|=\sqrt{x^2+y^2+z^2}=\sqrt{\frac{4}{9}+\frac{9}{9}+\frac{1}{9}}=\sqrt{\frac{14}{9}}$
$\therefore |{\overrightarrow{r}}_{CM}{|}^2=\frac{14}{9}=1.55{\text{m}}^2$
Hence, the correct answer is option (a).