The masses and positions (in rectangular coordinates) of four particles are as follows : $1 \mathrm{~kg}$ at $(a, a), 2 \mathrm{~kg}$ at $(-a, a), 3 \mathrm{~kg}$ at $(a, a)$ and $4 \mathrm{~kg}$ at $(a, a)$. The position vector of the centre of mass of the system of four particles is

• A

  $-4 a \hat{\mathbf{i}}$
• B

  $-4 a \hat{\mathbf{i}}-4 \hat{\mathbf{j}}$
• C

  0
• Correct answer
  D

  $-0.4 \hat{\mathbf{j}}$

Solution

Position vector of COM of system is $$ \begin{aligned} & \mathbf{r}_{\mathrm{COM}}=\frac{m_1 \mathbf{r}_1+m_2 \mathbf{r}_2+m_3 \mathbf{r}_3+m_4 \mathbf{r}_4}{m_1+m_2+m_3+m_4} \\ & =\frac{1(a \hat{\mathrm{i}}+a \hat{\mathrm{j}})+2(-a \hat{\mathbf{\mathrm { i }}}+a \hat{\mathrm{j}})+3(-a \hat{\hat{\mathrm{i}}}-a \hat{\mathrm{j}})+4(a \hat{\mathrm{i}}-\hat{a})}{10} \\ & =\frac{0 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}}{10}=-0.4 \hat{\mathrm{j}} \end{aligned} $$

Hence, the correct answer is option: 'D'.