Two moving particles P & Q are 10m apart at any instant. Velocity of P is 8 m/s and that of Q is 6 m/s at ${30}^{\circ }$ angle with the line joining P & Q.
Calculate the angular velocity P w.r.t. Q

The correct option is B 1.4 rad/s

$\begin{array}{c}\text{Distance between}P\text{and}\theta \text{is}10m\text{.}\\ \text{Given}\to \left|{\overrightarrow{V}}_p\right|=8m/s\text{and}\left|{\overrightarrow{V}}_B\right|=6m/s\end{array}$

$\begin{array}{cc}\text{Thul,}{\overrightarrow{V}}_p& =8(\cos {30}^{\circ }\hat{i}-\sin {30}^{\circ }\hat{j})m/s\\ & =4\left(\sqrt{3}\hat{i}\hat{j}\right)m/s\end{array}$

$\text{and,}\begin{array}{cc}\overrightarrow{V_B}& =6(-\cos {30}^{\circ }\hat{i}+\sin {30}^{\circ }j)m/s.\\ & =3(\sqrt{3}\hat{i}+\hat{j})m/s.\end{array}$

$\begin{array}{cc}\text{so,}{\overrightarrow{V}}_{P/\theta }& =(7\sqrt{3}\hat{ı}-7\hat{j})m/s.\\ & =7\left(\sqrt{3}\hat{i}\hat{j}\right)m/s.\end{array}$

$\text{and,}{\overrightarrow{r}}_{P/Q}=-10\hat{i}m$

$\text{Hence, we know,}\overrightarrow{V}=\overrightarrow{\omega }\times \overrightarrow{r}\Rightarrow |\overrightarrow{\omega }|=\frac{\left|{\overrightarrow{v}}^{}\right|}{\left|{\overrightarrow{r}}_{P/Q}\right|}=\frac{7\times 2}{10}=1.4\text{rad/s}$