Question

The resultant of the forces $P$ and $Q$ is equal to $2\sqrt{3}$ and makes angle $\frac{\pi }{6}$ with the direction of $P$ then $\frac{P}{Q}$

• A. $1$
• B. $2$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

Answer: (C)

$\frac{1}{2}$

$\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}$
$\Rightarrow \overrightarrow{R}+\overrightarrow{P}=\overrightarrow{Q}$
$\Rightarrow {|\overrightarrow{R}-\overrightarrow{P}|}^2={\left|\overrightarrow{Q}\right|}^2$
$\Rightarrow {\left|\overrightarrow{R}\right|}^2+{\left|\overrightarrow{P}\right|}^2-2\times \overrightarrow{R}\overrightarrow{P}={\left|\overrightarrow{Q}\right|}^2$ but $\left|\overrightarrow{R}\right|=\sqrt{3}\left|\overrightarrow{Q}\right|$
$\Rightarrow 3{\left|\overrightarrow{Q}\right|}^2+P^2-2\sqrt{3}\left|\overrightarrow{Q}\right|\overrightarrow{P}={\left|\overrightarrow{Q}\right|}^2$
$\Rightarrow (P-Q)(P-2\overrightarrow{Q})=0$ on factorizing
$\Rightarrow P=Q,P=2Q$ on simplification
$\therefore \frac{P}{Q}=\frac{1}{2}$