Question

When two forces of magnitude $P$ and $Q$ are perpendicular to each other, their resultant is of magnitude $R$. When they are at an angle of ${180}^{\circ }$ to each other their resultant is of magnitude $\frac{R}{\sqrt{2}}$. Find the ratio $\frac{P}{Q}$.

• A. $\sqrt{3}\pm 2$
• B. $5\pm \sqrt{3}$
• C. $3\pm \sqrt{5}$
• D. $2\pm \sqrt{3}$

Answer 1

The correct option is D $2\pm \sqrt{3}$


Using parallelogram law,

$R^2=P^2+Q^2+2PQ\cos {90}^{\circ }$

$R^2=P^2+Q^2$ ------------(I)


(Assuming $P>Q$)

$|\overrightarrow{R}|=|\overrightarrow{P}|-|\overrightarrow{Q}|$

$\frac{R}{\sqrt{2}}=P-Q$

$R=\sqrt{2}(P-Q)$ ------------(II)

Putting (II) in (I)

we get;

${\left[\sqrt{2}\right(P-Q\left)\right]}^2=P^2+Q^2$

$(\sqrt{2}{)}^2[P^2+Q^2-2PQ]=P^2+Q^2$

$2P^2+2Q^2-4PQ=P^2+Q^2$

$P^2+Q^2-4PQ=0$

$\frac{P^2}{Q^2}+1-4\left(\frac{P}{Q}\right)=0$

Let $\frac{P}{Q}=x$ , $x^2-4x+1=0$

$\therefore x=\frac{-(-4)\pm \sqrt{{(-4)}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}$

$=\frac{4\pm \sqrt{12}}{2}$

$x=2\pm \sqrt{3}=\frac{P}{Q}$

Hence, option $\left(B\right)$ is correct.