Statement I: Two forces $\left(\mathrm{P}+\mathrm{Q}\right)$ and $\left(\mathrm{P}-\mathrm{Q}\right)$ where $\mathrm{P}\perp \mathrm{Q}$, when act at an angle ${\mathrm{\theta }}_1$ to each other, the magnitude of their resultant is $\sqrt{3\left({\mathrm{P}}^2+{\mathrm{Q}}^2\right)}$, when they act at an angle ${\mathrm{\theta }}_2$ , the magnitude of their resultant becomes $\sqrt{2\left({\mathrm{P}}^2+{\mathrm{Q}}^2\right)}$. This is possible only when ${\mathrm{\theta }}_1<{\mathrm{\theta }}_2$.

Statement II: In the situation given above. ${\mathrm{\theta }}_1={60}^{\circ }\text{ and }{\mathrm{\theta }}_2={90}^{\circ }.$

In the light of the above statements, choose the most appropriate answer from the options given below.

Given, force vectors P ⊥ Q, i.e. θ = 90°

Let resultant of P + Q = x and resultant of P − Q = y

    $\begin{array}{l}\therefore {\mathrm{x}}^2={\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQcos}{90}^{\circ }={\mathrm{P}}^2+{\mathrm{Q}}^2 ...\left(\mathrm{i}\right)\\ {\mathrm{y}}^2={\mathrm{P}}^2+{\mathrm{Q}}^2-2\mathrm{PQcos}{90}^{\circ }={\mathrm{P}}^2+{\mathrm{Q}}^2 ...\left(\mathrm{ii}\right)\end{array}$

When θ₁ is the angle between (P + Q) and (P − Q), then, their resultant is given by

     $\begin{array}{l} |\mathrm{x}+\mathrm{y}|=\sqrt{3\left({\mathrm{P}}^2+{\mathrm{Q}}^2\right)}\\ \therefore \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{xycos}{\mathrm{\theta }}_1}=\sqrt{3{\mathrm{P}}^2+3{\mathrm{Q}}^2} ...\left(\mathrm{iii}\right)\end{array}$

Substituting the values of x² and y² in Eq. (iii), we get

$\begin{array}{l}{\mathrm{P}}^2+{\mathrm{Q}}^2+{\mathrm{P}}^2+{\mathrm{Q}}^2+2\sqrt{{\mathrm{P}}^2+{\mathrm{Q}}^2}\sqrt{{\mathrm{P}}^2+{\mathrm{Q}}^2}\cos {\mathrm{\theta }}_1=3{\mathrm{P}}^2+3{\mathrm{Q}}^2\\ 2{\mathrm{P}}^2+2{\mathrm{Q}}^2+\left(2{\mathrm{P}}^2+2{\mathrm{Q}}^2\right)\cos {\mathrm{\theta }}_1=3{\mathrm{P}}^2+3{\mathrm{Q}}^2\\ \Rightarrow \cos {\mathrm{\theta }}_1=\frac{{\mathrm{P}}^2+{\mathrm{Q}}^2}{2\left({\mathrm{P}}^2+{\mathrm{Q}}^2\right)}=\frac{1}{2}\\ \therefore {\mathrm{\theta }}_1={60}^{\circ }\end{array}$

When θ₂ is the angle between (P + Q) and (P − Q), then the magnitude of their resultant is given by

$\begin{array}{l}|\mathrm{x}+\mathrm{y}|=\sqrt{2\left({\mathrm{P}}^2+{\mathrm{Q}}^2\right)}\\ {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{xycos}{\mathrm{\theta }}_2=2{\mathrm{P}}^2+2{\mathrm{Q}}^2 ...\left(\mathrm{iv}\right)\end{array}$

Substituting the values, we get

$\begin{array}{l} {\mathrm{P}}^2+{\mathrm{Q}}^2+{\mathrm{P}}^2+{\mathrm{Q}}^2+2\left({\mathrm{P}}^2+{\mathrm{Q}}^2\right)\cos {\mathrm{\theta }}_2=2{\mathrm{P}}^2+2{\mathrm{Q}}^2\\ \cos {\mathrm{\theta }}_2=0\\ \Rightarrow {\mathrm{\theta }}_2={90}^{\circ }\\ \text{ So, }{\mathrm{\theta }}_1={60}^{\circ }\text{ and }{\mathrm{\theta }}_2={90}^{\circ }\text{ and }{\mathrm{\theta }}_1<{\mathrm{\theta }}_2\end{array}$

Hence, both statement I and statement II are true.