Statement-I : Two forces $\left(\overrightarrow{P}+\overrightarrow{Q}\right)$ and $\left(\overrightarrow{P}-\overrightarrow{Q}\right)$ where $\overrightarrow{P}\perp \overrightarrow{Q},$ when act at an angle ${\theta }_1$ each other, the magnitude of their resultant is $\sqrt{3\left(P^2+Q^2\right)},$ when they act at an angle ${\theta }_2,$ the magnitude of their resultant becomes $\sqrt{2\left(P^2+Q^2\right)}.$ This is possible only when ${\theta }_1<{\theta }_2.$
Statement-II : In the situation given above.
${\theta }_1=60°$ and ${\theta }_2=90°$
In the light of the above statement, choose the most appropriate answer from the options given below :

• A

  Statement I is false but Statement II is true.
• Correct answer
  B

  Both Statement I and Statement II are true.
• C

  Both Statement I and Statement II are false.
• D

  Statement I is true but Statement II is false.

Solution

${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ at ${\theta }_1$
$F_{net 1}=\sqrt{P^2+Q^2+P^2+Q^2+2\left(P^2+Q^2\right)\cos {\theta }_1}$
$F_{net 2}=\sqrt{P^2+Q^2+P^2+Q^2+2\left(P^2+Q^2\right)\cos {\theta }_2}$
If $F_{net 1}=\sqrt{3\left(P^2+Q^2\right)}=\sqrt{2\left(P^2+Q^2\right)+2\left(P^2+Q^2\right)\cos {\theta }_1}$
$\Rightarrow \cos {\theta }_1=\frac{P^2+Q^2}{2\left(P^2+Q^2\right)}$
$\Rightarrow {\theta }_1=60°$
$F_{net 2}=\sqrt{2\left(P^2+Q^2\right)}=\sqrt{2\left(P^2+Q^2\right)+2\left(P^2+Q^2\right)\cos {\theta }_2}$
$\Rightarrow \cos {\theta }_2=0 \Rightarrow {\theta }_2=90°$

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