Question 19 of 121

Statement-I : Two forces $\left(\vec{P}+\vec{Q}\right)$ and $\left(\vec{P}-\vec{Q}\right)$ where $\vec{P}\perp \vec{Q},$ 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 ${\theta }_{1}\lt {\theta }_{2}.$
Statement-II : In the situation given above.
${\theta }_{1}=60^{\circ}$ and ${\theta }_{2}=90^{\circ}$
In the light of the above statement, choose the most appropriate answer from the options given below :

AStatement I is false but Statement II is true.BBoth Statement I and Statement II are true.CBoth Statement I and Statement II are false.DStatement I is true but Statement II is false.

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Answer: (B) Both Statement I and Statement II are true.

${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ at ${\theta }_{1}$
${F}_{net1}=\sqrt{{P}^{2}+{Q}^{2}+{P}^{2}+{Q}^{2}+2\left({P}^{2}+{Q}^{2}\right)\cos {\theta }_{1}}$
${F}_{net2}=\sqrt{{P}^{2}+{Q}^{2}+{P}^{2}+{Q}^{2}+2\left({P}^{2}+{Q}^{2}\right)\cos {\theta }_{2}}$
If ${F}_{net1}=\sqrt{3\left({P}^{2}+{Q}^{2}\right)}=\sqrt{2\left({P}^{2}+{Q}^{2}\right)+2\left({P}^{2}+{Q}^{2}\right)\cos {\theta }_{1}}$
$\Rightarrow \cos {\theta }_{1}=\dfrac{{P}^{2}+{Q}^{2}}{2\left({P}^{2}+{Q}^{2}\right)}$
$\Rightarrow {\theta }_{1}=60^{\circ}$
${F}_{net2}=\sqrt{2\left({P}^{2}+{Q}^{2}\right)}=\sqrt{2\left({P}^{2}+{Q}^{2}\right)+2\left({P}^{2}+{Q}^{2}\right)\cos {\theta }_{2}}$
$\Rightarrow \cos {\theta }_{2}=0\Rightarrow {\theta }_{2}=90^{\circ}$