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3-1.Vectors
hard
Statement $I :$Two forces $(\overrightarrow{{P}}+\overrightarrow{{Q}})$ and $(\overrightarrow{{P}}-\overrightarrow{{Q}})$ where $\overrightarrow{{P}} \perp \overrightarrow{{Q}}$, when act at an angle $\theta_{1}$ to 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. $\theta_{1}=60^{\circ} \text { and } \theta_{2}=90^{\circ}$
In the light of the above statements, 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
CStatement$-I$ is true but Statement$-II$ is false
DBoth Statement$-I$ and Statement$-II$ are false.
(JEE MAIN-2021)
Solution
$\overrightarrow{{A}}=\overrightarrow{{P}}+\overrightarrow{{Q}}$
$\overrightarrow{{B}}=\overrightarrow{{P}}-\overrightarrow{{Q}} \quad \overrightarrow{{P}} \perp \overrightarrow{{Q}}$
$|\overrightarrow{{A}}|=|\overrightarrow{{B}}|=\sqrt{{P}^{2}+{Q}^{2}}$
$|\overrightarrow{{A}}+\overrightarrow{{B}}|=\sqrt{2\left({P}^{2}+{Q}^{2}\right)(1+\cos \theta)}$
$\text { For }|\overrightarrow{{A}}+\overrightarrow{{B}}|=\sqrt{3\left({P}^{2}+{Q}^{2}\right)}$
$\theta_{1}=60^{\circ}$
For $|\vec{A}+\vec{B}|=\sqrt{2\left(P^{2}+Q^{2}\right)}$
$\theta_{2}=90^{\circ}$
$\overrightarrow{{B}}=\overrightarrow{{P}}-\overrightarrow{{Q}} \quad \overrightarrow{{P}} \perp \overrightarrow{{Q}}$
$|\overrightarrow{{A}}|=|\overrightarrow{{B}}|=\sqrt{{P}^{2}+{Q}^{2}}$
$|\overrightarrow{{A}}+\overrightarrow{{B}}|=\sqrt{2\left({P}^{2}+{Q}^{2}\right)(1+\cos \theta)}$
$\text { For }|\overrightarrow{{A}}+\overrightarrow{{B}}|=\sqrt{3\left({P}^{2}+{Q}^{2}\right)}$
$\theta_{1}=60^{\circ}$
For $|\vec{A}+\vec{B}|=\sqrt{2\left(P^{2}+Q^{2}\right)}$
$\theta_{2}=90^{\circ}$
Standard 11
Physics
