# Statement I : Two forces

Question:

Statement I :

Two forces $(\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}})$ and $(\overrightarrow{\mathrm{P}}-\overrightarrow{\mathrm{Q}})$ where $\overrightarrow{\mathrm{P}} \perp \overrightarrow{\mathrm{Q}}$, when act at an angle $\theta_{1}$ 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 $\theta_{2}$, the magnitude of their resultant becomes $\sqrt{2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)}$. This is possible only when $\theta_{1}<\theta_{2}$.

Statement II :

In the situation given above.

$\theta_{1}=60^{\circ}$ and $\theta_{2}=90^{\circ}$

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

1. Statement-I is false but Statement-II is true

2. Both Statement-I and Statement-II are true

3. Statement-I is true but Statement-II is false

4. Both Statement-I and Statement-II are false.

Correct Option: 2,

Solution:

$\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}$

$\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{P}}-\overrightarrow{\mathrm{Q}}$                  $\overrightarrow{\mathrm{P}} \perp \overrightarrow{\mathrm{Q}}$

$|\overrightarrow{\mathrm{A}}|=|\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}}$

$|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)(1+\cos \theta)}$

For $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{3\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)}$

$\theta_{1}=60^{\circ}$

For $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)}$

$\theta_{2}=90^{\circ}$