JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 4)
Statement I :
Two forces $$\left( {\overrightarrow P + \overrightarrow Q } \right)$$ and $$\left( {\overrightarrow P - \overrightarrow Q } \right)$$ where $$\overrightarrow P \bot \overrightarrow Q $$, when act at an angle $$\theta$$1 to each other, the magnitude of their resultant is $$\sqrt {3({P^2} + {Q^2})} $$, when they act at an angle $$\theta$$2, the magnitude of their resultant becomes $$\sqrt {2({P^2} + {Q^2})} $$. 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 :-
Two forces $$\left( {\overrightarrow P + \overrightarrow Q } \right)$$ and $$\left( {\overrightarrow P - \overrightarrow Q } \right)$$ where $$\overrightarrow P \bot \overrightarrow Q $$, when act at an angle $$\theta$$1 to each other, the magnitude of their resultant is $$\sqrt {3({P^2} + {Q^2})} $$, when they act at an angle $$\theta$$2, the magnitude of their resultant becomes $$\sqrt {2({P^2} + {Q^2})} $$. 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 :-
Statement I is false but Statement II is true
Both Statement I and Statement II are true
Statement I is true but Statement II is false
Both Statement I and Statement II are false.
Explanation
$$\overrightarrow A = \overrightarrow P + \overrightarrow Q $$
$$\overrightarrow B = \overrightarrow P - \overrightarrow Q $$
$$\overrightarrow P \bot \overrightarrow Q $$
$$\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})(1 + \cos \theta )} $$
For $$\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {3({P^2} + {Q^2})} $$
$${\theta _1} = 60^\circ $$
For $$\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})} $$
$${\theta _2} = 90^\circ $$
$$\overrightarrow B = \overrightarrow P - \overrightarrow Q $$
$$\overrightarrow P \bot \overrightarrow Q $$
$$\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})(1 + \cos \theta )} $$
For $$\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {3({P^2} + {Q^2})} $$
$${\theta _1} = 60^\circ $$
For $$\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})} $$
$${\theta _2} = 90^\circ $$
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