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Question

The resultant of two vectors $$ \overrightarrow P $$ and $$ \overrightarrow Q $$ is $$ \overrightarrow R $$ . If $$ \overrightarrow Q $$ is doubled then the new resultant vector is perpendicular to $$ \overrightarrow P $$ . The magnitude of $$ \overrightarrow R $$ is :-


A
P2Q22PQ
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B
Q
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C
PQ
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D
P+QPQ
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Solution

The correct option is A $$\displaystyle Q $$
After the vector $$Q$$ is doubled ,
The resultant is $$90^{o}$$ with $$P$$ hence angle between $$P$$ and $$2Q$$ is obtuse 
Let the angle be $$90+\theta$$ and $$P$$ be along X-axis then resultant would be along Y-axis
Resolving $$2Q$$ along X-axis and Y-axis
$$Qcos\theta = $$resultant
$$2Qsin \theta = P$$
Hence, $$sin\theta =\dfrac{P}{2Q}$$
Initially the magnitude of resultant $$R$$ is 
$$=\sqrt{P^{2}+Q^{2}+2PQcos(90+\theta)}$$

$$=\sqrt{P^{2}+Q^{2}-2PQsin\theta}$$ 
                                                                     
$$=\sqrt{P^{2}+Q^{2}-2PQ\times \dfrac{P}{2Q}}$$
                                                                     
$$=\sqrt{Q^{2}}$$

$$=Q$$

Physics

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