AP EAMCET · PHYSICS · Mathematics in Physics
A vector \(\mathbf{P}\) directed along the \(X\)-axis is added to vector \(\mathbf{Q}\) which has a magnitude of \(10 \mathrm{~m}\). The resultant vector is directed along the \(Y\)-axis, with a magnitude that is 2 times that of \(\mathbf{P}\). The magnitude of \(\mathbf{P}\) is
- A \(\sqrt{10} \mathrm{~m}\)
- B \(5 \sqrt{2} \mathrm{~m}\)
- C \(6 \mathrm{~m}\)
- D \(2 \sqrt{5} \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(2 \sqrt{5} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
By parallelogram law of vector addition, angle made by resultant with \(\mathbf{P}\) is given by; \(\tan \alpha=\frac{Q \sin \theta}{P+Q \cos \theta}\) Here \(\alpha=90^{\circ}\) so ; \(\tan 90^{\circ}=\infty=\frac{Q \sin \theta}{P+Q \cos \theta}\)…
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