A) \[\frac{2\sqrt{2}}{\pi }\times \frac{1}{\sqrt{15LC}}\]
B) \[\frac{\pi }{2}\times \frac{1}{\sqrt{15LC}}\]
C) \[\mu \]
D) \[\frac{15\lambda }{(\mu -1)}\]
Correct Answer: C
Solution :
(c.)According to the question, \[|\mathbf{P}\times \mathbf{Q}|=\frac{\mathbf{P}.\mathbf{Q}}{\sqrt{3}}\] ??.(i) but \[|\mathbf{P}\times \mathbf{Q}|=|\mathbf{P}|\,|\mathbf{Q}|\,sin\theta =PQsin\theta \] and \[\mathbf{P}\,.\mathbf{Q}=|\mathbf{P}||\mathbf{Q}|\,\,cos\theta =PQcos\theta \] Making these substitutions in Eq. (i), we have \[\frac{PQ\,\,\cos \theta }{\sqrt{3}}=PQ\,\sin \theta =\tan \theta =\frac{1}{3}\] or, \[\theta ={{30}^{o}}\] Again we have. \[|\mathbf{P}+\mathbf{Q}|=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\,\cos \theta }\] \[=\sqrt{{{P}^{2}}+{{Q}^{2}}+2\times PQ\times \cos {{30}^{o}}}\] \[=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\frac{\sqrt{3}}{2}}\] \[\Rightarrow \] \[|\mathbf{P}+\mathbf{Q}|=\sqrt{{{P}^{2}}+{{Q}^{2}}+\sqrt{3}PQ}\]
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