A) \[0{}^\circ \]
B) \[30{}^\circ \]
C) \[60{}^\circ \]
D) \[90{}^\circ \]
Correct Answer: D
Solution :
Resultant of P and Q, \[R=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta }\] Here,\[R=P\] \[\therefore \] \[P=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta }\] \[{{P}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta \] \[{{Q}^{2}}+2PQ\cos \theta =0\] \[Q+2P\cos \theta =0\] \[(\because \,P\ne 0)\] Required angle \[\beta ={{\tan }^{-1}}\left[ \frac{p}{Q+p\cos \theta } \right]\] \[={{\tan }^{-1}}\left( \frac{2P}{Q+2P\cos \theta } \right)\] \[(\because \,P=2P)\] \[={{\tan }^{-1}}\left( \frac{2P}{0} \right)\] \[={{\tan }^{-1}}(\infty )={{90}^{o}}\]
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