question_answer

Forces \[P\] and \[Q\] acting at a point \[O\] make an angle \[{{150}^{o}}\] between them. Their resultant acts at \[O\], has magnitude \[2\] unit and is perpendicular to \[P\]. Then in the same unit, the magnitudes of \[P\] and \[Q\] are:

A) \[2\sqrt{3},\,\,4\]                           

B) \[\sqrt{\frac{3}{2}},\,\,2\]

C) \[3,\,\,4\]                                           

D) \[4,\,\,5\]

Correct Answer: A

Solution :

We have,\[R=2,\,\,\theta ={{90}^{o}}\] \[\therefore \]  \[{{2}^{2}}={{p}^{2}}+{{Q}^{2}}+2PQ\cos \alpha \] \[\Rightarrow \]               \[4={{p}^{2}}+{{Q}^{2}}+2P(-P)\] \[\Rightarrow \]               \[4={{Q}^{2}}-{{P}^{2}}\]                                              ? (i) Also,      \[P+Q\cos {{150}^{o}}=0\] \[\Rightarrow \]               \[P=\frac{Q\sqrt{3}}{2}\] From Eq. (i)                 \[4={{Q}^{2}}-{{\left( -\frac{Q\sqrt{3}}{2} \right)}^{2}}\]                 \[4={{Q}^{2}}-\frac{3{{Q}^{2}}}{4}\] \[\Rightarrow \]               \[4=\frac{{{Q}^{2}}}{4}\Rightarrow Q=4\] and        \[P=2\sqrt{3}\]