A) \[\theta ={{\cos }^{-1}}\left( \frac{P-Q}{Q} \right),\,\,R=\sqrt{{{P}^{2}}-{{Q}^{2}}+2PQ}\]
B) \[\theta ={{\cos }^{-1}}\left( \frac{Q-P}{Q} \right),\,\,R=\sqrt{{{Q}^{2}}-{{P}^{2}}+2PQ}\]
C) \[\theta ={{\cos }^{-1}}\left( \frac{P-Q}{P} \right),\,\,R=\sqrt{2{{P}^{2}}-{{Q}^{2}}+2PQ}\]
D) None of the above
Correct Answer: B
Solution :
Let \[\theta \] be the angle between the forces \[P\] and \[Q\], then \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta \] ... (i) Let the resultant \[R\] make an angle \[\alpha \] with the direction of force \[P\] and \[\beta \] with the direction of force\[Q\]. Then, \[\sin \alpha =\frac{Q\sin \theta }{R},\,\,\cos \alpha =\frac{P+Q\cos \theta }{R}\] Given, resolved part of \[R\] in the direction of \[P=Q\] \[\Rightarrow \] \[P+Q\cos \theta =Q\] \[\Rightarrow \] \[\cos \theta =\frac{Q-P}{Q}\] \[\Rightarrow \] \[\theta ={{\cos }^{-1}}\left( \frac{Q-P}{Q} \right)\] which is the required angle. Putting\[\cos \theta =\frac{Q-P}{Q}\]in Eq. (i), we get \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2P(Q-P)\] \[\Rightarrow \] \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ-2{{P}^{2}}\] \[\Rightarrow \] \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\] \[\Rightarrow \] \[R=\sqrt{{{Q}^{2}}-{{P}^{2}}+2PQ}\]You need to login to perform this action.
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