question_answer

If the resultant of two forces P and Q acting on a particle is at right angle to Q, then the angle between the forces is

A)  \[{{\cos }^{-1}}\left( \frac{P}{Q} \right)\]

B)  \[\pi -{{\cos }^{-1}}\left( \frac{Q}{P} \right)\]

C)  \[\pi -{{\cos }^{-1}}\left( \frac{P}{Q} \right)\]

D)  None of these

Correct Answer: B

Solution :

\[\tan \,{{90}^{o}}=\frac{P\,\,\sin \,\alpha }{Q+P\,\,\cos \,\,\alpha }\] \[\Rightarrow \] \[Q+P\,\,\cos \,\alpha =0\] \[\Rightarrow \] \[\cos \,\alpha =-\frac{Q}{P}\] \[\Rightarrow \] \[\alpha ={{\cos }^{-1}}\left( -\frac{Q}{P} \right)\] \[\Rightarrow \] \[\alpha =\pi -{{\cos }^{-1}}\left( \frac{Q}{P} \right)\]