A) along \[\vec{P}\]
B) along \[\vec{Q}\]
C) at \[{{60}^{o}}\]to\[\vec{Q}\]
D) at right angle to \[\vec{Q}\]
Correct Answer: D
Solution :
Let \[\alpha \] be the angle between the forces \[\vec{P}\]and \[\vec{Q}\] It is given that the resultant of \[\vec{P}\]and \[\vec{Q}\] is of magnitude P. \[\therefore \] \[{{P}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos \alpha \] \[\Rightarrow \] \[Q(Q+2P\cos \alpha )=0\] \[\Rightarrow \] \[Q+2P\cos \alpha =0\] ?(i) Let \[\theta \] be the angle between the forces \[\vec{Q}\] and the new resultant. \[\therefore \] \[\tan \theta =\frac{2P\sin \alpha }{Q+2P\cos \alpha }=\infty \][from(i)] \[\Rightarrow \] \[\theta =\frac{\pi }{2}\] \[\therefore \]The new resultant is at right angle to \[\vec{Q}.\]
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