BVP Medical BVP Medical Solved Paper-2008

  • question_answer
    The square of resultant of two equal forces is three times their product. Angle between the forces is

    A) \[\pi \]                                 

    B) \[\frac{\pi }{2}\]

    C) \[\frac{\pi }{4}\]                                              

    D) \[\frac{\pi }{3}\]

    Correct Answer: D

    Solution :

                    Let \[\theta \]  be the angle between vectors \[\overrightarrow{\mathbf{P}}\] and \[\overrightarrow{\mathbf{Q}}\] whose resultant is \[\overrightarrow{\mathbf{R}}\]. Here, \[P=Q\] and \[{{R}^{2}}=3PQ=3{{P}^{2}}\] As     \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\,\cos \theta \] \[\therefore \]  \[3{{P}^{2}}={{P}^{2}}+{{P}^{2}}+2{{P}^{2}}\,\cos \theta \] or            \[3{{P}^{2}}-2{{P}^{2}}=2{{P}^{2}}\,\cos \theta \] or            \[{{P}^{2}}=2{{P}^{2}}\,\cos \theta \] or            \[1=2\cos \theta \] \[\therefore \]  \[\cos \theta =\frac{1}{2},\]thus, \[\cos \theta =\cos {{60}^{o}}\] or            \[\theta ={{60}^{o}}=\frac{\pi }{3}\]


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