JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Equilibrium of Forces

  • question_answer
    Two forces with equal magnitudes F act on a body and the magnitude of the resultant force is F/3. The angle between the two forces is                                 [MP PMT 1999]

    A)                         \[{{\cos }^{-1}}\left( -\frac{17}{18} \right)\]

    B)                           \[{{\cos }^{-1}}\left( -\frac{1}{3} \right)\]            

    C)                         \[{{\cos }^{-1}}\left( \frac{2}{3} \right)\]            

    D)                           \[{{\cos }^{-1}}\left( \frac{8}{9} \right)\]

    Correct Answer: A

    Solution :

                    \[F_{_{net}}^{2}=F_{1}^{2}+F_{2}^{2}+2{{F}_{1}}{{F}_{2}}\cos \theta \]             Þ \[{{\left( \frac{F}{3} \right)}^{2}}\]\[={{F}^{2}}+{{F}^{2}}+2{{F}^{2}}\cos \theta \] Þ \[\cos \theta =\left( -\frac{17}{18} \right)\]          


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