JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    Two forces P and Q, of magnitude 2F and 3F, respectively; are at an angle 9 with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle \[\theta \] is- [JEE Main 10-Jan-2019 Evening]

    A) \[90{}^\circ \]

    B) \[60{}^\circ ~\]

    C) \[30{}^\circ ~~\]

    D) \[120{}^\circ \]

    Correct Answer: B

    Solution :

    [b] \[{{R}_{1}}\,\,=\,\,\sqrt{{{\left( 2F \right)}^{2}}+{{\left( 3F \right)}^{2}}+2.2F.3F\,cos\,\theta }\]
    \[{{R}_{2}}\,\,=\,\,\sqrt{{{\left( 2F \right)}^{2}}+{{\left( 6F \right)}^{2}}+2.2F.6F\,cos\,\theta }\]
    If \[{{R}_{2}}=2{{R}_{1}}\]
    \[\sqrt{{{\left( 2F \right)}^{2}}+{{\left( 3F \right)}^{2}}+2.2F.3Fcos\,\theta }\]
    \[=\,\,2\sqrt{{{\left( 2F \right)}^{2}}+{{\left( 6F \right)}^{2}}+2.2F.6\,F\,cos\,\theta }\]
    \[\cos \,\theta \,\,=\,\,-\frac{1}{2}=\cos \,120{}^\circ \]
     \[\theta =120{}^\circ \]


You need to login to perform this action.
You will be redirected in 3 sec spinner