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

  • question_answer
    A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of \[45{}^\circ \] at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is \[\left( g=10\text{ }m{{s}^{-}}^{2} \right)\] [JEE Main 09-Jan-2019 Evening]

    A) 100 N

    B) 70 N

    C) 140 N

    D) 200 N

    Correct Answer: A

    Solution :

    \[\frac{f}{\sin \,(180{}^\circ -45{}^\circ )}\,\,=\,\,\frac{Mg}{\sin \,(90{}^\circ +45{}^\circ )}\]
    \[\frac{f}{\sin \,45{}^\circ }\,\,=\,\,\frac{Mg}{\cos \,45{}^\circ }\]
    \[\Rightarrow \,\,\,f=Mg=10g=100\,N\]

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