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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Static and Limiting Friction

  • question_answer
    A block of 1 kg is stopped against a wall by applying a force F perpendicular to the wall. If \[\mu =0.2\] then minimum value of F will be                              [MP PMT 2003]

    A)                         980 N            

    B)                           49 N            

    C)                         98 N               

    D)                         490 N

    Correct Answer: B

    Solution :

                                \[F=\frac{W}{\mu }=\frac{1\times 9.8}{0.2}=49N\]            


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