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

  • question_answer
    Sand is being dropped on a conveyer belt at the rate of 2 kg per second. The force necessary to keep the belt moving with a constant speed of 3 \[\text{m}{{\text{s}}^{\text{-1}}}\]will be          [JEE ONLINE 19-05-2012]

    A) 12N

    B) 6N

    C) Zero

    D) 18N

    Correct Answer: B

    Solution :

    [b] Here, rate of fall of sand on conveyer belt,\[\frac{dm}{dt}=2kg\,{{s}^{-1}}\]
    Conveyer belt speed \[v=3\,m{{s}^{-1}}\]
    Force F = ?
                \[F=v\frac{dm}{dt}=3\times 2=6N\]

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