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

  • question_answer
    Two blocks A and B of masses \[{{m}_{A}}=1kg\]and \[{{m}_{B}}=3kg\]are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is : (Take \[g=10\text{ }m/{{s}^{2}}\])
    [JEE Main 10-4-2019 Afternoon]

    A) 16 N

    B) 40 N

    C) 12 N

    D) 8 N

    Correct Answer: A

    Solution :

    \[{{a}_{A\max }}=\mu g=2m/{{s}^{2}}\]
    \[F-8=4\times 2\]

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