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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    In the above problem if the lift moves up with an acceleration equal to the acceleration due to gravity, the reading on the spring balance will be    [NCERT 1977]

    A)                         2 kg               

    B)                         \[(2\times g)\,kg\]            

    C)                         \[(4\times g)\,kg\]            

    D)                           4 kg            

    Correct Answer: D

    Solution :

                                When lift moves upward then reading of the spring balance, \[R=m(g+a)=2(g+g)=4g\ N=4\ kg\]  [As \[a=g\]]


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