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

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    A mass of 10 gm is suspended by a string and the entire system is falling with a uniform acceleration of \[400\,cm/{{\sec }^{2}}.\]The tension in the string will be \[(g=980\,cm/{{\sec }^{2}})\]                                    [SCRA 1994]

    A)                              5,800 dyne            

    B)                           9,800 dyne          

    C)                         11,800 dyne                

    D)                         13,800 dyne

    Correct Answer: A

    Solution :

                                \[T=m(g-a)\]= 10(980 ? 400) = 5800 dyne            


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