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

  • question_answer
    Two blocks of mass \[{{M}_{1}}=20\,kg\] and \[{{\operatorname{M}}_{2}}=12\operatorname{K}\operatorname{g},\] are connected by a metal rod of mass 8 kg. The system is pulled vertically up applying a force of 480 N as shown. The tension at mid-point of the rod is:
                            [JEE ONLINE 22-04-2013]

    A)             \[144\] N

    B) \[96\] N

    C)             \[240\] N

    D) \[192\] N

    Correct Answer: D

    Solution :

    [d] Acceleration produced in upward direction
    \[\text{a =}\frac{\text{F}}{{{\text{M}}_{\text{1}}}\text{+}{{\text{M}}_{\text{2}}}\text{+ Mass of metal rod}}\]
    Tension at the mid-point
    \[\text{T =}\left( {{\text{M}}_{\text{2}}}\text{+}\frac{\text{Mass of rod}}{\text{2}} \right)\text{a}\]
    \[=(12+4)\times 12=192N\]
     [d] When the body has maximum speed then

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