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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Self Evaluation Test - Laws of Motion

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

    A) 144 N

    B) 96 N

    C) 240 N

    D) 192 N

    Correct Answer: D

    Solution :

    [d] Acceleration produced in upward direction \[a=\frac{F}{{{M}_{1}}+{{M}_{2}}+\text{ }\text{Mass of metal rod}}\] =\[\frac{480}{20+12+8}=12\text{ m}{{\text{s}}^{-2}}\] Tensions at the mid-point \[T=\left( {{M}_{2}}+\frac{\text{Mass of rod}}{2} \right)a\] \[=\left( 12+4 \right)\times 12=192N\]


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