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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Mock Test - Newtons Laws of Motion without Friction

  • question_answer
    A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force F at the other end. If the mass of the rope is half the mass of the block, the tension in the middle of the rope will be

    A) \[F\]                

    B) \[2\,F/3\]

    C) \[3\,F/5\]          

    D) \[5\,F/6\]

    Correct Answer: D

    Solution :

    [d] The acceleration of block-rope system is \[a=\frac{F}{(M+m)}\] Where M is the mass of block and m is the mass of rope. So the tension in the middle of the rope will be \[T=\{M+(m/2)\}a=\frac{M+(m/2)F}{M+m}\] Given that m=M/2 \[\therefore T=\left[ \frac{M+(M/4)}{M+(M/2)} \right]F=\frac{5F}{6}\]


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