JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of \[45{}^\circ \] with the vertical. Then F equals: (Take \[g=10\text{ }m{{s}^{2}}\] and the rope to be massless) [JEE MAIN Held on 07-01-2020 Evening]

    A) 75 N

    B) 70 N

    C) 90 N

    D) 100 N

    Correct Answer: D

    Solution :

    [d] \[{{T}_{2}}\text{ }cos\,45{}^\circ =100\text{ }N\]                  ...(i) \[{{T}_{2}}\text{ }sin\,45{}^\circ =F\]                              ...(ii) \[\Rightarrow \,\,F=100\text{ }N\]

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