MCQEasyJEE Main 2024 · 6 April, Shift 1Newton's Second Law & Force

Physics Question from JEE Main 2024 · 6 April, Shift 1

A 1kg1 \, \text{kg} mass is suspended from the ceiling by a rope of length 4m4 \, \text{m}. A horizontal force FF is applied at the midpoint of the rope so that the rope makes an angle of 4545^\circ with respect to the vertical axis as shown in the figure. The magnitude of FF is:

The figure shows: the rope hangs from a hatched ceiling. Its upper half runs from the ceiling attachment point down and to the left to the midpoint of the rope, making θ=45\theta = 45^\circ with the vertical, and the tension in this upper half is labelled T1T_1. At the midpoint the horizontal force F\vec{F} is applied, pointing to the left. The lower half of the rope hangs vertically from the midpoint down to the 1kg1 \, \text{kg} block, and the tension in it is labelled T2T_2.

  • A

    102N\frac{10}{\sqrt{2}} \, \text{N}

  • B

    1N1 \, \text{N}

  • C

    1102N\frac{1}{10\sqrt{2}} \, \text{N}

  • D

    10N10 \, \text{N}

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