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JEE Main 2020 (7 Jan, Shift 2) · Equilibrium, Third Law & Frames

Equilibrium, Third Law & Frames question from JEE Main 2020 (7 Jan, Shift 2)

Q18JEE Main 2020 · 7 JAN · SHIFT 2Medium

Equilibrium, Third Law & Frames

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° with the vertical. Then F equal: (Take g = 10 ms–2 and the rope to be massless)

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Answer: (a)

  1. At the midpoint: Tcos⁡45∘=mgT\cos 45^\circ = mg and Tsin⁡45∘=FT\sin 45^\circ = F.
  2. F=mgtan⁡45∘=10×10×1F = mg\tan 45^\circ = 10 \times 10 \times 1
  3. So the answer is (a) 100 N.
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