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JEE Main 2019 (9 Jan, Shift 2) · Equilibrium, Third Law & Frames

Equilibrium, Third Law & Frames question from JEE Main 2019 (9 Jan, Shift 2)

Q19JEE Main 2019 · 9 JAN · SHIFT 2Medium

Equilibrium, Third Law & Frames

A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviates at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms−2{}^{-2})

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Answer: (d)

  1. Where the force is applied: Tcos⁡45∘=mgT\cos 45^\circ = mg and Tsin⁡45∘=FT\sin 45^\circ = F.
  2. F=mgtan⁡45∘=100 NF = mg\tan 45^\circ = 100\text{ N}
  3. So the answer is (d) 100 N.
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