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JEE Main 2002 (AIEEE) · Equilibrium, Third Law & Frames

Equilibrium, Third Law & Frames question from JEE Main 2002 (AIEEE)

Q24JEE Main 2002 · AIEEEEasy

Equilibrium, Third Law & Frames

When forces F1, F2, F3F_1,\ F_2,\ F_3 are acting on a particle of mass mm such that F2F_2 and F3F_3 are mutually perpendicular, then the particle remains stationary. If the force F1F_1 is now removed then the acceleration of the particle is

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Answer: (a)

  1. In equilibrium, F⃗2+F⃗3=−F⃗1\vec{F}_2 + \vec{F}_3 = -\vec{F}_1.
  2. After removing F1F_1, the net force is ∣F⃗2+F⃗3∣=F1|\vec{F}_2 + \vec{F}_3| = F_1.
  3. a=F1ma = \frac{F_1}{m}
  4. So the answer is (a) F1/mF_1/m.
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