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NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice

Power of the only force acting on a particle of mass m = 1 kg moving in a straight line depends on its velocity as P = v 2 where v is in m s - 1 and P is in watt . If the initial velocity of the particle is 1 m s - 1 , then the displacement of the particle in ln ( 2 ) seconds will be

Options

  1. Aln 2 - 1 m
  2. Bln 2 2   m
  3. C1 m
  4. D2 m

Correct answer

C. 1 m

Step-by-step solution

P = Fv v 2 = Fv ( since P = v 2 ) therefore F = v ma = v 1 × v dv dx = v ∫ 1 v dv = ∫ 0 x dx v - 1 = x v = x + 1 dx dt = x + 1 ∫ 0 x dx x + 1 = ∫ 0 t dt ln x + 1 - ln 0 + 1 = t x + 1 = e t x = e t - 1 x = e t - 1 At t = ln 2 x = 2 - 1 = 1 m

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