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JEE Main 2016 (9 Apr, Shift 1) · Terminal Velocity

Terminal Velocity question from JEE Main 2016 (9 Apr, Shift 1)

Q52JEE Main 2016 · 9 APR · SHIFT 1Easy

Terminal Velocity

Which of the following option correctly describes the variation of the speed v and acceleration ‘a’ of a point mass falling vertically in a viscous medium that applies a force F = − kv, where ‘k’ is a constant, on the body? (Graphs are schematic and not drawn to scale)

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Answer: (b)

tvag
  1. mdvdt=mg−kv⇒v=mgk(1−e−kt/m)m\dfrac{dv}{dt} = mg - kv \Rightarrow v = \dfrac{mg}{k}\left(1 - e^{-kt/m}\right)
  2. a=g e−kt/ma = g\,e^{-kt/m}: starts at gg and decays to zero.
  3. So vv rises from zero and saturates, while aa falls exponentially towards zero.
  4. So the answer is (b).
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