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JEE Main 2016 (10 Apr, Shift 1) · Inertia, First & Second Laws

Inertia, First & Second Laws question from JEE Main 2016 (10 Apr, Shift 1)

Q34JEE Main 2016 · 10 APR · SHIFT 1Difficult

Inertia, First & Second Laws

A particle of mass m is acted upon by a force F given by the empirical law F = Rt2v(t)\frac{R}{t^2}v(t) If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot:

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Answer: (d)

  1. mdvdt=Rt2v⇒dvv=Rmdtt2m\frac{dv}{dt} = \frac{R}{t^2}v \Rightarrow \frac{dv}{v} = \frac{R}{m}\frac{dt}{t^2}
  2. Integrating: ln⁡v=−Rm⋅1t+C\ln v = -\frac{R}{m}\cdot\frac{1}{t} + C
  3. So log⁡v\log v is linear in 1t\frac{1}{t}; a straight line confirms the law.
  4. So the answer is (d) log⁡v(t)\log v(t) against 1t\frac{1}{t}.
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