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JEE Main 2021 (24 Feb, Shift 2) · Inertia, First & Second Laws

Inertia, First & Second Laws question from JEE Main 2021 (24 Feb, Shift 2)

Q25JEE Main 2021 · 24 FEB · SHIFT 2Difficult

Inertia, First & Second Laws

A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = −α-\alphax2. The distance at which the particle stops:

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Answer: (a)

  1. mvdvdx=−αx2⇒m∫v00v dv=−α∫0xx2dxmv\frac{dv}{dx} = -\alpha x^2 \Rightarrow m\int_{v_0}^{0} v\,dv = -\alpha\int_0^{x} x^2dx
  2. mv022=αx33\frac{mv_0^2}{2} = \frac{\alpha x^3}{3}
  3. x=[3mv022α]1/3x = \left[\frac{3mv_0^2}{2\alpha}\right]^{1/3}. So the answer is (a).
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