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JEE Main 2021 (25 Jul, Shift 1) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2021 (25 Jul, Shift 1)

Q49JEE Main 2021 · 25 JUL · SHIFT 1Medium

Angular Momentum & its Conservation

A particle of mass 'm' is moving in time 't' on a trajectory given by

r⃗=10αt2i^+5β(t−5)j^\vec{r} = 10\alpha t^{2}\hat{i} + 5\beta(t - 5)\hat{j}

Where α\alpha and β\beta are dimensional constants.

The angular momentum of the particle becomes the same as it was for t = 0 at time t = ____________ seconds.

Numerical answer type. Work it out, then open the answer.

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Answer: 10

  1. v⃗=20αti^+5βj^\vec v=20\alpha t\hat i+5\beta\hat j.
  2. Lz=m(xvy−yvx)=m[10αt2(5β)−5β(t−5)(20αt)]L_z=m(xv_y-yv_x)=m[10\alpha t^2(5\beta)-5\beta(t-5)(20\alpha t)]
  3. Lz=mαβ(500t−50t2)=50mαβ t(10−t)L_z=m\alpha\beta(500t-50t^2)=50m\alpha\beta\,t(10-t)
  4. L=0L=0 at t=0t=0, and again at t=10t=10 s.
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