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JEE Main 2023 (29 Jan, Shift 2) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2023 (29 Jan, Shift 2)

Q37JEE Main 2023 · 29 JAN · SHIFT 2Medium

Angular Momentum & its Conservation

A particle of mass 100 g is projected at time t = 0 with a speed 20 ms−1-1 at an angle 45∘^\circ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t = 2s is found to be K kg m2/s\sqrt{\mathrm{K}}\text{ kg m}^{2}\text{/s}. The value of K is ___________. (Take g = 10 ms−2-2)

Figure

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

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Answer: 800

  1. For a projectile, LL about the launch point =12mg uxt2=\frac12mg\,u_x t^2 (only gravity gives torque).
  2. ux=20cos⁡45∘=102u_x=20\cos45^\circ=10\sqrt2 m/s
  3. L=12×0.1×10×102×22=202=800L=\frac12\times0.1\times10\times10\sqrt2\times2^2=20\sqrt2=\sqrt{800}
  4. So K=800K=800.
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