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JEE Main 2024 (31 Jan, Shift 2) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2024 (31 Jan, Shift 2)

Q25JEE Main 2024 · 31 JAN · SHIFT 2Medium

Angular Momentum & its Conservation

A body of mass 'mm' is projected with a speed 'uu' making an angle of 45∘45^\circ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as 2mu3Xg\frac{\sqrt{2}mu^{3}}{Xg}. The value of 'XX' is _________.

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

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Answer: 8

  1. At the top: v=ucos⁡45∘=u2v=u\cos45^\circ=\frac{u}{\sqrt2}, H=u2sin⁡245∘2g=u24gH=\frac{u^2\sin^2 45^\circ}{2g}=\frac{u^2}{4g}.
  2. L=mvH=mu2⋅u24g=mu342g=2mu38gL=mvH=\frac{mu}{\sqrt2}\cdot\frac{u^2}{4g}=\frac{mu^3}{4\sqrt2g}=\frac{\sqrt2mu^3}{8g}
  3. So X=8X=8.
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