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JEE Main 2024 (29 Jan, Shift 2) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2024 (29 Jan, Shift 2)

Q26JEE Main 2024 · 29 JAN · SHIFT 2Medium

Angular Momentum & its Conservation

A body of mass 5 kg5\text{ kg} moving with a uniform speed 32 ms−13\sqrt{2}\text{ ms}^{-1} in X−YX - Y plane along the line y=x+4y = x + 4. The angular momentum of the particle about the origin will be _________ kg m2 s−1\text{kg m}^{2}\ \text{s}^{-1}.

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

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Answer: 60

  1. L=mv dL=mv\,d, with dd = perpendicular distance of line x−y+4=0x-y+4=0 from origin =42=22=\frac{4}{\sqrt2}=2\sqrt2 m.
  2. L=5×32×22L=5\times3\sqrt2\times2\sqrt2
  3. L=60L=60 kg m² s⁻¹
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