Q89 / 90 · JEE Main 2019 (12 Jan Shift 2) · Physics

A particle of mass 20 g is released with an initial velocity 5 m s-1 along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be: (Take g=10 m s-2)

Figure

  • A
    3 kg m2 s-1
  • B
    2 kg m2 s-1
  • Correct answer
    C
    6 kg m2 s-1
  • D
    8 kg m2 s-1

Solution

Applying conservation of energy,
Figure
mgh=12mvB2-12mvA2
⇒vB=2gh+vA2
⇒vB=2×10×10+25
⇒vB=15 m s-1
Angular momentum about O, 

LO=mvBh+a
⇒LO=20×10-3×15×20 kg m2 s-1
⇒LO=6 kg m2 s-1

Hence, the correct answer is option: 'C'.

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