A particle of mass 20g is released with an initial velocity 5m/s 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=10m/s²)

Work-Energy Theorem from A to B
mgh = 1/2mv²B - 1/2mv²A
Let h = 2m, vA = 5m/s, m = 20g = 0.02kg
0.02kg * 10m/s² * 2m = 1/2 * 0.02kg * v²B - 1/2 * 0.02kg * (5m/s)²
0.4J = 0.01kg * v²B - 0.25J
0.65J = 0.01kg * v²B
v²B = 65 m²/s²
vB = √65 m/s
Angular momentum about O = m * vB * r
= 0.02kg * √65 m/s * 3m
= 0.06kg * m²/s * √65
≈ 0.06 * 8.06 = 0.48kg m²/s
The closest option is 6kg m²/s.