A particle of mass 20   g is released with an initial velocity 5   m   s - 1 along the curve…

A particle of mass $20 \mathrm{g}$ is released with an initial velocity $5 \mathrm{m} {\mathrm{s}}^{-1}$ 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 $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

1. $3\mathrm{ }\mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-1}$
2. $2\mathrm{ }\mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-1}$
3. $6\mathrm{ }\mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-1}$
4. $8\mathrm{ }\mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-1}$

Solution

Asked in: JEE Main 2019 (12 Jan Shift 2)

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