| A particle of mass 20 g is released with an initial velocity 5 m/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\[g=10m/{{s}^{2}}\]) [JEE Main 12-Jan-2019 Evening] |
 |
A) \[6kg-{{m}^{2}}/s\]
B) \[8kg-{{m}^{2}}s\]
C) \[3kg-{{m}^{2}}s\]
D) \[2kg-{{m}^{2}}/s\]
Correct Answer: A
Solution :
| [a] Applying law of conservation of energy |
| \[\frac{1}{2}mv_{A}^{2}+mgh=\frac{1}{2}mv_{B}^{2}\] |
| \[\Rightarrow \]\[\frac{1}{2}(20\times {{10}^{-3}}){{(5)}^{2}}+(20\times {{10}^{-3}})(10)(10)\] |
| \[=\frac{1}{2}(20\times {{10}^{-3}})(V_{B}^{2})\]\[\Rightarrow \]\[{{v}_{B}}=15m/s\] |
So, the angular momentum of the particle about point O is  |
| \[=6kg-{{m}^{2}}/s\] |
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