A) \[6kg-{{m}^{2}}/s\]
B) \[8kg-{{m}^{2}}s\]
C) \[3kg-{{m}^{2}}s\]
D) \[2kg-{{m}^{2}}/s\]
Correct Answer: A
Solution :
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 \[=6kg-{{m}^{2}}/s\]You need to login to perform this action.
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