A) \[62.5T-s\]
B) \[70.4T-s\]
C) \[79.6T-s\]
D) \[60.5T-s\]
Correct Answer: A
Solution :
If \[I\] be the moment of inertia of the cylinder about its axis. then \[I=\frac{1}{2}M{{R}^{2}}=\frac{1}{2}\times 20\times {{(0.25)}^{2}}=0.625\text{kg-}{{\text{m}}^{\text{2}}}\] \[\therefore \]KE associated with the rotating cylinder is given by, \[KE=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\times 0.625\times {{(100)}^{2}}=3125J\] Also using the relation, \[L=\sqrt{2I(KE)}\] We have \[L=\sqrt{2\times 0.625\times 3125}\]\[L=62.5J-s\]You need to login to perform this action.
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