A) 0.763 J
B) 0.345 J
C) 1.5 J
D) Zero
Correct Answer: A
Solution :
For a thin ring \[KE=\frac{1}{2}I{{\omega }^{2}}\] Here, J = mr2 = 2.7 (0.08)2 = 0.0172 kg-m2\[\omega \] = 1.5 rev/s= 3 \[\pi \] rad/s After substituting the values, we get KE= 0.763 J
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