Question

The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum $ (\text{in kg}\,{{m}^{2}}{{s}^{-1}}) $ is approximately:

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

The Correct Option is C

The correct answer is C:4
Kinetic energy of rotating body \(K=\frac{1}{2}I{{\omega }^{2}}=62.8\,J\) 
Angular momentum, \(L=I\omega\) \(=\left( \frac{1}{2}I{{\omega }^{2}} \right)\times \frac{2}{\omega }\) 
\(=\frac{62.8\times 2}{(2\times \pi \times 300/60)}\) \(=4\,kg\,{{m}^{2}}{{s}^{-1}}\)