A) 10 J
B) 100 J
C) 500 J
D) 250 J
Correct Answer: D
Solution :
Moment of inertia of a ring about its diameter \[l=\frac{1}{2}m{{r}^{2}}\]and kinetic energy is given by \[{{E}_{k}}=\frac{1}{2}l{{\omega }^{2}}\] \[{{E}_{k}}=\frac{1}{4}m{{r}^{2}}{{\omega }^{2}}\] \[=\frac{1}{4}\times 10\times {{(0.5)}^{2}}\times {{(20)}^{2}}\] \[=250\,J\]
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