A) \[\frac{2}{7}\]
B) \[\frac{3}{7}\]
C) \[\frac{4}{7}\]
D) \[\frac{5}{7}\]
Correct Answer: A
Solution :
When a body rolls down an inclined plane, it makes rotational as well as translational motions. Thus, it is associated with rotational and translational kinetic energies. Hence, total kinetic energy of sphere \[K={{K}_{rot.}}+{{K}_{trans}}\] \[=\frac{1}{2}I{{\omega }^{2}}+\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}\frac{2}{5}m{{r}^{2}}\left( \frac{{{v}^{2}}}{{{r}^{2}}} \right)+\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{5}m{{v}^{2}}+\frac{1}{2}m{{v}^{2}}=\frac{7}{10}m{{v}^{2}}\] Also, \[{{K}_{rot.}}=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{5}m{{v}^{2}}\] Hence, \[\frac{{{K}_{rot.}}}{K}=\frac{\frac{1}{5}m{{v}^{2}}}{\frac{7}{10}m{{v}^{2}}}=\frac{2}{7}\]
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