A) \[\frac{3}{5}\]
B) \[\frac{2}{7}\]
C) \[\frac{2}{5}\]
D) \[\frac{3}{7}\]
Correct Answer: B
Solution :
[b] Total kinetic energy of a rolling ball K.E. = linear K.E. + rotational K.E. \[\text{K}\text{.E}\text{.}=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}\times \frac{2}{5}m{{r}^{2}}.\frac{{{v}^{2}}}{{{r}^{2}}}\] \[=\left( \frac{5+2}{10} \right)\,m{{v}^{2}}=\frac{7}{10}m{{v}^{2}}\] Fraction of total energy associated with rotational kinetic energy \[=\frac{\frac{1}{5}m{{v}^{2}}}{\frac{7}{10}m{{v}^{2}}}\] \[=\frac{1}{5}\times \frac{10}{7}=\frac{2}{7}\]
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