A) 7 : 10
B) 2 : 5
C) 10 : 7
D) 2 : 7
Correct Answer: D
Solution :
For a solid sphere rolling down Translation \[KE=\frac{1}{2}M{{v}^{2}}\] Rotational \[KE=\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}\left[ \frac{2}{5}M{{R}^{2}} \right]{{\left[ \frac{v}{R} \right]}^{2}}\] \[\left[ \begin{align} & As\text{ }for\,a\text{ }sphere \\ & \,\,\,\,\,\,I=\frac{2}{5}M{{R}^{2}} \\ \end{align} \right]\] \[\therefore \] Total kinetic energy of rolling is \[\frac{1}{2}M{{v}^{2}}+\frac{1}{5}M{{v}^{2}}=\frac{7}{10}M{{v}^{2}}\] \[\therefore \] Ratio of rotational kinetic energy to total kinetic energy \[=\frac{\frac{1}{5}M{{v}^{2}}}{\frac{7}{10}M{{v}^{2}}}=\frac{2}{7}\]
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