A) 2gh
B) \[\frac{10}{7}gh\]
C) \[\ge \,\,\sqrt{2gh}\]
D) \[\ge \,\,\sqrt{\frac{10}{7}gh}\]
Correct Answer: D
Solution :
To climb the inclined surface upto height h \[{{\left( kinetic energy \right)}_{transition}}+ {{\left( kinetic energy \right)}_{rotation}}\] \[\ge \,\,potential\text{ }energy\] \[\frac{1}{2}m{{v}^{2}}+\frac{1}{2}l{{\omega }^{2}}\ge mgh\] \[\frac{1}{2}m{{v}^{2}}+\frac{1}{2}\,\left( \frac{2}{5}m{{r}^{2}} \right)\,{{\omega }^{2}}\ge mgh\] \[\frac{7}{10}m{{v}^{2}}\ge mhg\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,v\ge \sqrt{\frac{10\,gh}{7}}\]
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