Answer:
When the cylinder slides without rolling, \[E=\frac{1}{2}m\upsilon {{'}^{2}}\] \[\therefore \] \[\upsilon '=\sqrt{\frac{2E}{m}}\] When the cylinder rolls without slipping, \[E=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}\left( \frac{1}{2}m{{r}^{2}} \right){{\omega }^{2}}\] \[=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{4}m{{\upsilon }^{2}}=\frac{3}{4}m{{\upsilon }^{2}}\] \[\therefore \] \[\upsilon =\sqrt{\frac{4E}{3m}}\] As \[\upsilon '>\upsilon ,\]therefore sliding cylinder will win the race.
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