A) \[\sqrt{3gh}\]
B) \[\sqrt{4g}\]
C) \[\sqrt{\frac{3gh}{4}}\]
D) \[\sqrt{\frac{4gh}{3}}\]
Correct Answer: D
Solution :
From law of conservation of energy Translational kinetic energy + rotational kinetic energy = potential energy of fall \[\therefore \] \[\frac{1}{2}m{{v}^{2}}+\frac{1}{2}I{{\omega }^{2}}=mgh\] ... (i) where \[I\] is moment of inertia, co the angular velocity and v the linear velocity. For a solid cylinder \[I=\frac{1}{2}m{{r}^{2}}\] ... (ii) and \[\omega =\frac{v}{r}\] ... (iii) From Eqs. (i), (ii) and (iii), we get \[\frac{1}{2}m{{v}^{2}}+\frac{1}{4}m{{r}^{2}}\times \frac{{{v}^{2}}}{{{r}^{2}}}=mgh\] \[\Rightarrow \] \[v=\sqrt{\frac{4gh}{3}}\]
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