A) \[\sqrt{2gh}\]
B) \[\sqrt{\frac{4}{3}gh}\]
C) \[\sqrt{\frac{3}{4}gh}\]
D) \[\sqrt{\frac{4g}{h}}\]
Correct Answer: B
Solution :
Let the sought speed be v. Now from conservation of mechanical energy, we get \[mgh=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}I{{\omega }^{2}}\] For pure rolling \[\omega =\frac{v}{R}.\] Putting this value of \[\omega \] in the above equation, we get \[v=\sqrt{\frac{2gh}{1+\frac{I}{M{{R}^{2}}}}}\] For cylinder, \[I=\frac{1}{2}M{{R}^{2}}\]\[\therefore \]\[v=\sqrt{\frac{2gh}{1+\frac{1}{2}}}=\sqrt{\frac{4gh}{3}}\] Hence, the correction option is (b).
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