A) \[2\sqrt{\frac{g}{5(R-r)}}\]
B) \[2\sqrt{\frac{g}{7(R-r)}}\]
C) \[\sqrt{\frac{2g}{5(R-r)}}\]
D) \[\sqrt{\frac{5g}{2(R-r)}}\]
Correct Answer: B
Solution :
K.E. of ball in position \[\operatorname{B} =mg\left( R-r \right)\] Here\[\operatorname{w}= mass of ball\]. Since it rolls without slipping the ratio of rotational to translational kinetic energy will be 2/5. \[\frac{{{K}_{R}}}{{{K}_{T}}}=\frac{2}{5}\] \[{{K}_{T}}=\frac{2}{7}mg(R-r)\] \[\frac{1}{2}m{{\nu }^{2}}=\frac{2}{7}mg(R-r)\] \[\nu =\frac{2}{\sqrt{7}}\sqrt{g(R-r)}\] \[\omega =\frac{\nu }{R-r}=2\sqrt{\frac{g}{7(R-r)}}\]
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