A) \[g/\sqrt{3}\]
B) \[g/\sqrt{3}\]
C) \[\sqrt{\frac{2g}{3}}\]
D) None of these
Correct Answer: B
Solution :
For cylinder, \[I=\frac{1}{2}M{{R}^{2}}\] =1 MR2 So, \[\frac{{{K}^{2}}}{{{R}^{2}}}=\frac{1}{2}\] So. acceleration \[a=\frac{g\,\sin \theta }{1+\frac{{{K}^{2}}}{{{R}^{2}}}}=\frac{g\,\sin \,{{60}^{o}}}{1+\frac{1}{2}}\] \[=\frac{g}{\sqrt{3}}\]
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