A) 2s
B) 3 s
C) 1.5 s
D) 4 s
Correct Answer: B
Solution :
Retardation, \[a=\frac{g\,\sin \theta }{1+\frac{I}{m{{R}^{2}}}}\] Let the cylinder be solid, then \[I=\frac{1}{2}m{{R}^{2}}\] \[a=\frac{-g\sin \theta }{1+\frac{1}{2}}=\frac{-2}{3}\times 9.8\times \frac{1}{2}\Rightarrow a=\frac{-9.8}{3}m/{{s}^{2}}\] Using the relation, \[{{v}^{2}}-{{u}^{2}}=2as\], we get \[s=\frac{{{v}^{2}}-{{u}^{2}}}{2a}\Rightarrow s=\frac{0-{{5}^{2}}}{2\left( -\frac{9.8}{3} \right)}\Rightarrow s=3.83\,m\] Let T be the time taken by the cylinder to return to the bottom, T = 2t, where, t = time of ascending or descending. Here, initial velocity =0 Using the relation, \[s=ut+\frac{1}{2}a{{t}^{2}}\], we get \[t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times 3.83}{\left( \frac{9.8}{3} \right)}}=1.53\,\,s\] \[T=2\times 1.53=3.06\,\,s\approx 3.0\,s\]
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