A) \[3{{R}_{0}}/2\]
B) \[3{{R}_{0}}/4\]
C) \[9{{R}_{0}}/16\]
D) \[3{{R}_{0}}\]
Correct Answer: B
Solution :
When body rolls dawn on inclined plane with velocity \[{{\text{V}}_{\text{0}}}\] at bottom then body has both rotational and translational kinetic energy Therefore, by law of conservation of energy, \[P.E.=K.{{E}_{trans}}+K.{{E}_{rotational}}\] \[\frac{1}{2}m{{V}_{0}}^{2}+\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}m{{V}_{0}}^{2}+\frac{1}{2}m{{k}^{2}}\frac{V_{0}^{2}}{R_{0}^{2}}\] ........(i) \[\left[ \because I=m{{k}^{2}},\omega =\frac{V}{{{R}_{0}}} \right]\] When body is sliding down then body has only translatory motion. \[\therefore \]\[P.E.=K.{{E}_{trans}}\] \[=\frac{1}{2}m{{\left( \frac{5}{4}{{v}_{0}} \right)}^{2}}\] ……..(ii) Dividing (i) by (ii) we get \[\frac{P.E.}{P.E.}=\frac{\frac{1}{2}mv_{0}^{2}\left[ 1+\frac{{{K}^{2}}}{R_{0}^{2}} \right]}{\frac{1}{2}\times \frac{25}{16}\times mV_{0}^{2}}\] \[=\frac{25}{16}=1+\frac{{{K}^{2}}}{R_{0}^{2}}\Rightarrow \frac{{{K}^{2}}}{R_{0}^{2}}=\frac{9}{16}\]or\[K=\frac{3}{4}{{R}_{0}}.\]
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