A) 3/4
B) 1/3
C) 2/3
D) 1/2
Correct Answer: C
Solution :
[c] Let v be the linear velocity of centre of mass of the spherical body and w its angular velocity about centre of mass. Then \[\omega =\frac{v}{2R}\] KE of spherical body \[{{K}_{1}}=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}I{{\omega }^{2}}\] \[{{K}_{1}}=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}{{(2mR)}^{2}}\left( \frac{{{v}^{2}}}{4{{R}^{2}}} \right)=\frac{3}{4}m{{v}^{2}}\] ?. (i) Speed of the block will be \[v'=(\omega )(3R)=3R\omega =(3R)\left( \frac{v}{2R} \right)=\frac{3}{2}v\] \[\therefore \,\,\,KE\,of\,block\,{{K}_{2}}=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}m{{\left( \frac{3}{2}v \right)}^{2}}=\frac{9}{8}m{{v}^{2}}\] From equations (i) and (ii), \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{2}{3}\]
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