A) 4 : 3 : 2
B) 14 : 15 : 20
C) 10 : 15 : 7
D) 2 : 3 : 4
Correct Answer: C
Solution :
\[\frac{1}{2}\left( m+\frac{I}{{{R}^{2}}} \right){{\text{v}}^{2}}=mgh\] if radius of gyration is k, then \[h=\frac{\left( 1+\frac{{{k}^{2}}}{{{R}^{2}}} \right){{\text{v}}^{2}}}{2g},\frac{{{k}_{ring}}}{{{R}_{ring}}}=1,\frac{{{k}_{solid\,cylinder}}}{{{R}_{solid\,cylinder}}}=\frac{1}{\sqrt{2}}\] \[\frac{{{k}_{solid\,sphere}}}{{{R}_{solid\,sphere}}}=\sqrt{\frac{2}{5}}\] \[{{h}_{1}}:{{h}_{2}}:{{h}_{3}}::(1+1):\left( 1+\frac{1}{2} \right):\left( 1+\frac{2}{3} \right)::20:15:14\] Therefor most appropriate option is [b] athough which in not in correct sequence
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