A) 2.5
B) 1.5
C) 1
D) 0.4
Correct Answer: A
Solution :
Translational kinetic energy, \[{{K}_{T}}=\frac{1}{2}m{{v}^{2}}\] Rotational kinetic energy\[{{V}_{1}}\] \[{{K}_{R}}=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\times \frac{2}{5}m{{R}^{2}}\times \frac{{{v}^{2}}}{{{R}^{2}}}=\frac{1}{5}m{{v}^{2}}\] \[\frac{{{K}_{T}}}{{{K}_{R}}}=\frac{\frac{1}{2}m{{v}^{2}}}{\frac{1}{5}m{{v}^{2}}}=\frac{5}{2}=25\]You need to login to perform this action.
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