A) 1 : 7
B) 1 : 2
C) 1 : 1
D) 5 : 7
Correct Answer: D
Solution :
\[\frac{{{k}_{T}}}{k}=\frac{\frac{1}{2}m{{u}^{2}}}{\frac{1}{2}m{{u}^{2}}+\frac{1}{2}I{{\omega }^{2}}}=\frac{\frac{1}{2}m{{u}^{2}}}{\frac{1}{2}m{{u}^{2}}+\frac{1}{2}\times m{{r}^{2}}\times \frac{{{u}^{2}}}{{{r}^{2}}}}\] \[\left[ \because \,i=\frac{1}{5}m{{r}^{2}} \right]\] \[\therefore \] \[\frac{{{k}_{T}}}{k}=\frac{1}{2}=\frac{5}{7}\]or \[\frac{{{k}_{T}}}{k}=\frac{5}{7}\] Hence, the correction option is [d].
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