A) 3.5 m
B) 7.0 m
C) 10.5 m
D) 14.0 m
Correct Answer: B
Solution :
\[mgh=\frac{1}{2}m{{v}^{2}}\left( 1+\frac{{{K}^{2}}}{{{R}^{2}}} \right)\] \[\Rightarrow \]\[\frac{7}{10}m{{v}^{2}}=mgh\] \[\left( \because \frac{{{K}^{2}}}{{{R}^{2}}}=\frac{2}{5} \right)\] \[\Rightarrow \] \[h=\frac{7}{10}\left( \frac{{{v}^{2}}}{g} \right)\] Given, \[v=10\,m\text{/}s,\,g=10\,m\text{/}{{s}^{2}}.\] \[\therefore \] \[h=\frac{7}{10}\times \frac{10\times 10}{10}\] \[\Rightarrow \] \[h=7\,m\]
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