A) 15 : 14
B) \[{{15}^{2}}:{{14}^{2}}\]
C) \[\sqrt{14} :\sqrt{15}\]
D) 14:15
Correct Answer: C
Solution :
\[t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}\left( 1+\frac{{{K}^{2}}}{{{R}^{2}}} \right)}\] \[\Rightarrow \] \[\frac{{{t}_{S}}}{{{t}_{D}}}=\sqrt{\frac{1+{{\left( \frac{{{K}^{2}}}{{{R}^{2}}} \right)}_{S}}}{1+{{\left( \frac{{{K}^{2}}}{{{R}^{2}}} \right)}_{D}}}}\] \[=\sqrt{\frac{1+\frac{2}{5}}{1+\frac{1}{2}}}=\sqrt{\frac{14}{15}}\]
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