A) 15 : 14
B) \[\sqrt{14}:\sqrt{15}\]
C) 14 : 15
D) \[{{15}^{2}}:{{14}^{2}}\]
Correct Answer: B
Solution :
By the relation, \[t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}\left( 1+\frac{{{K}^{2}}}{{{R}^{2}}} \right)}\] We have \[\frac{{{t}_{s}}}{{{t}_{D}}}=\sqrt{\frac{{{\left( 1+\frac{{{K}^{2}}}{{{R}^{2}}} \right)}_{s}}}{{{\left( 1+\frac{{{K}^{2}}}{{{R}^{2}}} \right)}_{D}}}}\] \[=\sqrt{\frac{1+\frac{2}{5}}{1+\frac{1}{2}}}=\sqrt{\frac{\frac{7}{5}}{\frac{3}{2}}}\] \[=\sqrt{\frac{14}{15}}\] Hence \[{{t}_{s}}:{{t}_{D}}=\sqrt{14}:\sqrt{15}\]
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