A) 0.23 m/s
B) 0.46 m/s
C) 0.69 m/s
D) 0.92 m/s
Correct Answer: A
Solution :
\[{{v}_{\min }}=\sqrt{2{{\left( \frac{Tg}{\rho } \right)}^{1/2}}}=1.414{{\left( \frac{Tg}{\rho } \right)}^{1/4}}\] \[=1.414\,{{\left( \frac{7.2\times {{10}^{-2}}\times 9.8}{{{10}^{3}}} \right)}^{1/4}}\] = 0.23 m/s
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