A) \[\frac{1}{2\sqrt{2}}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{\sqrt{2}}\]
Correct Answer: D
Solution :
\[{{\operatorname{v}}_{1}}=\sqrt{2g\times \frac{h}{2}}\] Since density is different, apply Bernoulli theorem for \[{{\operatorname{v}}_{2}}\] Since holes are open to atmosphere pressure \[={{P}_{0}}\] (atmosphere) \[={{P}_{0}}+\rho gh+2\rho \frac{h}{2}={{P}_{0}}+\frac{1}{2}2\rho {{\operatorname{v}}^{2}}_{2}\] \[{{\operatorname{v}}_{2}}=\sqrt{2gh}\] \[\frac{{{\operatorname{v}}_{1}}}{{{\operatorname{v}}_{2}}}=\frac{1}{\sqrt{2}}\]You need to login to perform this action.
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