An open container is filled with an ideal fluid. If a small hole is poked in the side of the container at a depth $ h $ below the surface, the speed of the fluid exiting the hole is $ v $. If the container is then pressurized so that the air above the fluid is at $ 2 P_{\text{atm}} $ while the outside remains at $ P_{\text{atm}} $, how does the new exit speed $ v’ $ compare to $ v $?

Question

An open container is filled with an ideal fluid. If a small hole is poked in the side of the container at a depth $ h $ below the surface, the speed of the fluid exiting the hole is $ v $. If the container is then pressurized so that the air above the fluid is at $ 2 P_{\text{atm}} $ while the outside remains at $ P_{\text{atm}} $, how does the new exit speed $ v&#8217; $ compare to $ v $?(a) $$ v&#8217; = v + \sqrt{ \dfrac{ 2 P_{\text{atm}} }{ \rho } } $$
, (b) $$ v&#8217; = v \sqrt{2} $$
, (c) $$ v&#8217; = 2 v $$
, (d) $$ v&#8217; = \sqrt{ v^{2} + \dfrac{ 2 P_{\text{atm}} }{ \rho } } $$
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