---
title: "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 \\)?"
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url: "https://nerd-notes.com/ubq/107123/"
date_modified: "2026-08-18T02:38:28+00:00"
---

# 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 \( v \)?

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 \( v \)?

- **A.** \[ v’ = v + \sqrt{ \dfrac{ 2 P_{\text{atm}} }{ \rho } } \]
- **B.** \[ v’ = v \sqrt{2} \]
- **C.** \[ v’ = 2 v \]
- **D.** \[ v’ = \sqrt{ v^{2} + \dfrac{ 2 P_{\text{atm}} }{ \rho } } \]

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/107123/*