---
title: "A large open-top tank is filled with water to a height \\( H \\). A small hole is poked in the side at a height \\( h \\) from the bottom. If the tank is now sealed with a pressurized air pocket at the top such that the pressure at the surface is \\( 2P_{atm} \\) instead of \\( P_{atm} \\), how does the efflux speed \\( v \\) change?"
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url: "https://nerd-notes.com/ubq/107050/"
date_modified: "2026-08-18T02:38:33+00:00"
---

# A large open-top tank is filled with water to a height \( H \). A small hole is poked in the side at a height \( h \) from the bottom. If the tank is now sealed with a pressurized air pocket at the top such that the pressure at the surface is \( 2P_{atm} \) instead of \( P_{atm} \), how does the efflux speed \( v \) change?

A large open-top tank is filled with water to a height \( H \). A small hole is poked in the side at a height \( h \) from the bottom. If the tank is now sealed with a pressurized air pocket at the top such that the pressure at the surface is \( 2P_{atm} \) instead of \( P_{atm} \), how does the efflux speed \( v \) change?

- **A.** The speed \( v \) remains the same because it only depends on \( H – h \).
- **B.** The speed \( v \) decreases because the air pushes down on the water, increasing viscosity.
- **C.** The speed \( v \) increases by a factor of exactly \( \sqrt{2} \).
- **D.** The speed \( v \) increases.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/107050/*