Question

Consider a \(2-\mathrm{m}\) deep swimming pool. The pressure difference between the top and bottom of the pool is \((a) 12.0 \mathrm{kPa}\) \((b) 19.6 \mathrm{kPa}\) \((c) 38.1 \mathrm{kPa}\) \((d) 50.8 \mathrm{kPa}\) \((e) 200 \mathrm{kPa}\)

Short answer

Answer: (b) 19.6 kPa

Step-by-step solution

Identify the Known and Unknown Quantities

Here, the known quantities are: 1. Depth (h) = 2 meters 2. Density of water (ρ) = 1000 kg/m³ 3. Acceleration due to gravity (g) = 9.81 m/s² The unknown is the pressure difference between the top and bottom of the swimming pool.

Calculate the Pressure at the Surface

The pressure at the surface (\(P_0\)) is equal to the atmospheric pressure. Thus we will assume \(P_0 = 101325\,\text{Pa}\) (approximately equal to atmospheric pressure at sea level).

Calculate the Pressure at the Bottom

We'll use the formula for pressure at a depth in a fluid, \(P = P_0 + \rho gh\), to calculate the pressure at the bottom of the pool. \(P = 101325\,\text{Pa} + (1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(2\,\text{m})\) \(P = 101325\,\text{Pa} + 19620\,\text{Pa}\) \(P = 120945\,\text{Pa}\)

Calculate the Pressure Difference

Now, we will find the pressure difference between the top and bottom of the pool by subtracting the pressure at the surface (\(P_0\)) from the pressure at the bottom (P). Pressure difference (ΔP) = \(P - P_0\) ΔP = \((120945 - 101325)\,\text{Pa}\) ΔP = \(19620\,\text{Pa}\) ΔP = \(19.6\,\mathrm{kPa}\)

Compare to Given Options

Comparing our calculated pressure difference to the given options: (a) 12.0 kPa (b) 19.6 kPa (c) 38.1 kPa (d) 50.8 kPa (e) 200 kPa Our calculated pressure difference of \(19.6\,\mathrm{kPa}\) matches option (b), so the correct answer is \((b) 19.6 \mathrm{kPa}\).

Key concepts

Fluid Mechanics

When diving into the studies of fluid mechanics, you're exploring the behavior of fluids (liquids and gases) and the forces on them. It's a branch of physics that has significant applications ranging from hydraulic engineering to aeronautics. Understanding fluid mechanics is crucial for a wide variety of scientific and engineering disciplines.

In the context of our swimming pool example, fluid mechanics provides the foundation to calculate how the pressure varies with depth. The principles dictate that in a state of rest, a fluid's pressure increases with depth due to the weight of the fluid above. This helps explain why the pressure at the bottom of the pool is greater than at the top.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at any point of time due to the gravitational pull of the Earth. This concept is a cornerstone in calculations involving fluids at rest. According to the hydrostatic pressure formula, \(P = P_0 + \rho gh\), where \(P\) is the pressure at a certain depth, \(P_0\) is the surface pressure, \(\rho\) is the fluid density, \(g\) is the acceleration due to gravity, and \(h\) is the depth below the surface.

For a swimming pool, the hydrostatic pressure at the bottom includes the atmospheric pressure acting on the surface plus the weight of the water above. In practice, this means the deeper you dive, the greater the pressure, which is felt by swimmers as an increasing force on their eardrums.

Fluid Density

Fluid density, represented by the symbol \(\rho\), is a measure of how much mass a fluid contains within a unit volume. It is typically expressed in kilograms per cubic meter (kg/m³) for the metric system. The density of water, a common fluid in hydrostatic pressure problems, is approximately \(1000\, \text{kg/m}^3\) at standard room temperature and pressure.

Density is a key variable in calculating hydrostatic pressure because the weight of the fluid, contributing to the pressure, depends on it. A fluid with higher density will exert more pressure at the same depth compared to a fluid with lower density. Thus, understanding fluid density is essential for accurate pressure calculations in any fluid mechanics problem, like the one involving our 2-meter deep swimming pool.