Pressure Calculation: Water In A Test Tube

by Rajiv Sharma 43 views

Hey everyone! Let's dive into a fascinating physics problem that involves calculating the pressure exerted by water at the bottom of a test tube. This is a classic example that combines concepts of fluid mechanics, geometry, and basic physics principles. So, grab your calculators and let's get started!

Problem Statement

We're given a scenario where water is added to a test tube, and we need to figure out the pressure it exerts at the bottom. Here's the specific information:

  • Volume of water added: 0.05 L
  • Radius of the test tube: 0.50 cm

Our goal is to determine the pressure generated at the bottom of the test tube due to this volume of water. To solve this, we'll need to understand the relationship between pressure, force, and area, as well as how the volume of a liquid column relates to its height.

Understanding the Key Concepts

Before we jump into the calculations, let's refresh some key physics concepts that are crucial for solving this problem.

Pressure

Pressure is defined as the force acting perpendicularly on a surface per unit area. Mathematically, it's expressed as:

Pressure (P) = Force (F) / Area (A)

The standard unit of pressure in the International System of Units (SI) is the Pascal (Pa), which is equivalent to one Newton per square meter (N/m²).

Hydrostatic Pressure

When dealing with fluids, especially liquids, the pressure exerted by the fluid at a certain depth is known as hydrostatic pressure. This pressure is due to the weight of the fluid column above that point. The hydrostatic pressure is given by:

P = ρgh

Where:

  • P is the hydrostatic pressure
  • ρ (rho) is the density of the fluid (for water, approximately 1000 kg/m³)
  • g is the acceleration due to gravity (approximately 9.81 m/s²)
  • h is the height of the fluid column

Volume of a Cylinder

Since our test tube is cylindrical, we need to know how to calculate the volume of a cylinder. The volume (V) of a cylinder is given by:

V = πr²h

Where:

  • π (pi) is a mathematical constant approximately equal to 3.14159
  • r is the radius of the cylinder's base
  • h is the height of the cylinder

Step-by-Step Solution

Now that we have the problem statement and the necessary physics concepts, let's break down the solution step by step.

Step 1: Convert Units

First, we need to ensure all our units are consistent. We'll convert the given values to SI units:

  • Volume of water: 0.05 L = 0.05 × 10⁻³ m³ = 5 × 10⁻⁵ m³
  • Radius of the test tube: 0.50 cm = 0.50 × 10⁻² m = 5 × 10⁻³ m

Step 2: Calculate the Height of the Water Column

We can use the formula for the volume of a cylinder to find the height (h) of the water column in the test tube. Rearranging the formula, we get:

h = V / (πr²)

Plugging in the values:

h = (5 × 10⁻⁵ m³) / (π × (5 × 10⁻³ m)²) ≈ (5 × 10⁻⁵ m³) / (3.14159 × 2.5 × 10⁻⁵ m²) ≈ 0.6366 m

So, the height of the water column in the test tube is approximately 0.6366 meters.

Step 3: Calculate the Hydrostatic Pressure

Now that we have the height of the water column, we can calculate the pressure at the bottom of the test tube using the hydrostatic pressure formula:

P = ρgh

Where:

  • ρ (density of water) ≈ 1000 kg/m³
  • g (acceleration due to gravity) ≈ 9.81 m/s²
  • h (height of the water column) ≈ 0.6366 m

Plugging in the values:

P = (1000 kg/m³) × (9.81 m/s²) × (0.6366 m) ≈ 6245.95 Pa

Therefore, the pressure at the bottom of the test tube due to the 0.05 L of water is approximately 6245.95 Pascals.

Alternative Approach: Force and Area

We can also calculate the pressure by first finding the force exerted by the water and then dividing it by the area of the test tube's base.

Step 1: Calculate the Weight of the Water

The weight of the water is the force it exerts due to gravity. We can calculate it using:

Weight (F) = mass (m) × g

First, we need to find the mass of the water. We know the density and volume, so:

Density (ρ) = mass (m) / Volume (V)

m = ρV = (1000 kg/m³) × (5 × 10⁻⁵ m³) = 0.05 kg

Now we can calculate the weight:

F = 0.05 kg × 9.81 m/s² ≈ 0.4905 N

Step 2: Calculate the Area of the Test Tube's Base

The base of the test tube is a circle, so its area is given by:

Area (A) = πr²

Where r = 5 × 10⁻³ m

A = π × (5 × 10⁻³ m)² ≈ 3.14159 × 2.5 × 10⁻⁵ m² ≈ 7.854 × 10⁻⁵ m²

Step 3: Calculate the Pressure

Now we can find the pressure by dividing the force by the area:

P = F / A = 0.4905 N / (7.854 × 10⁻⁵ m²) ≈ 6245.95 Pa

As you can see, we arrive at the same result using both methods.

Conclusion

In this article, we've successfully calculated the pressure exerted by 0.05 L of water at the bottom of a test tube with a radius of 0.50 cm. We found that the pressure is approximately 6245.95 Pascals. We achieved this by understanding the concepts of pressure, hydrostatic pressure, and the volume of a cylinder. We also demonstrated an alternative approach using the relationship between force and area.

This problem illustrates how fundamental physics principles can be applied to solve practical problems. Understanding these concepts is crucial for anyone studying physics or engineering. Keep exploring, keep learning, and keep applying your knowledge to the world around you!

I hope this explanation was helpful and clear. If you have any more physics questions, feel free to ask. Keep exploring the fascinating world of physics, guys! There's always something new to discover and learn.