Problem 27 - University Physics with Modern Physics

Question

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 11.20 N. Find the total volume and the density of the sample.

Step-by-Step Solution

Answer

The volume is approximately \(6.42 \times 10^{-4} \, \text{m}^3\) and the density is approximately \(2779 \, \text{kg/m}^3\).

Step 1: Understand the Problem

We are given that the ore sample weighs 17.50 N in air and that the tension in the cord when the sample is immersed in water is 11.20 N. We need to find the volume and density of the ore sample.

Step 2: Calculate Buoyant Force

The buoyant force can be found using the difference in the weight of the sample in air and the tension in the cord when immersed in water. The formula is:\[ F_b = W_{air} - T_{water} \]Substitute the given values:\[ F_b = 17.50 \, \text{N} - 11.20 \, \text{N} = 6.30 \, \text{N} \]

Step 3: Use Archimedes' Principle to Find Volume

According to Archimedes' principle, the buoyant force is equal to the weight of the water displaced by the sample. The weight of the displaced water can be expressed as:\[ F_b = \rho_{water} \cdot V \cdot g \]Assuming the density of water, \(\rho_{water} = 1000 \, \text{kg/m}^3\) and \(g = 9.81 \, \text{m/s}^2\), we can solve for the volume \(V\):\[ V = \frac{F_b}{\rho_{water} \cdot g} = \frac{6.30}{1000 \cdot 9.81} \, \text{m}^3 \]\[ V \approx 6.42 \times 10^{-4} \, \text{m}^3 \]

Step 4: Calculate Density of the Sample

The density of the sample can be calculated using its mass and the volume found in the previous step. First, find the mass:\[ m = \frac{W_{air}}{g} = \frac{17.50}{9.81} \, \text{kg} \]\[ m \approx 1.784 \text{ kg} \]Now calculate the density \(\rho\) using the formula:\[ \rho = \frac{m}{V} = \frac{1.784}{6.42 \times 10^{-4}} \, \text{kg/m}^3 \]\[ \rho \approx 2779 \, \text{kg/m}^3 \]

Key Concepts

Understanding Buoyant Force

Buoyant force is a core concept in physics related to floating and sinking objects in fluids. Archimedes' Principle helps us understand this phenomenon. It states that any object submerged in a fluid experiences an upward, or buoyant, force equal to the weight of the fluid displaced by the object.
In practical terms, like in the example of the ore sample, you can calculate the buoyant force by subtracting the tension in the cord when the sample is immersed in water from its weight in air:

• Weight in air: 17.50 N
• Tension when immersed: 11.20 N

So, the buoyant force is found using the formula:\[ F_b = W_{air} - T_{water} = 17.50 \, \text{N} - 11.20 \, \text{N} = 6.30 \, \text{N} \]This buoyant force helps determine other factors like the volume of the displaced fluid, assisting in calculations about the object's density.

Mastering Density Calculation

Think of density as how much stuff is packed into a given space. It’s pivotal in identifying material properties. Mathematically, density \(\rho\) is the object's mass divided by its volume:\[ \rho = \frac{m}{V} \]
In the case of the ore sample, using the given gravitational force, the object's mass \(m\) is found as follows:\[ m = \frac{W_{air}}{g} = \frac{17.50}{9.81} \, \text{kg} \approx 1.784 \, \text{kg} \]
After the volume is determined through buoyant force calculations, the density is calculated:\[ \rho = \frac{1.784}{6.42 \times 10^{-4}} \, \text{kg/m}^3 \approx 2779 \, \text{kg/m}^3 \]
This value shows us how dense the material of our ore sample is compared to other substances.

Exploring Volume Calculation

The volume of an object measures the space it occupies. In our exercise, we used Archimedes' Principle to find the volume of water displaced when the ore sample is submerged. This displaced volume is equal to the volume of the ore itself.
The formula to derive the volume \(V\) using the buoyant force \(F_b\) is:\[ V = \frac{F_b}{\rho_{water} \cdot g} \]
Where:

• \(\rho_{water}\) is the density of water, typically \(1000 \, \text{kg/m}^3\)
• \(g\) is the acceleration due to gravity, \(9.81 \, \text{m/s}^2\)
• \(F_b\) is the buoyant force, \(6.30 \, \text{N}\)

Let's put these values into action:\[ V = \frac{6.30}{1000 \times 9.81} \, \text{m}^3 \approx 6.42 \times 10^{-4} \, \text{m}^3 \]
This calculation provides a precise measurement of the sample's volume, crucial for subsequent density calculations.