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A copper sphere with a mass of 0.20 \(\mathrm{g}\) and a density of 8900 \(\mathrm{kg} / \mathrm{m}^{3}\) is observed to fall with a terminal speed of 6.0 \(\mathrm{cm} / \mathrm{s}\) in an unknown liquid. Find the viscosity of the unknown liquid if its buoyancy can be neglected.

Short Answer

Expert verified
The viscosity of the unknown liquid is 0.0106 Pa·s.

Step by step solution

01

Understand the Known Values

First, identify the values we know from the problem statement: the mass of the sphere is 0.20 g or 0.0002 kg, the density of copper is 8900 kg/m³, the terminal speed is 6.0 cm/s or 0.06 m/s, and buoyancy can be neglected.
02

Calculate the Volume of the Sphere

Use the formula for density, which is density = mass/volume, to find the volume. Rearrange it to find volume: \[\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{0.0002 \text{ kg}}{8900 \text{ kg/m}^3} = 2.24719 \times 10^{-8} \text{ m}^3\]
03

Determine the Radius of the Sphere

Since volume is also given by \( V = \frac{4}{3} \pi r^3 \), solve for the radius \( r \):\[\text{r} = \left(\frac{3V}{4\pi}\right)^{1/3} = \left(\frac{3 \times 2.24719 \times 10^{-8}}{4\pi}\right)^{1/3} = 0.00177 \text{ m}\]
04

Use Stokes' Law to Find Viscosity

Stokes' Law for an object moving through a viscous fluid is \( F_d = 6 \pi \eta r v_t \), where \( \eta \) is the viscosity, \( r \) is the radius, and \( v_t \) is the terminal velocity. The drag force \( F_d \) is equal to the gravitational force \( F_g = mg \) at terminal velocity.\[m \cdot g = 6 \pi \eta r v_t\]Solve for \( \eta \):\[\eta = \frac{m g}{6 \pi r v_t} = \frac{0.0002 \times 9.81}{6 \pi \times 0.00177 \times 0.06} = 0.0106 \text{ Pa~s}\]
05

Conclusion

The calculated viscosity of the liquid is 0.0106 Pa·s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stokes' Law
Stokes' Law is a fundamental principle in fluid dynamics that relates the drag force experienced by spherical objects moving through a viscous fluid. It is described by the formula:
\[ F_d = 6 \pi \eta r v_t \]
where:
  • \( F_d \) is the drag force,
  • \( \eta \) is the viscosity of the fluid,
  • \( r \) is the radius of the sphere,
  • \( v_t \) is the terminal velocity.
Understanding this equation helps us see how different factors affect the movement of a sphere through a liquid.
As the sphere reaches terminal velocity, the gravitational force is balanced by the drag force. This allows us to use Stokes' Law to solve for unknowns, like fluid viscosity, knowing the other variables.
Terminal Velocity
Terminal velocity occurs when a falling object reaches such a velocity that the downward gravitational force is equaled by the upward drag force created by the fluid through which it is moving. At this point, the object stops accelerating and continues to move at a constant speed.
For a small copper sphere in a liquid, reaching terminal velocity means:
  • The net force acting on the sphere is zero, allowing us to simplify calculations.
  • The balancing of forces provides an opportunity to determine the fluid's viscosity, by applying equations like Stokes' Law.
Recognizing terminal velocity in real-world applications helps us predict the behavior of particles in fluids, key for fields like aerodynamics and engineering.
Copper Sphere
In the context of this problem, a copper sphere is an ideal choice because its properties are quite well known and consistent. It has a density of 8900 kg/m³, which is essential for calculating volume when mass is given:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Since the volume of a sphere is determined by the formula:\[ V = \frac{4}{3} \pi r^3 \]
we can also solve for the radius given either the volume or the mass and density. The mass of the copper sphere here is 0.20 grams, which converts to 0.0002 kilograms for use in calculations.
The uniformity and reality-based properties of spheres make it straightforward to apply theoretical concepts to practical scenarios in physics and engineering.
Fluid Dynamics
Fluid dynamics is the study of fluids (liquids and gases) in motion. It encompasses various forces and the effects these forces have on fluids' behavior. In this particular problem, fluid dynamics comes into play as we assess how the copper sphere moves through a liquid.
Neglecting buoyancy simplifies the calculation, allowing us to focus on the liquid's viscosity as the primary resistance to the sphere's motion through it.
  • The sphere displaces a volume of liquid that equals its own volume.
  • The interaction between the sphere and the fluid involves complex equations, but for small and steady velocities, Stokes' Law provides an accurate approximation.
Fluid dynamics is essential in many scientific fields, from meteorology to civil engineering, as understanding flow characteristics is critical for design and analysis.
By employing principles from fluid dynamics, we can solve practical problems, like finding the viscosity of an unknown fluid.

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Most popular questions from this chapter

An electrical short cuts off all power to a submersible diving vehicle when it is 30 m below the surface of the ocean. The crew must push out a hatch of area 0.75 \(\mathrm{m}^{2}\) and weight 300 \(\mathrm{N}\) on the bottom to escape. If the pressure inside is 1.0 atm, what downward force must the crew exert on the hatch to open it?

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