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Eight equal drops of water each of radius \(r=2 \mathrm{~mm}\) are falling through air with a terminal velocity of \(16 \mathrm{~cm} / \mathrm{s}\). The eight drops combine to form a big drop. The terminal velocity of big drop is: (a) \(16 \mathrm{~cm} / \mathrm{s}\) (b) \(32 \mathrm{~cm} / \mathrm{s}\) (c) \(64 \mathrm{~cm} / \mathrm{s}\) (d) none of these

Short Answer

Expert verified
The terminal velocity of the big drop is 64 cm/s.

Step by step solution

01

Understanding Initial Conditions

We are given that there are eight small drops of water, each with a radius of \( r = 2 \) mm, and a terminal velocity of \( 16 \) cm/s. We need to find the terminal velocity when these drops merge into one large drop.
02

Volume Relation of Drops

First, note that the volume of a single small drop is \( V = \frac{4}{3} \pi r^3 \). If eight droplets combine to form a larger drop, the total volume of the large drop will be \( 8V = \frac{4}{3} \pi (2)^3 \times 8 \).
03

Calculating Radius of Big Drop

Let the radius of the big drop be \( R \). The volume of the big drop is equal to the combined volume of the eight smaller drops: \[ \frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi (2)^3 \].
04

Solving for Big Drop Radius

Cancel the common factors and solve for \( R \): \( R^3 = 8 \times 2^3 = 8 \times 8 = 64 \) which gives \( R = 4 \) mm.
05

Relationship Between Velocity and Radius

For a sphere moving at terminal velocity through a fluid, the velocity is proportional to the square of its radius. This can be expressed as \( V \propto R^2 \).
06

Calculating Terminal Velocity of Big Drop

The new terminal velocity \( V' \) for the big drop is calculated using the ratio of the radii squared: \[ \frac{V'}{V} = \left(\frac{R}{r}\right)^2 = \left(\frac{4}{2}\right)^2 = 4 \].
07

Finding the Terminal Velocity

Thus, \( V' = 4 imes V = 4 \times 16 = 64 \) cm/s.

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is the branch of physics concerned with the behavior of fluids and their interactions with surrounding forces. It is crucial for understanding how objects like droplets move through the air. When we discuss terminal velocity in the context of fluid dynamics, we refer to the constant speed that a falling object reaches when the drag force from the air equals the gravitational force pulling it down. This balance of forces results in no net acceleration, hence a steady terminal velocity.
  • Air provides resistance, creating a drag force that opposes gravitational pull.
  • Terminal velocity is achieved when these forces balance each other out.
This understanding of fluid dynamics allows us to calculate how fast an object, like a drop of water, will fall through a fluid, leading to important applications in various fields ranging from meteorology to engineering.
Spherical Objects
Spherical objects, such as water droplets, have a simple, uniform shape that simplifies calculations considerably. In physics, many problems involving fluids can be solved more easily using spheres due to their symmetrical properties. When droplets merge in fluid dynamics problems, it’s essential to treat them as small spheres. This simplification allows us to focus on how their volumes add up and how their shapes influence their behavior when falling through a fluid.
  • Spheres provide a symmetrical shape, simplifying volume and surface area calculations.
  • The behavior of spheres in fluids is a core part of studying fluid dynamics.
By understanding spherical objects' behavior, we can predict how they will interact with forces such as gravity and drag, leading to a better grasp of their motion and final velocities.
Volume Calculation
Volume calculation is a key process in determining how merged droplets behave when they combine into larger ones. For a sphere, the volume is given by the formula: \[ V = \frac{4}{3} \pi r^3 \]where \( r \) is the radius of the sphere. To find the volume of multiple spheres merging into one, we sum their individual volumes. In our exercise, eight small droplets combined. The new spherical volume becomes the sum of the eight smaller droplet volumes:
  • If each droplet has a volume \( V = \frac{4}{3} \pi (2)^3 \), then total volume for eight drops is \( 8V \).
  • The volume of the resulting big droplet is also calculated using the same formula but involves solving for a new, larger radius.
Volume calculations are critical for understanding how changing the size of droplets affects their properties and behaviors in fluid dynamics scenarios.
Velocity-Radius Relationship
The velocity-radius relationship is crucial when assessing how merged droplets behave in terms of their terminal velocities. In spherical drops moving through a fluid, terminal velocity varies with the square of their radius. This concept is expressed as:\[ V' = \left(\frac{R}{r}\right)^2 \times V \]Where:
  • \( V' \) is the terminal velocity of the larger drop.
  • \( R \) is the radius of the bigger, merged droplet.
  • \( r \) is the radius of one original smaller droplet.
  • \( V \) is the terminal velocity of one small droplet.
In our example, when the droplets combine, increasing their radius, the terminal velocity calculation reflects a significant increase, showcasing how a seemingly simple change in size influences the physics of falling objects in a fluid.

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