Question

8000 identical water drops are combined to form a bigdrop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is (A) \(1: 10\) (B) \(1: 15\) (C) \(1: 20\) (D) \(1: 25\)

Short answer

The ratio of the final surface energy to the initial surface energy of all the drops together is 1:20, which corresponds to answer choice (C).

Step-by-step solution

Write the formula for the surface area of a sphere

We will use the formula for the surface area of a sphere to compare the surface energies. The surface area of a sphere with radius r is given by: \[A_s = 4\pi r^2\]

Find the initial surface area of the 8000 individual drops

Assuming each drop has a radius of r, the total surface area of the 8000 individual drops would be: \[A_{initial} = 8,000 \times 4\pi r^2\]

Find the volume of the big drop

Since the volumes of the drops are conserved when combined, we can find the volume of the big drop by multiplying the volume of a single drop by 8000. The volume of a sphere with radius r is given by: \[V = \frac{4}{3} \pi r^3\] So the volume of the big drop will be: \[V_{big} = 8,000 \times \frac{4}{3} \pi r^3\]

Find the radius of the big drop

To find the radius of the big drop, we can set the volume of the big drop equal to the volume of a sphere with radius R: \[8,000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3\] Solving for R, we get: \[R = \sqrt[3]{8,000 \times r^3}\]

Find the final surface area of the big drop

We can now use the radius R of the big drop to find its surface area: \[A_{final} = 4\pi R^2\] Substitute the expression for R from step 4: \[A_{final} = 4\pi (\sqrt[3]{8,000 \times r^3})^2\]

Find the ratio of the final surface energy to the initial surface energy

Now we can find the ratio of the surface energies by taking the ratio of the final surface area to the initial surface area: \[\frac{A_{final}}{A_{initial}} = \frac{4\pi (\sqrt[3]{8,000 \times r^3})^2}{8,000 \times 4\pi r^2}\] Simplify the expression, and we get: \[\frac{A_{final}}{A_{initial}} = \frac{(\sqrt[3]{8,000})^2}{8,000}\] Since \(\sqrt[3]{8,000} = 20\), the ratio becomes: \[\frac{A_{final}}{A_{initial}} = \frac{20^2}{8,000} = \frac{1}{20}\] So the ratio of the final surface energy to the initial surface energy of all the drops together is 1:20, which corresponds to answer choice (C).

Key concepts

Understanding the Surface Area of a Sphere

The surface area of a sphere is a key geometric concept crucial in solving this exercise. The formula to calculate it is \(A_s = 4\pi r^2\), where \(r\) represents the radius of the sphere. This means that the surface area is directly dependent on the square of the radius.
The more the radius increases, the larger the surface area will become. In the context of this exercise, we are looking at comparing the surface areas of many small spheres (water droplets) and one large sphere (the big drop formed by combining them). Understanding this principle helps establish how changes in radius affect the overall surface composition, making the computation of surface energy more comprehensible.

Exploring Volume Conservation

The concept of volume conservation signifies that the total volume remains constant throughout the process of recombining the droplets into a bigger drop. In mathematical terms, it implies that the sum of the volumes of all the small drops equals the volume of the large drop they form. For a single spherical droplet, the volume is given by \(V = \frac{4}{3} \pi r^3\).
In this problem, 8000 identical drops coalesce, and their combined volume is \(8000 \times \frac{4}{3} \pi r^3\). This entire volume transforms into one single larger sphere. Conservation of volume is a crucial physics concept as it assists in calculating the new dimensions of the larger sphere, ensuring that we don’t lose any liquid during the process. This aids in finding the new radius, which is essential for determining the new surface area in subsequent calculations.

Determining the Radius of a Sphere

Finding the radius is pivotal when moving from a multitude of smaller spheres to one larger sphere. The radius is essential in computing both surface area and volume. Once we understand that the volume is conserved, we can equate the total volume of small drops to that of the big drop, forming an equation \(8000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3\). Solving this gives us \(R = \sqrt[3]{8000 \times r^3}\).
Calculating this expression is crucial, as it helps find the new radius \(R\) of the larger sphere formed. This radius will further assist in comparing initial and final surface energies, showcasing how the radius transformation due to volume conservation plays a role.

Calculating the Ratio of Surface Energies

To determine the ratio of surface energies, one must find how the surface areas transition from initial droplets to the final big sphere. The final surface energy is proportional to \(A_{final} = 4\pi R^2\), while the initial surface energy is proportional to \(8,000 \times 4\pi r^2\).
The ratio \(\frac{A_{final}}{A_{initial}}\) reveals how much change occurs in surface area terms. Upon simplifying, the formula becomes \(\frac{(\sqrt[3]{8000})^2}{8000}\).
Recognizing that \(\sqrt[3]{8000} = 20\), this transforms the expression to \(\frac{20^2}{8000} = \frac{1}{20}\). Thus, demonstrating that the transformation of water drops into a larger drop results in a ratio of surface energies of 1:20, providing a deeper understanding of the geometric and physical implications through this ratio calculation.