Question 65: Properties of Matter

Question

A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become :

Accepted Answer

To answer this question, we need to understand the relationship between the surface area of the droplets and the surface energy involved.
Surface energy is directly proportional to the surface area of the liquid. The surface energy, $ E $, for a droplet is given by:
$ E = \gamma 	imes A $
Where:
$ E $ is the surface energy,
$ \gamma $ is the surface tension of the liquid, and
$ A $ is the surface area of the droplet.
When multiple droplets coalesce, they form a larger droplet with a certain volume. Since the volume is conserved, the volume of the large droplet will be equal to the sum of the volumes of the small droplets.
Let's denote:
$ r $ as the radius of a small droplet,
$ R $ as the radius of the large droplet,
$ V_{	ext{small}} $ as the volume of a small droplet, and
$ V_{	ext{large}} $ as the volume of the large droplet.
The volume of one small droplet is:
$ V_{	ext{small}} = \frac{4}{3}\pi r^3 $
The total volume of 1000 small droplets is:
$ 1000 	imes V_{	ext{small}} = 1000 	imes \frac{4}{3}\pi r^3 $
Since the volume is conserved, the volume of the large droplet formed by the coalescence of 1000 small droplets is:
$ V_{	ext{large}} = 1000 	imes \frac{4}{3}\pi r^3 $
Now, if $ R $ is the radius of the large droplet, then:
$ V_{	ext{large}} = \frac{4}{3}\pi R^3 $
Equating the volumes, we have:
$ \frac{4}{3}\pi R^3 = 1000 	imes \frac{4}{3}\pi r^3 $
$ R^3 = 1000 	imes r^3 $
$ R = 10r $
Now, let's look at the surface area. The surface area for a small droplet is $ A_{	ext{small}} = 4\pi r^2 $ and for a large droplet is $ A_{	ext{large}} = 4\pi R^2 $. Substitute $ R = 10r $:
$ A_{	ext{large}} = 4\pi (10r)^2 $
$ A_{	ext{large}} = 4\pi 	imes 100r^2 $
$ A_{	ext{large}} = 100 	imes 4\pi r^2 $
$ A_{	ext{large}} = 100 	imes A_{	ext{small}} $
So, the large droplet has 100 times the surface area of one small droplet.
The surface energy of 1000 small droplets would be $ 1000 	imes E_{	ext{small}} $ because each small droplet has an energy $ E_{	ext{small}} = \gamma 	imes A_{	ext{small}} $.
The surface energy of the big droplet is $ E_{	ext{large}} = \gamma 	imes A_{	ext{large}} $. But we have just shown that $ A_{	ext{large}} = 100 	imes A_{	ext{small}} $, so:
$ E_{	ext{large}} = \gamma 	imes 100 	imes A_{	ext{small}} $
This means the surface energy of the big droplet is 100 times the surface energy of one small droplet. Since there were 1000 small droplets originally, the surface energy of the big droplet is:
$ \frac{E_{	ext{large}}}{1000 	imes E_{	ext{small}}} = \frac{\gamma 	imes 100 	imes A_{	ext{small}}}{1000 	imes \gamma 	imes A_{	ext{small}}} = \frac{1}{10} $
Therefore, the correct answer is:
Option B: The surface energy will become $\frac{1}{10}$ th of the original.

Answer

$\frac{1}{100}$ th

Answer

$\frac{1}{10}$ th

Answer

100 times

Answer

10 times

Properties of Matter

A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become :

Choose the correct answer: