QJEEM24Ph7ee4: JEE Main 2024 Physics - Class 11 JEEM24Ph7ee4 Solution: A big drop is formed by coalescing 1000 small droplets of…

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:

$E = \gamma \times A$

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,
• $V_{\text{small}}$ as the volume of a small droplet, and
• $V_{\text{large}}$ as the volume of the large droplet.

The volume of one small droplet is:

$V_{\text{small}} = \frac{4}{3}\pi r^3$

The total volume of 1000 small droplets is:

$1000 \times V_{\text{small}} = 1000 \times \frac{4}{3}\pi r^3$

Since the volume is conserved, the volume of the large droplet formed by the coalescence of 1000 small droplets is:

$V_{\text{large}} = 1000 \times \frac{4}{3}\pi r^3$

Now, if $R$ is the radius of the large droplet, then:

$V_{\text{large}} = \frac{4}{3}\pi R^3$

Equating the volumes, we have:

$\frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3$

$R^3 = 1000 \times r^3$

$R = 10r$

Now, let's look at the surface area. The surface area for a small droplet is $A_{\text{small}} = 4\pi r^2$ and for a large droplet is $A_{\text{large}} = 4\pi R^2$ . Substitute $R = 10r$ :

$A_{\text{large}} = 4\pi (10r)^2$

$A_{\text{large}} = 4\pi \times 100r^2$

$A_{\text{large}} = 100 \times 4\pi r^2$

$A_{\text{large}} = 100 \times A_{\text{small}}$

So, the large droplet has 100 times the surface area of one small droplet.

The surface energy of 1000 small droplets would be $1000 \times E_{\text{small}}$ because each small droplet has an energy $E_{\text{small}} = \gamma \times A_{\text{small}}$ .

The surface energy of the big droplet is $E_{\text{large}} = \gamma \times A_{\text{large}}$ . But we have just shown that $A_{\text{large}} = 100 \times A_{\text{small}}$ , so:

$E_{\text{large}} = \gamma \times 100 \times A_{\text{small}}$

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:

$\frac{E_{\text{large}}}{1000 \times E_{\text{small}}} = \frac{\gamma \times 100 \times A_{\text{small}}}{1000 \times \gamma \times A_{\text{small}}} = \frac{1}{10}$

Therefore, the correct answer is:

Option B: The surface energy will become $\frac{1}{10}$ th of the original.