Question

When a big drop of water is formed by combining $n$ small drops of water, the energy loss is $3E{\text{,}}$ where $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then find the number of smaller drops $(n)$ .
a. $\dfrac{{4R}}{{{r^3}}}$
b. $\dfrac{{4R}}{r}$
c. $\dfrac{{2{R^2}}}{r}$
d. $\dfrac{{4{R^2}}}{{{r^2}}}$

Answer 1

Hint: Assume the shape of the water drops to be spherical. The energy released in the formation of the bigger drop can be expressed in terms of its surface tension and the change in the surface area. The change in the surface area will be the difference between the surface area of the big drop and that of $n$ smaller drops.

Formulas used:
- Surface area $A$ of a sphere of radius $r$ is $A = n \times 4\pi {r^2}$
- The energy released in the formation of a drop is $E = T \times \Delta A$ , where $E$ is the energy of the formed drop, $T$ is its surface tension and $\Delta A$ is the change in its surface area

Complete step by step answer:
Step 1: List the data given in the question.
The energy of the bigger drop is $E$ .
The energy released on the formation of the bigger drop $ = 3E$ .
$n$ represents the number of smaller drops that combined to form the big drop of water.
$R$ and $r$ represent the radius of the big drop and one small drop respectively.
Let $T$ be the surface tension of the big drop.

Step 2: Express the relation for the change in the surface area involved in the formation of the bigger drop.
The water drops are assumed to be spherical in shape.
Let ${A_1}$ be the total surface area of $n$ smaller drops and ${A_2}$ be the surface area of the big drop.
Now, the total surface area of $n$ smaller drops $ = {A_1} = n \times 4\pi {r^2}$ where $4\pi {r^2}$ is the surface area of one small drop of radius $r$ .
The surface area of the big drop of water with a radius $R$ is given by, ${A_2} = 4\pi {R^2}$ where $R$ is the radius of the bigger drop.
The change in surface area is given by, $\Delta A = {A_1} - {A_2}$ (since, ${A_1} > {A_2}$ )
Substituting for ${A_1} = n \times 4\pi {r^2}$ and ${A_2} = 4\pi {R^2}$ , we get $\Delta A = (n \times 4\pi {r^2}) - (4\pi {R^2})$
On simplifying, $\Delta A = 4\pi (n{r^2} - {R^2})$

Step 3: Express the energy released in terms of surface tension $T$ and change in area $\Delta A$ .
It is given that the energy of the bigger drop is $E$ and the energy released in the formation of the bigger drop is $3E$ .
We can express $E$ and $3E$ in terms of surface tension and surface area
${\text{Energy of big drop}} = {\text{surface tension}} \times {\text{surface area of big drop}}$
i.e., $E = T \times {A_2} = T \times 4\pi {R^2}$
${\text{Energy released}} = {\text{surface tension}} \times {\text{change in area}}$
i.e., $3E = T \times \Delta A = T \times 4\pi (n{r^2} - {R^2})$

Step 4: Substitute for $E$ in the left-hand-side of the equation of $3E$
We have, $3E = T4\pi (n{r^2} - {R^2})$
On substituting, $3 \times T4\pi {R^2} = T4\pi (n{r^2} - {R^2})$
Cancel out the similar terms to get $3{R^2} = n{r^2} - {R^2}$
On simplifying, $n{r^2} = 4{R^2}$
Finally, $n = \dfrac{{4{R^2}}}{{{r^2}}}$

Hence, the correct answer is option (D).

Note: When $n$ small drops of water combine to form one big drop of water, energy is released or lost in the formation. This is because the process involves a decrease in the surface area of the drop.