What happens to the final drop when ${10^6}$ small drops coalesce to make a larger drop?
A. Density increases
B. Density increases
C. Temperature increases
D. Temperature decreases

Hint: The given question is related to the property of surface tension in fluids. To solve it, find out whether the surface area decreases or increases when a huge number of drops coalesce to make a larger drop.

Complete answer:
Let the radius of ${10^6}$ small drops be $r$ and the radius of the one large drop be $R$ .
As it is given that these ${10^6}$ coalesce to make one large drop, the total initial and final volume will be the same.

Considering the drops to be spherical,
Initial volume $ = {10^6} \times \dfrac{4}{3}\pi {r^3}$ … (1)
Final volume $ = \dfrac{4}{3}\pi {R^3}$ … (2)

As both the initial and final volume are same,
${10^6} \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}$

On simplifying further, we get
$R = 100r$ … (3)

Calculating total surface area of the small drops,
Initial surface area $ = {10^6} \times 4\pi {r^2}$ … (4)

Calculating total surface area of the large drop,
Final surface area $ = 4\pi {R^2}$

Substituting the value of $R$ from equation (3), we get:
Final surface area $ = {10^4} \times 4\pi {r^2}$ … (5)

Comparing equation (4) and (5),
Initial surface area $ > $ Final surface area

From the above comparison, we can conclude that the surface area decreased, that is, the surface area of the final drop is less than the surface area of the initial small drops.

We know that when surface area decreases, energy is released. As energy is released, the temperature of the drop will increase. Thus, the correct option is C.

Note: When two or more drops merge to form a single drop, the surface area of the final drop is less than the surface area of the initial small drops. When surface area is reduced, energy is released. This liberation of energy ends up increasing the temperature.