Question

Given,$8000$ Identical water drops are combined to form a big drop. Then the ratio of finals energy to the initial surface energy of all drops together
A.$1:10$
B.$1:15$
C.$1:20$
D.$1:25$

Answer 1

As we know that the water is made up of oxygen and hydrogen gases. It is prepared by mixing two gases in the presence of sufficient heat. The water has a simple molecular structure which contains one oxygen atom and two hydrogen atoms. The water is polar compound which is used as an excellent solvent and the particles present in the liquid are far apart.

Complete answer:
The ratio of $S:s$is not equal to$1:10$. Hence, option (A) is incorrect.
The ratio of final energy to the initial surface energy of all drops together is not equal to$1:15$. Hence, option (B) is incorrect.
Given n is equal to $8000$
The volume of the liquid increases when the number of little liquid drops combine together. Because, there is a formation of a big liquid drop. Let’s see the formula,
$\dfrac{4}{3}\pi {R^3} = n\left( {\dfrac{4}{3}\pi {r^3}} \right)$
Where, ${R^3} = 8000{r^3}$
Therefore, $R = 20r$
The surface energy of the big drop, $S = 4\pi {R^2}T$. Here, T is equal to surface temperature.
And the surface tension of small$8000$drop, $s = 8000\left( {4\pi {r^2}T} \right)$
Compare the ratio of surface energy of both big drop and small drop.
$Ratio,\dfrac{S}{s} = \dfrac{{4\pi {R^2}T}}{{8000\left( {4\pi {r^2}T} \right)}} = \dfrac{{{R^2}}}{{8000{r^2}}}$
By simplification, will get;
$Ratio,\dfrac{S}{s} = \dfrac{{{{\left( {20r} \right)}^2}}}{{8000{r^2}}} = \dfrac{{400{r^2}}}{{8000{r^2}}} = \dfrac{1}{{20}}$
Therefore, the ratio of$S:s = 1:20$. Hence, option (C) is correct.
The ratio of final energy to the initial surface energy of all drops together is not equal to$1:10$. Hence, option (D) is incorrect.

Hence, option (C) is correct.

Note:
We have to know that the surface energy can be calculated by multiplying the surface tension with increase in the surface area. The surface tension of a liquid is equal to the force applied in a unit length of the surface of a liquid and it will be perpendicular to that liquid surface. The surface energy always determines the amount of work which is required to be done in a unit area.