Question

The total energy of the free surface of a liquid drop is $2\pi$ times the surface tension of the liquid. What is the diameter of the drop? [Assume all terms in S.I units].

Answer 1

First of all, find the expression of total energy of a liquid drop which relates energy, surface area and surface tension. We have another relation of energy given in the question. Compare these two equations and manipulate it accordingly.

Complete step by step answer:
Given that the total energy of the free surface of a liquid drop is $2\pi$ times the surface tension of the liquid. So, we can write:
$E = 2\pi \times S$ …… (1)
Where,
$E$ indicates the total energy of the free surface.
$S$ indicates the surface tension of the liquid.
Again, we know the relation between total energy and the surface tension as:
$E = A \times S$ …… (2)
Where,
$E$ indicates the total energy of the free surface.
$S$ indicates the surface tension of the liquid.
$A$ indicates the total surface area of the drop.
Again, it's known that the shape of the liquid drop is spherical. Hence, its total surface area is given by the formula:
$A = 4\pi {r^2}$
Where,
$A$ indicates the total surface area of the drop, which is spherical in shape.
$r$ indicates the radius of the liquid drop.
Now, we compare the two equations numbered as (1) and (2), and we get:
$E = 2\pi \times S = A \times S$
Manipulating the above equation, we get:
$2\pi \times S = A \times S \\\ \Rightarrow 2\pi \times S = 4\pi {r^2} \times S \\\ \Rightarrow 2\pi = 4\pi {r^2} \\\ \Rightarrow{r^2} = \dfrac{1}{2} \\\$
Simplifying again to find the value of radius:
$r = \dfrac{1}{{\sqrt 2 }}$
Hence, the radius of the drop comes out to be $\dfrac{1}{{\sqrt 2 }}\,{\text{m}}$ .
We know, diameter $\left( D \right)$ is twice the radius, so we get:
$D = 2r \\\ \Rightarrow D= 2 \times \dfrac{1}{{\sqrt 2 }} \\\ \Rightarrow D= \dfrac{2}{{\sqrt 2 }} \\\ \therefore D= \sqrt 2 \,{\text{m}} \\\$
Hence, the diameter of the bubble is $\sqrt 2 \,{\text{m}}$.

Note: In the given problem, it is impossible to find the radius, from one equation itself. Hence, you need energy equations. One is already given in the question. Find another one. After finding the radius you need to find the diameter. It should be remembered that diameter is half the radius.