Question

The energy of the free surface of a liquid drop is $ 5\pi $ times the surface tension of the liquid. Find the diameter of the drop in the C.G.S. system.

Answer 1

Hint : To solve this question, we need to use the formula of the surface energy of a liquid in terms of its surface tension. From there we will get the diameter in the SI units. Finally we have to convert it into the C.G.S. unit to get the final answer.

Formula used: The formulae used for solving this question are given by
 $ E = SA $ , here $ E $ is the surface energy of a surface of a liquid of surface area equal to $ A $ , and which has a surface tension of $ S $ .
 $ A = 4\pi {R^2} $ , here $ A $ is the total surface area of a sphere of radius $ R $ .

Complete step by step answer:
Let $ S $ be the surface tension of the given liquid drop.
We know that the surface tension of a liquid is equal to its surface energy per unit area. So the surface energy becomes equal to the surface tension times the surface area of the liquid, that is,
 $ E = SA $ (1)
According to the question, the energy of the free surface of the liquid drop is $ 5\pi $ times the surface tension of the liquid. So this means that
 $ E = 5\pi S $ (2)
The shape of the drop of a liquid is approximately the same as that of a sphere. Now, we know that the surface area of a sphere is given by
 $ A = 4\pi {R^2} $ (3)
Substituting (2) and (3) in (1) we get
 $ 5\pi S = 4\pi {R^2}S $
Cancelling out $ S $ from both the sides, we get
 $ 5\pi = 4\pi {R^2} $
Dividing by $ \pi $ on both the sides, we get
 $ 4{R^2} = 5 $
 $ \Rightarrow {R^2} = \dfrac{5}{4} $
Taking square root both the sides, we get
 $ R = \dfrac{{\sqrt 5 }}{2}m $ (Assuming the given quantities to be in SI units)
The diameter is equal to twice the radius, that is
 $ D = 2R $
 $ \Rightarrow D = 2 \times \dfrac{{\sqrt 5 }}{2} = \sqrt 5 m $
In the C.G.S. system of units, the unit of length is equal to centimeter. Also we know that $ 1m = 100cm $ . So the diameter of the sphere in the C.G.S. system is given by
 $ D = 100\sqrt 5 cm $
 $ \Rightarrow D = 223.6cm $
Hence, the required value of the diameter of the drop is equal to $ 223.6cm $ .

Note:
The surface tension discussed in this question is responsible for many interesting phenomena. For example, a mosquito is able to sit on the surface of a liquid due to the surface tension offered as a force of reaction which balances its weight. Also, a needle can float on a liquid despite being denser due to the surface tension.