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Question: The energy of the free surface of a liquid drop is \( 5\pi \) times the surface tension of the liqui...

The energy of the free surface of a liquid drop is 5π5\pi times the surface tension of the liquid. Find the diameter of the drop in the C.G.S. system.

Explanation

Solution

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=SAE = SA , here EE is the surface energy of a surface of a liquid of surface area equal to AA , and which has a surface tension of SS .
A=4πR2A = 4\pi {R^2} , here AA is the total surface area of a sphere of radius RR .

Complete step by step answer:
Let SS 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=SAE = SA (1)
According to the question, the energy of the free surface of the liquid drop is 5π5\pi times the surface tension of the liquid. So this means that
E=5πSE = 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=4πR2A = 4\pi {R^2} (3)
Substituting (2) and (3) in (1) we get
5πS=4πR2S5\pi S = 4\pi {R^2}S
Cancelling out SS from both the sides, we get
5π=4πR25\pi = 4\pi {R^2}
Dividing by π\pi on both the sides, we get
4R2=54{R^2} = 5
R2=54\Rightarrow {R^2} = \dfrac{5}{4}
Taking square root both the sides, we get
R=52mR = \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=2RD = 2R
D=2×52=5m\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=100cm1m = 100cm . So the diameter of the sphere in the C.G.S. system is given by
D=1005cmD = 100\sqrt 5 cm
D=223.6cm\Rightarrow D = 223.6cm
Hence, the required value of the diameter of the drop is equal to 223.6cm223.6cm .

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.