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The total energy of free surface of a liquid drop is 2π times the surface tension of the liquid. What is the diameter of the drop? (Assume all terms in SI unit).

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Question

The total energy of free surface of a liquid drop is 2π times the surface tension of the liquid. What is the diameter of the drop? (Assume all terms in SI unit).

Sum
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Solution

Given: E = 2πT 

To find: Diameter of drop (d)

Formula: E = TΔA

`Delta A = E/T`

`Delta A = (2piT)/T`

`Delta A = 2pi`      ...(i)

we know, `Delta A=4pir^2`

Substituting in equation (i)

`2pi = 4pir^2`

`4r^2 = 2`

`r^2 = 2/4`

`r^2 = 1/2`

r2 = 0.5

`r = sqrt(0.5)`

= 0.71m

∴ `d = 2r`

= 2(0.71)

d = 1.42 m

The diameter of the drop is 1.42 m.

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2016-2017 (March)

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