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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).
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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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