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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).
Solution
Given: E = 2πT
To find: Diameter of drop (d)
Formula: E = TΔA
Calculation: From formula,
`Delta A=E/T`
`Delta A=(2piT)/T`
`Delta A=2pi`
we know ,`Delta A=4pir^2`
Substituting in equation (1)
`2pi=4pir^2`
`4r^2=2`
`r^2=2/4=1/2`
`r^2=0.5`
`r=sqrt(0.5)=0.71m`
`d=2r`
=2(0.71)
d=1.414 m
The diameter of the drop is 1.414 m.
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