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Question
The energy of the free surface of a liquid drop is 5π times the surface tension of the liquid. Find the diameter of the drop in C.G.S. system.
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Solution
Given that E = 5πT
Surface energy E = T dA -------- (Equation 1)
dA = 4πr2 -------- (where r is the radius of the liquid drop)
Substituting in Equation 1, we get
E = T * 4πr2
5πT = T* 4πr2 -------- (since E = 5πT)
`thereforer^2=5/4`
`thereforer=sqrt5/2`
Diameter, `d=2r=2*sqrt5/2`
`therefored=sqrt5=2.23cm`
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