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`

Concept: Surface Tension

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