The time period (T)of a small drop of liquid under surface tension depends on the density(d ), radius (r) and surface tension (S) .Prove that Tµ ( hint :S is force acting per unit length)

Question

Arun · Answer

Let expression is T =
Here, r is the radius of drop, S is the surface tension and d is the Density of liquid .
Dimension of T = [ T ]
dimension of r = [ L ]
dimension of S = [MT⁻²]
dimension of d = [ ML⁻³]

[ T ] = [L]^a [MT⁻²]^b [ML⁻³]^c
[T] = [L]^(a -3c) [M]^(b + c) [T]^(-2b)]
Compare both sides,
b = -1/2 , c = 1/2 and a = 3c = 3/2

Hence, expression is T = r^(3/2)S^(-1/2)d^(1/2)

T = sqrt (r³d / S)

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The time period (T)of a small drop of liquid under surface tension depends on the density(d ), radius (r) and surface tension (S) .Prove that Tµ ( hint :S is force acting per unit length)

The time period (T)of a small drop of liquid under surface tension depends on the density(d ), radius (r)
and surface tension (S) .Prove that Tµ ( hint :S is force acting per unit length)

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