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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)
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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