Question

The time period of oscillation t of a small drop of liquid under surface tension depends upon the density , surface tension s of the liquid and the radius r of the drop. Find the relation for the time period using dimensional analysis.

Answer 1

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)