Question

The time of oscillation 't' of a small drop of a liquid under surface tension (s) depends upon the density d, the radius r, and the surface tension s. Prove dimensionally that  t ∝ √dr³/s.

Answer 1

The time of oscillation t depends upon (i) d, (ii) r, and (iii) s.

i.e., t ∝ d^a r^b s^c

or, t = k d^a r^b s^c ......(iii)

Where k is the constant of proportionality, (dimensionless).

Writing the dimensional formulae of the various physical quantities in equation (i):

Dimensional formula for 't' = [M⁰L⁰T¹]

Dimensional formula for 'd' = [ML^-3T⁰]

Dimensional formula for 'r' = [M⁰L¹T⁰]

Dimensional formula for 's' = [M¹L⁰T^-2]

Therefore, equation (i) becomes,

[M⁰L⁰T] = [M¹L^-3T⁰]^a [M⁰L¹T⁰]^b [M¹L⁰T^-2]a

= [M^a+c L^{-3a +b} T^-2c]

Using the principle of homogeneity:

a + c = 0 ........(ii)

-3a +b = 0 ....(iii)

-2c = 1 ........(iv)

Form equation (iv), c = -1/2

Form equation (ii) a = -c = -(-1/2) = 1/2

Form equation (iii) b = 3a =3 x 1/2 = 3/2

Substituting the values of a, b and c in equation (i) we get,