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in JEE by kratos

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

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

i.e., t ∝ da rb sc

or, t = k da rb sc ......(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 '*' = [M1L0T-2]

Therefore, equation (i) becomes,

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

= [Ma+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,

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