By the method of dimensions, obtain an expression for the surface tension S of a liquidk rising in a capillary tube. Assume that S depends on mass m of liquied, Pressure p of liquid and radius r of the capillary tube. Take `K = 1//2`.

Explanation

Let `S = K m^a p^b r^c ….(i)`
where K is dimensionless constant of proportionality.
Writing the dimensions in (i),
`[M^1L^0T^(-2)] = M^a [ML^(-1)T^(-2)]^bL^c`
`=M^(a+b) L^(-b +c) t^(-2b)`
Applying principle of homogeneity of
dimensions, `a +b =1, - b+c = 0 or c =b`,
`-2b = - 2,b =1
a = 1 -b =1 -1 =0`
From (i), `S = K m^0 p^1 r^1 = (1)/(2) pr`