By the method of dimensions, obtain an expression for the surface tension S of a liquid 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`.

Answer

S = Kpr

Explanation

Let `S = Km^a p^b r^c …(i)`
Where a, b, c are the dimensions and K is
dimensionless constant. Writing the dimensions
on either side of (i), we get
`M^1 T^(-2) =M^a(ML^(-1) T^(-2))^b L^c`
`[M^1 L^0 T^(-2)] =M^(a+b) L^(-b+c) t^(-2b)`
Applying the principle of homogeneity of
dimensions, we get
a+b =1 , -b+c =0 or c =b, -2b =- 2,b=1
a =1 -b =1-1 =0
Putting these values is (i), we get
`S = K m^0 pr =Kpr`