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Obtain by dimensional analysis an expression for the surface tension of a liquid rising in a capillary tube. Assume that the surface tension [tex]T[/tex] depends on mass [tex]m[/tex] of the liquid, pressure [tex]P[/tex] of the liquid and radius [tex]r[/tex] of the capillary tube (Take the constant [tex]k = \frac{1}{2}[/tex]).

Answer :

so let
 [tex]T [tex] \alpha [/tex]  M^{a}P^{b}r^{c} [/tex]
[tex]T =k  M^{a}P^{b}r^{c} [/tex]
we know that dimensions of
 [tex]T = M^{1}L^{0}T^{-2} [/tex]
[tex]P (pressure)= M^{1}L^{-1}T^{-2}[/tex]
[tex]M(mass) = M^{1}[/tex]
[tex]R(radius)= L^{1}[/tex]
where l = length,m = mass t = time sice k is constant and have no dimension
so  [tex] M^{1}L^{0}T^{-2} [/tex] ={[tex] M^{1}[/tex]}^a {[tex] M^{1}L^{-1}T^{-2}[/tex]}^b {[tex] L^{1}[/tex]}^c
⇒ [tex] M^{1}L^{0}T^{-2} [/tex]= [tex]M^{a +b} [/tex] [tex]L^{c - b}[/tex] [tex]T^{-2b}[/tex]
using priciple dimensional homogeinity
we get
a + b = 1 ⇒a = 1 - b
c - b = 0 ⇒c= b
-2b = -2 ⇒b = 1
a = 1 - 1 
c = 1 
so [tex]T =k M^{a}P^{b}r^{c} [/tex]
⇒[tex]T =k M^{0}P^{1}r^{1} [/tex]
⇒[tex]T =k Pr [/tex]
given k = 1/2
so
[tex]T =(Pr)/2[/tex]

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