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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

T depends on mass m of the liquid, pressure P of the liquid and radius r of the capillary tube (Take the constant k = \frac{1}{2}).



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So let
 T [tex] \alpha   M^{a}P^{b}r^{c} [/tex]
T =k  M^{a}P^{b}r^{c}
we know that dimensions of
 T = M^{1}L^{0}T^{-2}

P (pressure)= M^{1}L^{-1}T^{-2}
M(mass) = M^{1}
R(radius)= L^{1}
where l = length,m = mass t = time sice k is constant and have no dimension
so   M^{1}L^{0}T^{-2} ={ M^{1}}^a { M^{1}L^{-1}T^{-2}}^b { L^{1}}^c
⇒  M^{1}L^{0}T^{-2} M^{a +b}  L^{c - b} T^{-2b}
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 T =k M^{a}P^{b}r^{c}
T =k M^{0}P^{1}r^{1}
T =k Pr
given k = 1/2
T =(Pr)/2

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