Question

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 $k = \frac{1}{2}$).

Answer 1

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
 [tex]T = M^{1}L^{0}T^{-2} [/tex]
$P (pressure)= M^{1}L^{-1}T^{-2}$
$M(mass) = M^{1}$
$R(radius)= L^{1}$
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
so
$T =(Pr)/2$

Answer 2

Answer:

here is your ans

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