Question

obtain by the method of dimentional analysis and expression for the surface tension of liquid rising in a capillary tube assumed that surface tension depends on mass of the liquid pressure radius of the capillary tube the constant k=1/2

Answer 1

Explanation:

so let

T $\alphaT[tex]α M^{a}P^{b}r^{c}$

T =k M^{a}P^{b}r^{c}T=k M

a

P

b

r

c

we know that dimensions of

T = M^{1}L^{0}T^{-2}T=M

1

L

0

T

−2

P (pressure)= M^{1}L^{-1}T^{-2}P(pressure)=M

1

L

−1

T

−2

M(mass) = M^{1}M(mass)=M

1

R(radius)= L^{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

L

0

T

−2

={M^{1}M

1

}^a {M^{1}L^{-1}T^{-2}M

1

L

−1

T

−2

}^b {L^{1}L

1

}^c

⇒ M^{1}L^{0}T^{-2}M

1

L

0

T

−2

= M^{a +b}M

a+b

L^{c - b}L

c−b

T^{-2b}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=kM

a

P

b

r

c

⇒T =k M^{0}P^{1}r^{1}T=kM

0

P

1

r

1

⇒T =k PrT=kPr

given k = 1/2

so

T =(Pr)/2T=(Pr)/2