Question

Which one of the following relations is dimensionally consistent where $h$ is height to which a liquid of density $\rho $ rises in a capillary tube of radius$r$. $T$ is the surface tension of the liquid, $\theta $ of the angle of contact and $g$ the acceleration due to gravity?
(a) $h = \dfrac{{2T\cos \theta }}{{r\rho g}}$
(b) $h = \dfrac{{2Tr}}{{\rho g\cos \theta }}$
(c) $h = \dfrac{{2\rho g\cos \theta }}{{2Tr}}$
(d) $h = \dfrac{{2Tr\rho g}}{{\cos \theta }}$

Answer 1

Hint: Here we have to find which one is dimensionally consistent among the given options as we know that dimensional analysis is analyzing the relationship between the different physical quantities by identifying its base quantity such as length, mass , time etc... And tracking these dimensions as calculations or comparisons are performed.

Complete step by step solution:
Now Let us check whether the following relations is c or not
$h = \dfrac{{2T\cos \theta }}{{r\rho g}}$
First we have to take the dimension of L.H.S so we get
$L.H.S = [L]$
Here $h$ is height to which a liquid of density $\rho $ rises in a capillary tube of radius $r$. $T$ is the surface tension of the liquid, $\theta $ of the angle of contact and $g$ be the acceleration due to gravity
Now we have to take individual physical quantities dimension
$T = [M]{[T]^{ - 2}}$
And now we have to take for height $h$ density $\rho $ and acceleration due to gravity $g$
$r = \left[ M \right]$
Then for density $\rho $
$\rho = [M]{[L]^{ - 3}}$
And now for the acceleration due to gravity $g$
$g = [L]{[T]^{ - 2}}$
Now we have to substitute the value of surface tension $T$ height $h$ density $\rho $ and acceleration due to gravity $g$ in the R.H.S so we get
$R.H.S = \dfrac{{2T\cos \theta }}{{\rho rg}}$
Then after substituting we get
$R.H.S = \dfrac{{[M{T^{ - 2}}]}}{{[M{L^{ - 3}}T{}^{}][L][L{T^{ - 2}}]}}$
Now after simplifying the above equation we get
$R.H.S = [{M^0}{L^1}{T^0}]$
Hence $R.H.S = [L]$
Therefore here we got $L.H.S = R.H.S$
So it is dimensionally consistent.

Hence. the correct answer is option (a)

Note:An equation in which each term has the same dimension is said to be dimensionally correct. All equations used in any field should be dimensionally correct. If the dimension of the quantities is equal on both sides of the equation, then it is said to be dimensionally consistent