Question

A large fluid oscillates in shape under the influence of its own gravitational field. Using dimensional analysis find the expression for periodic oscillation $\left( T \right)$ in terms of radius of star $\left( R \right)$ , mass density of fluid $\left( \rho \right)$ and universal gravitational constant $\left( G \right)$.

Answer 1

Dimensions of a physical quantity are the powers to which the fundamental quantities must be raised to represent the given physical quantity. The principle of Homogeneity of Dimension states that the dimensions of all the terms in a physical expression should be the same.

Complete step by step answer:
We are given, periodic oscillation is a function of the radius of the star, the mass density of the fluid, and the universal gas constant. Thus let’s assume,
$T = K{R^a}{\rho ^b}{G^c}$ where $K$ is a constant
Dimension of the period of oscillation is $\left[ T \right]$.
Dimension of radius of star $\left[ L \right]$.
Dimension of mass density of fluid is $\left[ {M{L^{ - 3}}} \right]$.
Dimension of the universal gravitational constant is $\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$.

Now, the principle of Homogeneity of Dimension states that the dimensions of all the terms in a physical expression should be the same.Thus equating the dimensions of both sides we get,
${M^0}{L^0}{T^1} = {L^a}{\left( {M{L^{ - 3}}} \right)^b}{\left( {{M^{ - 1}}{L^3}{T^ - }^2} \right)^c}$
Simplifying the powers we get,
$\Rightarrow T = {M^{\left( {b - c} \right)}}{L^{\left( {a - 3b + 3c} \right)}}{T^{ - 2c}}$
On comparing the two sides we get,
$- 2c = 1$
$\Rightarrow c = - \dfrac{1}{2}$
$\Rightarrow b - c = 0$
$\Rightarrow b = c = - \dfrac{1}{2}$
$a - 3b + 3c = 0$
$\Rightarrow a = 0$
Thus from the first relation we get
$\therefore T = K{\rho ^{\left( { - 1/2} \right)}}{G^{ - 1/2}}$

Thus, periodic oscillation of the fluid can be expressed as $T = \dfrac{K}{{\sqrt {\rho G} }}$.

Note: The method dimension is useful when a physical quantity depends on other quantities by multiplication and power relations. It cannot be used to find out a physical quantity that depends on the sum or difference of two quantities. For example, we cannot get the relation $s = ut + \dfrac{1}{2}a{t^2}$ from dimensional analysis. Also, the value of the dimensional constant cannot be calculated using dimensional analysis.