Question

A gas bubble, from an explosion under water oscillates with a period $\ T \propto {P^a}{d^b}{E^c}$, where P is the static pressure, d is the density of water and E is the total energy of the explosion. Find the values of a, b and c.

Answer 1

Hint:Here we will use the concept of the dimensional analysis. The dimensional analysis is the mathematical tool to calculate the physical parameters and its analysis involves the fundamental units of the dimensions M (mass), Length (L) and T (Time). It is used to determine the relationships between the numbers of variables.

Complete step by step answer:
Given that T proportional to PadbEc
$\therefore T \propto {P^a}{d^b}{E^c}$
Let us suppose that K be the constant of proportionality and dimensionless quantity.
$\therefore T = {P^a}{d^b}{E^c}$
Place the dimensions of Time, Pressure, density and energy in the above equation – (Not placing dimensions for the constant K as its dimensionless $[{M^0}{L^0}{T^0}]$
$[{M^0}{L^0}{T^1}] = {[{M^1}{L^{ - 1}}{T^{ - 2}}]^a}{[{M^1}{L^{ - 3}}]^b}{[{M^1}{L^2}{T^{ - 2}}]^c}$
Simplify the above equation-
$[{M^0}{L^0}{T^1}] = {[{M^{1a}}{L^{ - 1a}}{T^{ - 2a}}]^{}}{[{M^{1b}}{L^{ - 3b}}]^{}}{[{M^c}{L^{2c}}{T^{ - 2c}}]^{}}$
As per the property- Powers are added when the terms are with the same base and in multiplication.
That is ${2^a} \times {2^b} = {2^{a + b}}$
$[{M^0}{L^0}{T^1}] = [{M^{a + b + c}}{L^{ - a - 3b + 2c}}{T^{ - 2a - 2c}}]$
Again, as per the mathematical property- The powers on the left hand side of the equation is always equal to the powers on the right hand side of the equation.
Therefore, accordingly equating powers of M, L, T on both the sides of the equation.
$\eqalign{
& a + b + c = 0{\text{ }}......{\text{(1)}} \cr
& {\text{ - a - 3b + 2c = 0 }}......{\text{(2)}} \cr
& {\text{ - 2a - 2c = 1 }}.......{\text{(3)}} \cr} $
Simplify equation $(3)$
$\eqalign{
& - 2a - 2c = 1 \cr
& - 2(a + c) = 1 \cr
& a + c = \dfrac{{ - 1}}{2}{\text{ }}......{\text{(4)}} \cr} $
Place the value of equation $(4){\text{ in the equation (1)}}$
$\eqalign{
& a + b + c = 0 \cr
& b - \dfrac{1}{2} = 0 \cr} $
Make “b” the subject. When the term is moved from one side to another side then the sign changes from positive to negative and vice-versa.
$\therefore b = \dfrac{1}{2}{\text{ }}...........{\text{(5)}}$
Using equation $(3)$
$ - 2a - 2c = 1$
Here, make “c” as the subject-
$ - 2a - 1 = 2c$
Place the above value in equation $(2)$
$\eqalign{
& - a - 3b + 2c = 0 \cr
& - a - 3b - 2a - 1 = 0 \cr} $
Place value of “b” using equation $(5)$ and simplify-
$ - 3a - 3\left( {\dfrac{1}{2}} \right) - 1 = 0$
Make variable “a” as the subject
$\eqalign{
& \therefore - 3a = \dfrac{3}{2} + 1 \cr
& \therefore - 3a = \dfrac{{3 + 2}}{2} \cr
& \therefore - 3a = \dfrac{5}{2} \cr} $
The term in multiplication changes its side goes to the (division) denominator and vice-versa.
$a = \dfrac{{ - 5}}{6}{\text{ }}.........{\text{(6) }}$
Place value of “a” in equation $(3)$
$ - 2a - 2c = 1$
Simplify and make “c” as the subject –
$c = \dfrac{{ - 2a - 1}}{2}$
Place, $a = \dfrac{{ - 5}}{6}{\text{ }}$in the above equation.
$\eqalign{
& c = \dfrac{{ - 2(\dfrac{{ - 5}}{6}) - 1}}{2} \cr
& c = \dfrac{{\dfrac{5}{3} - 1}}{2} \cr
& c = \dfrac{{\dfrac{{5 - 3}}{3}}}{2} \cr
& c = \dfrac{{\dfrac{2}{3}}}{2} \cr
& c = \dfrac{2}{{3 \times 2}} \cr
& c = \dfrac{1}{3}{\text{ }}.....{\text{(7)}} \cr} $ (Numerator’s denominator goes to the denominator)
Therefore, the required solution is $a = \dfrac{{ - 5}}{6}{\text{ , b = }}\dfrac{1}{2}{\text{ and c = }}\dfrac{1}{3}{\text{ }}$

Note: The mathematical expression which shows the powers to which the fundamental units are to be raised to get one unit of the derived quantity is called the dimensional formula of that quantity. If “x” is the unit of the derived quantity represented by $x = {M^a}{L^b}{T^c}$ where ${M^a}{L^b}{T^c}$is called the dimensional formula and the exponents a, b and c are called the dimensions.