Question

A gas bubble, from an explosion under water oscillates with a period $T$ proportional to ${P^a}{d^b}{E^c}$, 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

In order to solve this question, we have to unit dimensional analysis for that by using the basic physical quantities and then we have to equate them with the given quantities by breaking them to basic quantities. Unit dimensions do not calculate the constant so we have to assign it ourselves.

Complete step by step answer:
We are given that the time period is proportional to static pressure, density of water and total energy.
$T\alpha {P^a}{d^b}{E^c}$
Static pressure has power of a, density of water has power of b and total energy has power of now we have to remove the proportionality sign so we have to assign a constant for that so in this we are taking $K$ as the constant, where $K$ is a constant of proportionality and dimensionless quantity.
$T = K{\text{ }}{P^a}{d^b}{E^c}$

Now we will break the dimensional quantities static pressure, density of water and total energy into basic quantities such as mass, length and time. We will denote mass by $M$, length by $L$ and time by $T$. Static pressure in basic quantity
$\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
Density of water in basic quantity
$\left[ {ML{T^{ - 3}}} \right]$
Total energy in basic quantity
$\left[ {M{L^2}{T^{ - 2}}} \right]$
As we know static pressure, density of water and total energy are raised to power (a, b, c) so now dimensions will be ${\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]^a}$ , ${\left[ {ML{T^{ - 3}}} \right]^b}$ , ${\left[ {M{L^2}{T^{ - 2}}} \right]^c}$

Now we have to equate time period with static pressure, density of water and total energy
$\left[ T \right] = {\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]^a} \times {\left[ {ML{T^{ - 3}}} \right]^b} \times {\left[ {M{L^2}{T^{ - 2}}} \right]^c}$
Equating powers of Mass on both side
${\left[ M \right]^0} = {\left[ M \right]^a}{\left[ M \right]^b}{\left[ M \right]^c}$
Equating powers of length on both side
${\left[ L \right]^0} = {\left[ L \right]^a}{\left[ L \right]^{ - 3b}}{\left[ L \right]^{2c}}$
Equating powers of time on both side
${\left[ T \right]^1} = {\left[ T \right]^{ - 2a}}{\left[ T \right]^0}{\left[ T \right]^{ - 2c}}$
Equating powers of M, L, T on both sides we get
$0 = a + b + c$
$\Rightarrow 0 = - a + - 3b + 2c$
$\Rightarrow 1 = - 2a + - 2c$
By solving all the equations, we get
$\therefore a = \dfrac{{ - 5}}{6}{\text{ b = }}\dfrac{1}{2}{\text{ c = }}\dfrac{1}{3}$

Hence we got the answer $a = \dfrac{{ - 5}}{6}{\text{ b = }}\dfrac{1}{2}{\text{ c = }}\dfrac{1}{3}$.

Note: A very common mistake which happen while doing these type of question is by not assuming the constant as here we assumed $K$ sa the constant if we will not do so then in the end the formula which we will derive will be incomplete along with this masses , length and time should be solved separately to avoid the confusion in the superscripts.