Question

An air bubble inside the water oscillates due to some explosion with period T. If $T\propto {{P}^{a}}{{d}^{b}}{{E}^{c}}$ then determine the values of a, b and c. Here P, d and E are static pressure, density and total energy of explosion of water, respectively.
a) $a=\dfrac{5}{6},\text{ b}=1,\text{ }c=\dfrac{1}{3}\text{ }$
b)$a=-\dfrac{5}{6},\text{ b}=\dfrac{1}{2},\text{ }c=\dfrac{1}{3}\text{ }$
c)$a=1,\text{ b}=1,\text{ }c=1\text{ }$
d)$a=0,\text{ b}=0,\text{ }c=1\text{ }$

Answer 1

In the above question it is given that the time period of oscillation of the air bubble is related to static pressure, density and the energy of explosion, all raised to some power. The net dimension of all these quantities should be equal to that of the time period. Hence using the laws of exponents we can determine the respective values of a, b and c.

Formula used:
$T(d)=\left[ {{M}^{0}}{{L}^{0}}T \right]$
$P(d)=\left[ M{{L}^{-1}}{{T}^{-2}} \right]$
$d(d)=\left[ M{{L}^{-3}}{{T}^{0}} \right]$
$E(d)=\left[ M{{L}^{2}}{{T}^{-2}} \right]$

Complete step-by-step answer:
To begin with the question first let us write all the above physical quantities in the above relation given in terms of their fundamental dimensions. The dimension of time period T(d) is $\left[ {{M}^{0}}{{L}^{0}}T \right]$, that of pressure is $P(d)=\left[ M{{L}^{-1}}{{T}^{-2}} \right]$, that of density is $d(d)=\left[ M{{L}^{-3}}{{T}^{0}} \right]$ and similarly that of energy is $E(d)=\left[ M{{L}^{2}}{{T}^{-2}} \right]$.
In the above question the relation between the above quantities is given by $T\propto {{P}^{a}}{{d}^{b}}{{E}^{c}}$. Let us say the proportionality constant for the above expression be ‘k’. Therefore the above relation can be written as
$T=k{{P}^{a}}{{d}^{b}}{{E}^{c}}$
Substituting the dimensions of the respective physical quantities in the above relation we get,
$\begin{aligned} & T=k{{P}^{a}}{{d}^{b}}{{E}^{c}} \\\ & \Rightarrow \left[ {{M}^{0}}{{L}^{0}}T \right]=k{{\left[ M{{L}^{-1}}{{T}^{-2}} \right]}^{a}}{{\left[ M{{L}^{-31}}{{T}^{0}} \right]}^{b}}{{\left[ M{{L}^{2}}{{T}^{-2}} \right]}^{c}} \\\ & \Rightarrow \left[ {{M}^{0}}{{L}^{0}}T \right]=k\left[ {{M}^{a}}{{L}^{-a}}{{T}^{-2a}} \right]\left[ {{M}^{b}}{{L}^{-3b}}{{T}^{0}} \right]\left[ {{M}^{c}}{{L}^{2c}}{{T}^{-2c}} \right] \\\ & \therefore \left[ {{M}^{0}}{{L}^{0}}T \right]=k\left[ {{M}^{a+b+c}}{{L}^{-a-3b+2c}}{{T}^{-2a-2c}} \right] \\\ \end{aligned}$
From the above equation comparing the powers of the respective dimensions, we obtain
$a+b+c=0.....(1)$
$-a-3b+2c=0.....(2)$
$-2a-2c=1.....(3)$
From equation 1 and 2 we obtain,
$\begin{aligned} & a+b+c=0 \\\ & \because a+c=-\dfrac{1}{2} \\\ & \Rightarrow b-\dfrac{1}{2}=0 \\\ & \therefore b=\dfrac{1}{2} \\\ \end{aligned}$
Substituting the value of ‘b’ in 2 we get,
$\begin{aligned} & -a-3\left( \dfrac{1}{2} \right)+2c=0 \\\ & \therefore a-2c=-\dfrac{3}{2}....(4) \\\ \end{aligned}$
Subtracting equation 4 from 3 we obtain,
$\begin{aligned} & -2a-2c-\left( a-2c \right)=1-\left( -\dfrac{3}{2} \right) \\\ & \Rightarrow -a-2a=\dfrac{5}{2} \\\ & \therefore a=-\dfrac{5}{6} \\\ \end{aligned}$
Hence the value of ‘c’ from equation 3 we get,
$\begin{aligned} & -2a-2c=1 \\\ & \because a=-\dfrac{5}{6} \\\ & \Rightarrow -2\left( -\dfrac{5}{6} \right)-2c=1 \\\ & \Rightarrow -2c=1-\dfrac{5}{3} \\\ & \therefore c=\dfrac{1}{3} \\\ \end{aligned}$

So, the correct answer is “Option b”.

Note: It is to be noted that a constant is always dimensionless. In the above question we have used the law of exponent i.e. if the base is the same than the power of the base gets added up. In the above question we have to determine 3 variables hence we require 3 equations.