Question

If velocity (V), force (F), and energy (E) are taken as fundamental units, then find the dimensional formula for mass.

Answer 1

Hint
The given question is based on the conversion of the units to their dimensional formulae. Using the units of the force, velocity and that of the energy, their dimensional formulae can be obtained. Then by using their dimensional formula we can obtain the formula for mass.
In this solution we will be using the following formula,
$\Rightarrow v = {\raise0.7ex\hbox{$d$} \!\mathord{\left/
 {\vphantom {d t}}\right.}
\!\lower0.7ex\hbox{$t$}}$
where $v$ is velocity, $d$ is distance and $t$ is time
$\Rightarrow F = ma$
where $F$ is force, $m$ is mass and $a$ is acceleration.
$\Rightarrow E = m{c^2}$
where $E$ is energy and $c$ is the speed of light.

Complete step by step answer
The unit of the velocity is meter per second, as the velocity is the ratio of displacement by time. $v = {\raise0.7ex\hbox{$d$} \!\mathord{\left/
 {\vphantom {d t}}\right.}
\!\lower0.7ex\hbox{$t$}}$. The dimensional formula of this unit is, $V = [{\text{L}}{{\text{T}}^{ - 1}}]$.
The unit of the force is kg meter per square second, as the force is the product of the mass and acceleration. The dimensional formula of this unit is ${\text{F}} = [{\text{ML}}{{\text{T}}^{ - 2}}]$.
The unit of the energy is kg meter square per second square, as the energy is the product of mass and the velocity of light in the air, that is, $E = m{c^2}$. The dimensional formula of this unit is, ${\text{E}} = [{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}]$
Therefore, the dimensional formulae are: The dimensional formula of velocity is ${\text{V}} = [{\text{L}}{{\text{T}}^{ - 1}}]$. The dimensional formula of force is ${\text{F}} = [{\text{ML}}{{\text{T}}^{ - 2}}]$. The dimensional formula of energy is ${\text{E}} = [{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}]$
The unit of the mass is kg (in terms of SI units).
Thus, the dimensional formula of the mass is, ${\text{M}} = [{\text{M}}{{\text{L}}^0}{{\text{T}}^0}]$…… (1)
Let the dimensional formula of the mass in terms of the energy, force and velocity be represented as follows,
$\Rightarrow {\text{M}} = [{{\text{V}}^{\text{a}}}{{\text{F}}^{\text{b}}}{{\text{E}}^c}]$ …… (a)
Now, substituting the dimensional formulae of the energy, force and velocity in the above equation we get,
$\Rightarrow {\text{M}} = [{[{\text{L}}{{\text{T}}^{ - 1}}]^{\text{a}}}{[{\text{ML}}{{\text{T}}^{ - 2}}]^{\text{b}}}{[{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}]^c}]$
Upon further removing the brackets and adding the like terms, we get,
$\Rightarrow {\text{M}} = [{{\text{M}}^{b + c}}{{\text{L}}^{a + b + 2c}}{{\text{T}}^{ - a - 2b - 2c}}]$…… (b)
Compare the equations (a) and (b), so we can write,
$\Rightarrow [{\text{M}}{{\text{L}}^0}{{\text{T}}^0}] = [{{\text{M}}^{b + c}}{{\text{L}}^{a + b + 2c}}{{\text{T}}^{ - a - 2b - 2c}}]$
We can now compare and equate the components accordingly. Therefore, we can write,
$\Rightarrow 1 = b + c$…… (1)
$\Rightarrow 0 = a + b + 2c$ …… (2)
$\Rightarrow 0 = - a - 2b - 2c$ …… (3)
Now let us consider the equation (3) again,
$\Rightarrow 0 = - a - 2(b + c)$
Substituting the equation (1) in the above equation, we get,
$\Rightarrow 0 = - a - 2(1)$
Calculating we get,
$\Rightarrow a = - 2$
Now let us consider the equation (2),
$\Rightarrow 0 = a + b + 2c$
We can break the c as,
$\Rightarrow 0 = a + (b + c) + c$
Substituting the values of a and (b+c) in the above equation, we get,
$\Rightarrow 0 = - 2 + (1) + c$
Calculating we get,
$\Rightarrow c = 1$
Now considering the equation (1) again we can substitute the value of c and get,
$\Rightarrow 1 = b + 1$
Hence the value of b is,
$\Rightarrow b = 0$
Substitute the values of a, b and c in the equation (b) we get,
$\Rightarrow {\text{M}} = [{{\text{V}}^{{\text{ - 2}}}}{{\text{F}}^{\text{0}}}{{\text{E}}^1}]$
$\therefore $ The dimensional formula mass, if velocity (V), force (F), and energy (E) are taken as fundamental units is ${\text{M}} = [{{\text{V}}^{{\text{ - 2}}}}{{\text{F}}^0}{{\text{E}}^{\text{1}}}]$.

Note
Dimensional analysis is a simple technique used in science for anticipating the unit of a physical quantity. We can also use the technique of dimensional analysis for checking the correctness of an equation.