Question

If force $\mathrm{F}$, time $\mathrm{t}$ and velocity $\mathrm{V}$ are taken as fundamental quantities and $\mathrm{K}$ is the dimensionless constant of proportionality. What is the dimensional formula for mass?

Answer 1

Step 1. Given data

The force $\mathrm{F}$, time $\mathrm{t}$ and velocity $\mathrm{V}$ are taken as fundamental quantities and $\mathrm{K}$ is the dimensionless constant of proportionality.

We have to find the dimensional formula for mass.

Step 2. Concept used

1. The rate of change of the object’s position with respect to a frame of reference and time is called velocity.
2. The product of the component of the force in the direction of the displacement and the magnitude of this displacement is called work.
3. The push or pull on an object with mass that causes it to change its velocity is called force.
4. Time in physics is defined as the progression of events from the past to the present into the future. Time is considered the fourth dimension of reality that is used to describe events in three-dimensional space.

Step 3. Find the dimensional formula for the mass.

Consider the mass is,

$\mathrm{m}\propto {\mathrm{F}}^{\mathrm{a}}{\mathrm{V}}^{\mathrm{b}}{\mathrm{T}}^{\mathrm{c}}$
or

$\mathrm{m}={\mathrm{kF}}^{\mathrm{a}}{\mathrm{V}}^{\mathrm{b}}{\mathrm{T}}^{\mathrm{c}}$ ------- $\left(1\right)$
Here, $\mathrm{k}$ is a dimensionless constant and $\mathrm{a},\mathrm{b}$ and $\mathrm{c}$ are the exponents.
Equate dimensions on both sides.
$\Rightarrow$ $\left[{\mathrm{ML}}^0{\mathrm{T}}^0\right]={\left[{\mathrm{MLT}}^{-2}\right]}^{\mathrm{a}}{\left[{\mathrm{LT}}^{-1}\right]}^{\mathrm{b}}{\left[\mathrm{T}\right]}^{\mathrm{c}}$

$\Rightarrow$ $\left[{\mathrm{ML}}^0{\mathrm{T}}^0\right]=\left[{\mathrm{M}}^{\mathrm{a}}{\mathrm{L}}^{\mathrm{a}+\mathrm{b}}{\mathrm{T}}^{-2\mathrm{a}-\mathrm{b}+\mathrm{c}}\right]$
Use the principle of homogeneity of dimensions.
$\Rightarrow$ $\mathrm{a}=1$ ---------- $\left(2\right)$
$\Rightarrow$ $\mathrm{a}+\mathrm{b}=0$ --------- $\left(3\right)$
$\Rightarrow$ $-2\mathrm{a}-\mathrm{b}+\mathrm{c}=0$ -------- $\left(4\right)$

Step 4. Solve the equations
Solve the equations. $\left(2\right)$, $\left(3\right)$ and $\left(4\right)$.
$\Rightarrow$$\mathrm{a}=1$

put the above value in equation $\left(3\right)$, we get,

$1+\mathrm{b}=0$
$\mathrm{b}=-1$,

Put above calculated values in equation $\left(4\right)$, we get,

$-2\mathrm{a}-\mathrm{b}+\mathrm{c}=0$

$-2\left(1\right)-\left(-1\right)+\mathrm{c}=0$
$\mathrm{c}=1$

Substitute the value of $\mathrm{a},\mathrm{b}$ and $\mathrm{c}$ in the equation $\left(1\right)$

$\Rightarrow$ $\mathrm{m}=\left[{\mathrm{F}}^1{\mathrm{V}}^{-1}{\mathrm{T}}^1\right]$
Hence, the dimension formula for mass is $\left[{\mathrm{F}}^1{\mathrm{V}}^{-1}{\mathrm{T}}^1\right]$.