If time (t), velocity (v) and angular momentum (I) are taken as the fundamental units, then the dimension of mass (m) in terms of t, v and l is

Let us suppose,

$\begin{array}{r}\mathrm{m}\propto {\mathrm{t}}^{\mathrm{a}}\\ \mathrm{m}\propto {\mathrm{v}}^{\mathrm{b}}\\ \text{ and }\mathrm{m}\propto {\mathrm{I}}^{\mathrm{c}}\\ \Rightarrow \mathrm{m}\propto {\mathrm{t}}^{\mathrm{a}}{\mathrm{v}}^{\mathrm{b}}{\mathrm{I}}^{\mathrm{c}}\end{array}$

or       $\mathrm{m}={\mathrm{kt}}^{\mathrm{a}}{\mathrm{v}}^{\mathrm{b}}{\mathrm{I}}^{\mathrm{c}}$          ...(i)

where, k is any constant and k = 1

$\because$ We know that dimensions of

Mass, m = [M¹ L⁰ T⁰]

Time, t = [M⁰ L⁰ T¹]

Velocity, v = [M⁰ L¹ T^-1]

Angular momentum, l = [M¹ L² T^-1]

Substituting all these dimensions in Eq. (i), we get

$\begin{array}{l}\left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^0\right]={\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^1\right]}^{\mathrm{a}}{\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-1}\right]}^{\mathrm{b}}{\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-1}\right]}^{\mathrm{c}}\\ \left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^0\right]=\left[{\mathrm{M}}^{\mathrm{c}}{\mathrm{L}}^{\mathrm{b}+2\mathrm{c}}{\mathrm{T}}^{\mathrm{a}-\mathrm{b}-\mathrm{c}}\right]\end{array}$

On comparing powers, we get

c = 1,

b + 2c = 0

b + 2 = 0

$\begin{array}{l}\Rightarrow \mathrm{b}=-2\\ \mathrm{a}-\mathrm{b}-\mathrm{c}=0\\ \Rightarrow \mathrm{a}-(-2)-1=0\\ \Rightarrow \mathrm{a}=-1\\ \therefore \text{ We can write }\mathrm{m}\propto {\mathrm{t}}^{-1}{\mathrm{v}}^{-2}{\mathrm{I}}^1\end{array}$