Given: Surface tension (S), moment of inertia (I) and Planck's constant $(h)$ are fundamental units.

To find: Dimensional formula for linear momentum.

Surface tension

$S = \frac{\text{~Force~}}{\text{~Length~}}\text{,~}$

Moment of inertia $=$ Mass $\times (\text{distance})^{2}$,

dimensions of moment of inertia:

$\lbrack I\rbrack = \left\lbrack M^{1}{\text{\ }L}^{2} \right\rbrack$

Unit of Planck's constant $=$ Joule $\times$ sec,

dimensional formula of Planck's constant:

$\lbrack h\rbrack = \left\lbrack M^{1}{\text{\ }L}^{2}{\text{\ }T}^{- 1} \right\rbrack$

linear momentum $=$ mass $\times$ velocity,

Dimensional formula of linear momentum:

$\lbrack p\rbrack = \left\lbrack M^{1}{\text{\ }L}^{1}{\text{\ }T}^{- 1} \right\rbrack$

Expressing dimension of $\lbrack p\rbrack$ in terms of $\lbrack S\rbrack,\lbrack I\rbrack$ and $\lbrack h\rbrack$, let:

$p = (S)^{x}(I)^{y}(h)^{z}$

$\left\lbrack M^{1}{\text{\ }L}^{1}{\text{\ }T}^{- 1} \right\rbrack = \left\lbrack M^{1}{\text{\ }T}^{- 2} \right\rbrack^{x}\left\lbrack M^{1}{\text{\ }L}^{2} \right\rbrack^{y}\left\lbrack M^{1}{\text{\ }L}^{2}{\text{\ }T}^{- 1} \right\rbrack^{z}$

Equating the exponents: $x + y + z = 1$

$2y + 2z = 1$

$\begin{matrix} & \ - 2x - z = - 1 \\ \end{matrix}$

Solving the above set of equations for $x,y$ and $z$, we get:

$x = \frac{1}{2},y = \frac{1}{2}\text{~and~}z = 0.$

Putting values of $x,y$ and $z$ in eqn. (v):

$p \propto (S)^{\frac{1}{2}}(I)^{\frac{1}{2}}(h)^{0}\text{.~}$