If Surface tension (S), Moment of Inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be:

We start by recalling the ordinary dimensional symbols for the physical quantities involved.

Surface tension $$S$$ is defined as force per unit length. Since the dimensional formula of force is $$[F]=M L T^{-2}$$, we have

$$[S]=\frac{[F]}{L}=M\,T^{-2}\;.$$

The moment of inertia $$I$$ of a body is mass times the square of distance, so

$$[I]=M\,L^{2}\;.$$

Planck’s constant $$h$$ represents action (or angular momentum), whose dimensions are

$$[h]=M\,L^{2}\,T^{-1}\;.$$

Linear momentum $$p$$ is the product of mass and velocity, giving

$$[p]=M\,L\,T^{-1}\;.$$

Now we assume that linear momentum can be expressed as a product of the chosen fundamental quantities raised to unknown powers:

$$p = S^{a}\;I^{b}\;h^{c}\;.$$

Writing the dimensional formula of the right-hand side, we substitute each quantity’s dimensions:

$$ \begin{aligned} [S^{a}] &= (M\,T^{-2})^{a}=M^{a}\,T^{-2a},\\[4pt] [I^{b}] &= (M\,L^{2})^{b}=M^{b}\,L^{2b},\\[4pt] [h^{c}] &= (M\,L^{2}\,T^{-1})^{c}=M^{c}\,L^{2c}\,T^{-c}. \end{aligned} $$

Multiplying these three results gives the overall dimensions of the right-hand side:

$$ M^{a+b+c}\,L^{\,2b+2c}\,T^{-2a-c}. $$

For the equality to hold, these exponents must match those of linear momentum, viz. $$M^{1}L^{1}T^{-1}$$. Therefore we equate the exponents term by term:

$$ \begin{aligned} \text{Mass (}M\text{):}&\quad a+b+c &=&\;1,\\[4pt] \text{Length (}L\text{):}&\quad 2b+2c &=&\;1,\\[4pt] \text{Time (}T\text{):}&\quad -2a-c &=&\;-1. \end{aligned} $$

From the length equation we immediately get

$$2b+2c = 1 \;\Longrightarrow\; b+c = \tfrac12.$$

Substituting this into the mass equation gives

$$ a + (b+c) = 1 \;\Longrightarrow\; a + \tfrac12 = 1 \;\Longrightarrow\; a = \tfrac12. $$

Next we use the time equation, inserting $$a=\tfrac12$$:

$$ -2\left(\tfrac12\right) - c = -1 \;\Longrightarrow\; -1 - c = -1 \;\Longrightarrow\; c = 0. $$

Finally, with $$c=0$$ in the earlier relation $$b+c=\tfrac12$$, we obtain

$$b = \tfrac12.$$

Thus the required powers are

$$a=\tfrac12,\quad b=\tfrac12,\quad c=0.$$

Putting these values back into $$S^{a}I^{b}h^{c}$$ we find

$$p = S^{1/2}\,I^{1/2}\,h^{0}.$$

Since $$h^{0}=1,$$ the dimensional formula of linear momentum in the new fundamental system is indeed $$S^{1/2}I^{1/2}h^{0}$$, exactly matching Option A.

Hence, the correct answer is Option A.