1. Write dimensions of the given fundamental quantities

We take surface tension $S$, moment of inertia $I$, and Planck's constant $h$ as fundamental quantities.

Their dimensions in terms of $M, L, T$ are:

• Surface tension: $[S] = \frac{\text{force}}{\text{length}} = \frac{MLT^{-2}}{L} = MT^{-2}$

• Moment of inertia: $[I] = ML^2$

• Planck's constant: $[h] = \text{energy} \times \text{time} = (ML^2T^{-2})\cdot T = ML^2T^{-1}$

2. Write dimensions of linear momentum

Linear momentum $p$ has dimensions: $[p] = MLT^{-1}$

3. Assume momentum in terms of $S, I, h$

Let $[p] = [S]^a [I]^b [h]^c$

Substitute the dimensions: $MLT^{-1} = (MT^{-2})^a (ML^2)^b (ML^2T^{-1})^c$

Expanding, $MLT^{-1} = M^{a+b+c} L^{2b+2c} T^{-2a-c}$

4. Compare powers of $M, L, T$

So we get the equations:

• For $M$: $a+b+c=1$
• For $L$: $2b+2c=1$
• For $T$: $-2a-c=-1$

5. Solve the equations

From $2b+2c=1$ we get $b+c=\frac{1}{2}$

From $-2a-c=-1$ we get $2a+c=1$

Now use $a+b+c=1$ Substitute $b+c=\frac{1}{2}$: $a+\frac{1}{2}=1$ $a=\frac{1}{2}$

Then from $2a+c=1$ $2\left(\frac{1}{2}\right)+c=1$ $1+c=1$ $c=0$

Now $b+c=\frac{1}{2} \Rightarrow b=\frac{1}{2}$

Thus, $[p] = S^{1/2} I^{1/2} h^0$

6. Match with options

This corresponds to:

Option A: $S^{1/2} I^{1/2} h^0$

Final Answer

$\boxed{S^{1/2} I^{1/2} h^0}$ So, Option A is correct.