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Step-by-Step Solution
Step 1: Identify the dimensional formulas of the given physical quantities
1. Surface tension (S): It is defined as force per unit length. Mathematically,
$S = \dfrac{F}{L}$.
Since $F$ (force) has dimensions $[MLT^{-2}]$ and $L$ has dimension $[L]$, the dimension of $S$ will be:
$[S] = \left[\dfrac{MLT^{-2}}{L}\right] = [M T^{-2}].$
2. Moment of inertia (I): It is defined as $mr^2$, where $m$ is mass and $r$ is the distance from the axis of rotation. Thus,
$[I] = [M L^2].$
3. Planck's constant (h): It can be expressed as $h = E \times t$ or $h = \dfrac{E}{\nu}$, where $E$ is energy and $\nu$ is frequency. The dimension of energy ($E$) is $[ML^2 T^{-2}]$, and multiplying by time ($[T]$) yields:
$[h] = [M L^2 T^{-1}].$
4. Linear momentum (p): Defined as $p = mv$, so its dimension is:
$[p] = [M L T^{-1}].$
Step 2: Set up the dimensional equation
We want to express $p$ in terms of $S$, $I$, and $h$. Suppose
$[p] = [S^a \, I^b \, h^c].$
Substituting the dimensions of each quantity on the right-hand side:
$[M L T^{-1}] = [M T^{-2}]^a \; [M L^2]^b \; [M L^2 T^{-1}]^c.
$
Step 3: Combine the exponents of M, L and T on the right-hand side
Combining powers of $M$, $L$, and $T$ on the right-hand side, we get:
Power of $M$: $a + b + c$
Power of $L$: $2b + 2c$
Power of $T$: $-2a - c$
Thus, the dimension on the right-hand side becomes
$[M^{(a+b+c)} \, L^{(2b + 2c)} \, T^{(-2a-c)}].$
Step 4: Equate the exponents to those of $[p] = [M^1 L^1 T^{-1}]$
We match each dimension’s exponent on both sides:
$a + b + c = 1$ (for M)
$2b + 2c = 1$ (for L)
$-2a - c = -1$ (for T)
Step 5: Solve the system of equations
From $2b + 2c = 1$, we have $b + c = \tfrac{1}{2}.$
From $-2a - c = -1$, we have $2a + c = 1.$
From $a + b + c = 1$, combine the above equations suitably to find $a, b,$ and $c$.
Let us isolate them in a simpler manner:
Equation (2): $b + c = \tfrac{1}{2}.$
Equation (3): $2a + c = 1.$
Equation (1): $a + b + c = 1.$
From (2) we get $b = \tfrac{1}{2} - c.$
Substitute $b$ in (1):
$a + \left(\tfrac{1}{2} - c\right) + c = 1 \quad \Longrightarrow \quad a + \tfrac{1}{2} = 1 \quad \Longrightarrow \quad a = \tfrac{1}{2}.
Now from (3): $2 \left(\tfrac{1}{2}\right) + c = 1 \quad \Longrightarrow \quad 1 + c = 1 \quad \Longrightarrow \quad c = 0.
Finally, $b = \tfrac{1}{2} - 0 = \tfrac{1}{2}.$
Step 6: Write the final expression for linear momentum
We get $a = \tfrac{1}{2},\, b = \tfrac{1}{2},\, c = 0.$ Thus,
$[p] = [S^{\frac{1}{2}} \, I^{\frac{1}{2}} \, h^0] = [S^{1/2} I^{1/2}].
This confirms the correct dimensional form is:
$\displaystyle S^{1/2} \, I^{1/2} \, h^{0}.$
