Animated Solution for Physics - Physics and Measurement: 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

The problem asks us to find the dimensional formula of linear momentum ($p$) if surface tension ($S$), moment of inertia ($I$), and Planck's constant ($h$) are chosen as the new fundamental base units. This is a classic application of the principle of dimensional homogeneity.

Whenever we are asked to express one physical quantity in terms of a new set of fundamental quantities, we start by assuming a power-law relationship. We can say that momentum $p$ depends on $h$, $S$, and $I$ raised to some unknown powers $a$, $b$, and $c$.

Mathematically, we write this as:

Here, $k$ is just a dimensionless proportionality constant. Our goal is to find the exact values of $a$, $b$, and $c$. To do this, we need to write down the standard dimensional formulas for all these quantities in terms of Mass ($\text{M}$), Length ($\text{L}$), and Time ($\text{T}$).

Let's recall the dimensions of each term: 1. Linear Momentum ($p$): Mass $\times$ Velocity $⟹[{\text{MLT}}^{-1}]$ 2. Surface Tension ($S$): Force / Length $⟹[{\text{MT}}^{-2}]$ 3. Moment of Inertia ($I$): Mass $\times$ Distance${}^2⟹[{\text{ML}}^2]$ 4. Planck's Constant ($h$): Energy $\times$ Time (or Angular Momentum) $⟹[{\text{ML}}^2{\text{T}}^{-1}]$

Now, we substitute these dimensional formulas into our assumed relationship:

Next, we group the powers of $\text{M}$, $\text{L}$, and $\text{T}$ on the right-hand side:

According to the principle of dimensional homogeneity, the powers of corresponding base quantities on both sides must be equal. This gives us a system of three linear equations:

For $\text{M}$:

For $\text{L}$:

For $\text{T}$:

Now, let's solve this system. It's actually quite simple! If we substitute $a+c=\frac{1}{2}$ into the first equation, we get:

Now, substitute $b=\frac{1}{2}$ into the third equation:

Finally, substitute $a=0$ into the second equation:

We have found our powers! $a=0$, $b=\frac{1}{2}$, and $c=\frac{1}{2}$.

Substituting these back into our original expression, we get:

This means that linear momentum can be expressed purely in terms of surface tension and moment of inertia; it doesn't even depend on Planck's constant in this specific system! The correct option is (c).