Calculate the dimensions of linear momentum and surface tension in terms of velocity `(upsilon),` density `(rho)` and frequency (V) as fundamental units.

Answer

To solve the problem, we need to calculate the dimensions of linear momentum and surface tension in terms of velocity (ν), density (ρ), and frequency (V).

### Step 1: Dimensions of Linear Momentum

Linear momentum (p) is defined as the product of mass (m) and velocity (v). The formula for linear momentum is:

\[ p = m \cdot v \]

**Step 1.1: Identify the dimensions of mass and velocity.**

- The dimension of mass (m) is denoted as \([M]\).
- The dimension of velocity (v) is given by the formula \( \text{velocity} = \frac{\text{distance}}{\text{time}} \). The dimensions of distance are \([L]\) (length) and time are \([T]\). Thus, the dimension of velocity is:

\[ [v] = \frac{[L]}{[T]} = [L][T]^{-1} \]

**Step 1.2: Combine the dimensions to find linear momentum.**

Now, substituting the dimensions of mass and velocity into the equation for momentum:

\[ [p] = [M] \cdot [L][T]^{-1} = [M][L][T]^{-1} \]

### Step 2: Dimensions of Surface Tension

Surface tension (σ) is defined as the force (F) per unit length (L). The formula for surface tension is:

\[ \sigma = \frac{F}{L} \]

**Step 2.1: Identify the dimensions of force and length.**

- The dimension of force (F) can be expressed using Newton's second law, \( F = m \cdot a \), where \( a \) (acceleration) is \( \frac{v}{t} \). Thus, the dimensions of force are:

\[ [F] = [M] \cdot [L][T]^{-2} \]

- The dimension of length (L) is simply \([L]\).

**Step 2.2: Combine the dimensions to find surface tension.**

Now substituting the dimensions of force and length into the equation for surface tension:

\[ [\sigma] = \frac{[F]}{[L]} = \frac{[M][L][T]^{-2}}{[L]} = [M][T]^{-2} \]

### Step 3: Expressing in terms of velocity, density, and frequency

**Step 3.1: Expressing Linear Momentum in terms of ν, ρ, and V**

- Velocity (ν) has dimensions \([L][T]^{-1}\).
- Density (ρ) has dimensions \([M][L]^{-3}\).
- Frequency (V) has dimensions \([T]^{-1}\).

To express momentum in terms of these quantities, we can relate mass to density and volume:

\[ [M] = [\rho][L]^3 \]

Substituting this into the momentum equation:

\[ [p] = [\rho][L]^3 \cdot [L][T]^{-1} = [\rho][L]^4[T]^{-1} \]

**Step 3.2: Expressing Surface Tension in terms of ν, ρ, and V**

Using the expression for surface tension:

\[ [\sigma] = [M][T]^{-2} \]

Substituting mass in terms of density:

\[ [\sigma] = [\rho][L]^3[T]^{-2} \]

### Final Results

1. Dimensions of Linear Momentum:
\[ [p] = [\rho][L]^4[T]^{-1} \]

2. Dimensions of Surface Tension:
\[ [\sigma] = [\rho][L]^3[T]^{-2} \]