To solve the problem of obtaining the dimensions of linear momentum and surface tension in terms of velocity (v), density (ρ), and frequency (ν), we will follow a systematic approach.
### Step 1: Identify the Dimensional Formulas
1. **Linear Momentum (p)**: The formula for linear momentum is given by:
\[
p = mv
\]
where \(m\) is mass and \(v\) is velocity.
- Dimensions of mass (m) = [M]
- Dimensions of velocity (v) = [L][T]^{-1}
- Therefore, the dimensions of linear momentum are:
\[
[p] = [M][L][T]^{-1} = [M L T^{-1}]
\]
2. **Surface Tension (S)**: Surface tension is defined as force per unit length.
- Dimensions of force (F) = [M][L][T]^{-2]
- Dimensions of length (L) = [L]
- Therefore, the dimensions of surface tension are:
\[
[S] = \frac{[F]}{[L]} = \frac{[M L T^{-2}]}{[L]} = [M L^{0} T^{-2}] = [M T^{-2}]
\]
### Step 2: Express Linear Momentum in Terms of v, ρ, and ν
We need to express the dimensions of linear momentum in terms of the given quantities:
\[
p \propto v^A \cdot \rho^B \cdot \nu^C
\]
Substituting the dimensional formulas:
- Dimensions of velocity (v) = [L T^{-1}]
- Dimensions of density (ρ) = [M L^{-3}]
- Dimensions of frequency (ν) = [T^{-1}]
Thus, we have:
\[
[M L T^{-1}] \propto ([L T^{-1}])^A \cdot ([M L^{-3}])^B \cdot ([T^{-1}])^C
\]
### Step 3: Equate the Dimensions
Now we equate the dimensions:
\[
[M L T^{-1}] = [L^A T^{-A}][M^B L^{-3B}][T^{-C}]
\]
This gives us:
- For mass (M): \(1 = B\)
- For length (L): \(1 = A - 3B\)
- For time (T): \(-1 = -A - C\)
### Step 4: Solve the Equations
From \(1 = B\), we have:
\[
B = 1
\]
Substituting \(B\) into the second equation:
\[
1 = A - 3(1) \implies A = 4
\]
Substituting \(A\) into the third equation:
\[
-1 = -4 - C \implies C = -3
\]
### Step 5: Final Expression for Linear Momentum
Substituting \(A\), \(B\), and \(C\) back into the expression:
\[
p = k \cdot v^4 \cdot \rho^1 \cdot \nu^{-3}
\]
Thus, the dimensions of linear momentum in terms of \(v\), \(\rho\), and \(\nu\) are:
\[
p = k v^4 \rho \nu^{-3}
\]
### Step 6: Express Surface Tension in Terms of v, ρ, and ν
Now, we express surface tension in a similar manner:
\[
S \propto v^A \cdot \rho^B \cdot \nu^C
\]
Substituting the dimensional formulas:
\[
[M T^{-2}] \propto ([L T^{-1}])^A \cdot ([M L^{-3}])^B \cdot ([T^{-1}])^C
\]
### Step 7: Equate the Dimensions for Surface Tension
This gives us:
- For mass (M): \(1 = B\)
- For length (L): \(0 = A - 3B\)
- For time (T): \(-2 = -A - C\)
### Step 8: Solve the Equations for Surface Tension
From \(1 = B\), we have:
\[
B = 1
\]
Substituting \(B\) into the second equation:
\[
0 = A - 3(1) \implies A = 3
\]
Substituting \(A\) into the third equation:
\[
-2 = -3 - C \implies C = 1
\]
### Step 9: Final Expression for Surface Tension
Substituting \(A\), \(B\), and \(C\) back into the expression:
\[
S = k \cdot v^3 \cdot \rho^1 \cdot \nu^1
\]
Thus, the dimensions of surface tension in terms of \(v\), \(\rho\), and \(\nu\) are:
\[
S = k v^3 \rho \nu
\]
### Summary of Results
- Dimensions of Linear Momentum:
\[
p = k v^4 \rho \nu^{-3}
\]
- Dimensions of Surface Tension:
\[
S = k v^3 \rho \nu
\]
