Young-Laplace law states that the excess pressure inside a soap bubble of radius R is given by `DeltaP = 4 sigma //R` where `sigma` is the coefficient of surface tension of the soap. The `Eoverset(* *)(o) tv overset(* *)(o)s` number `E_0` is a dimensionless number that is used to describe the shape of bubbles rising through a surrounding fluid. It is a combination of g, the acceleration due to gravity, `rho`, the density of the surrounding fluid, `sigma` and a characterstic length scale L which could be the radius of the bubble. A possible expression for `E_0` is

Options

A. `(rhog)/( sigma L^(3))`
B. `(rho L^(2))/( sigma g)`
C. `(rho gL^(2))/(sigma)`
D. `(gL^(2))/(sigma rho )`

Answer

To solve the problem regarding the dimensionless number \( E_0 \) and its expression in terms of gravitational forces, surface tension forces, and other parameters, we will follow these steps:

### Step 1: Understand the parameters involved
We have the following parameters:
- \( g \): acceleration due to gravity (dimension: \( LT^{-2} \))
- \( \rho \): density of the surrounding fluid (dimension: \( ML^{-3} \))
- \( \sigma \): coefficient of surface tension (dimension: \( MT^{-2} \))
- \( L \): characteristic length scale (which could be the radius of the bubble, dimension: \( L \))

### Step 2: Formulate the expression for \( E_0 \)
The dimensionless number \( E_0 \) is often expressed as the ratio of gravitational forces to surface tension forces. A possible expression can be formulated as:

\[ E_0 = \frac{\rho g L^3}{\sigma} \]

### Step 3: Check the dimensions of \( E_0 \)
Now we will check the dimensions of the expression we formulated:

1. **Density \( \rho \)**:
- Dimension: \( ML^{-3} \)

2. **Acceleration due to gravity \( g \)**:
- Dimension: \( LT^{-2} \)

3. **Characteristic length \( L \)**:
- Dimension: \( L \)

4. **Surface tension \( \sigma \)**:
- Dimension: \( MT^{-2} \)

Now substituting these dimensions into the expression for \( E_0 \):

\[ E_0 = \frac{\rho g L^3}{\sigma} = \frac{(ML^{-3})(LT^{-2})(L^3)}{MT^{-2}} \]

### Step 4: Simplify the expression
Now we simplify the dimensions:

\[ E_0 = \frac{M L^{-3} L T^{-2} L^3}{M T^{-2}} = \frac{M L^{1}}{M} = L^0 \]

### Step 5: Conclusion
Since all the mass and length dimensions cancel out, we find that:

\[ E_0 = L^0 = 1 \]

Thus, \( E_0 \) is dimensionless, confirming that our expression is valid.

### Final Answer
The possible expression for the dimensionless number \( E_0 \) is:

\[ E_0 = \frac{\rho g L^3}{\sigma} \]