Two vertical parallel glass plates are partially submerged in water. The distance between the plates is d and the length is `l`. Assume that the water between the plates does not reach the upper edges of the plates and the wetting is complete. The water will rise to height (`rho=` density of water and `alpha =` surface tension of water)

Options

A. `(2sigma)/(rhogd)`
B. `(sigma)/(2rhogd)`
C. `(4sigma)/(rhogd)`
D. `(5sigma)/(rhogd)`

Explanation

To solve the problem of how high the water will rise between two vertical parallel glass plates that are partially submerged in water, we can use the principles of capillarity and surface tension. Here’s a step-by-step solution:

### Step 1: Understand the forces acting on the water column
When the water rises between the plates, it is influenced by two main forces: the upward force due to surface tension and the downward force due to the weight of the water column.

### Step 2: Calculate the upward force due to surface tension
The upward force due to surface tension can be expressed as:
\[ F_{\text{up}} = 2 \alpha \cdot l \]
where:
- \( \alpha \) is the surface tension of water,
- \( l \) is the length of the plates.

The factor of 2 accounts for the surface tension acting on both sides of the water column between the plates.

### Step 3: Calculate the downward force due to the weight of the water column
The weight of the water column can be expressed as:
\[ F_{\text{down}} = \rho \cdot g \cdot V \]
where:
- \( \rho \) is the density of water,
- \( g \) is the acceleration due to gravity,
- \( V \) is the volume of the water column, which can be expressed as \( V = A \cdot h \) (where \( A \) is the cross-sectional area and \( h \) is the height of the water column).

For the area between the plates, we have:
\[ A = l \cdot d \]
Thus, the volume becomes:
\[ V = l \cdot d \cdot h \]

So, the downward force becomes:
\[ F_{\text{down}} = \rho \cdot g \cdot (l \cdot d \cdot h) \]

### Step 4: Set the forces equal to each other
At equilibrium, the upward force due to surface tension equals the downward force due to the weight of the water column:
\[ 2 \alpha \cdot l = \rho \cdot g \cdot (l \cdot d \cdot h) \]

### Step 5: Solve for the height \( h \)
We can simplify the equation by canceling \( l \) from both sides (assuming \( l \neq 0 \)):
\[ 2 \alpha = \rho \cdot g \cdot (d \cdot h) \]
Now, rearranging to solve for \( h \):
\[ h = \frac{2 \alpha}{\rho \cdot g \cdot d} \]

### Final Result
Thus, the height to which the water will rise between the plates is given by:
\[ h = \frac{2 \alpha}{\rho \cdot g \cdot d} \]