A capillary due sealed at the top has an internal radius of `r = 0.05 cm`. The tube is placed vertically in water, with its open end dipped in water. Find greatest interger corresponding to the length (in water) of such a tube be for the water in it to rise in these conditions to a height `h = 1 cm` ? The pressure The pressure of the air is `P_(0) = 1 atm`. `= 7 cm` of `Hg` , density of `Hg = 13.6 //cm^(3)`, `g = 9.8 m//sec^(2)` The surface tension of water is `sigma = 70 "dyne"//"cm"`. (assume temperature of air in the tube is constant)

Answer

To solve the problem, we need to analyze the forces acting on the water column in the capillary tube and apply the principles of hydrostatics and surface tension. Here’s a step-by-step solution:

### Step 1: Understand the forces acting on the water column
In the capillary tube, the water is held up by two main forces:
1. The weight of the water column (mg) acting downward.
2. The upward force due to surface tension (σL) and the buoyant force (ρgV).

### Step 2: Define the variables
- Let \( r = 0.05 \, \text{cm} = 0.0005 \, \text{m} \) (internal radius of the tube).
- Let \( h = 1 \, \text{cm} = 0.01 \, \text{m} \) (height of water column).
- Let \( \sigma = 70 \, \text{dyne/cm} = 70 \times 10^{-5} \, \text{N/m} \) (surface tension of water).
- Let \( \rho = 1000 \, \text{kg/m}^3 \) (density of water).
- Let \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

### Step 3: Calculate the volume of water in the tube
The volume \( V \) of the water column in the tube can be expressed as:
\[ V = \pi r^2 h \]
Substituting the values:
\[ V = \pi (0.0005)^2 (0.01) = \pi (2.5 \times 10^{-9}) \approx 7.85 \times 10^{-9} \, \text{m}^3 \]

### Step 4: Calculate the weight of the water column
The weight \( W \) of the water column is given by:
\[ W = mg = \rho V g \]
Substituting the values:
\[ W = 1000 \times (7.85 \times 10^{-9}) \times 9.8 \approx 7.69 \times 10^{-5} \, \text{N} \]

### Step 5: Calculate the surface tension force
The surface tension force \( F_{\sigma} \) can be calculated as:
\[ F_{\sigma} = \sigma L \]
Where \( L \) is the length of the water surface in contact with the air. Since the tube is sealed at the top, \( L \) is equal to the height of the water column \( h \) plus the length of the air column above it. For simplicity, we will consider the effective length to be \( h \) for the calculation of surface tension.

### Step 6: Set up the equilibrium equation
At equilibrium, the forces acting on the water column can be expressed as:
\[ W + F_{\sigma} = \rho g V \]
Substituting the known values:
\[ 7.69 \times 10^{-5} + 70 \times 10^{-5} L = 1000 \times (7.85 \times 10^{-9}) \times 9.8 \]

### Step 7: Solve for \( L \)
Rearranging the equation gives:
\[ 70 \times 10^{-5} L = 7.69 \times 10^{-5} + 7.69 \times 10^{-5} \]
\[ L = \frac{2 \times 7.69 \times 10^{-5}}{70 \times 10^{-5}} = \frac{15.38 \times 10^{-5}}{70 \times 10^{-5}} \approx 0.2197 \, \text{m} \approx 21.97 \, \text{cm} \]

### Step 8: Find the greatest integer corresponding to the length
The greatest integer corresponding to the length of the tube for the water to rise to a height of 1 cm is:
\[ \lfloor L \rfloor = 21 \]

### Final Answer
The greatest integer corresponding to the length of the tube is **21 cm**.