A glass rod of radius `1 mm` is inserted symmetrically into a glass capillary tube with inside radius `2 mm`. Then the whole arrangement is brought in contact of the surface of water. Surface tension of water is `7 xx 10^(-2) N//m`. To what height will the water rise in the capillary? (`theta = 0^@`)

Options

A. `1.4cm`
B. `4.2cm`
C. `2.1cm`
D. `6.8cm`

Explanation

To solve the problem of how high water will rise in a capillary tube when a glass rod is inserted, we can follow these steps:

### Step 1: Understand the Setup
We have a glass rod of radius \( r_1 = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \) inserted into a glass capillary tube of radius \( r_2 = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \). The surface tension of water is given as \( T = 7 \times 10^{-2} \, \text{N/m} \).

### Step 2: Calculate the Length of Contact
The total length of the contact line where the surface tension acts is given by:
\[ L = 2 \pi r_1 + 2 \pi r_2 \]
Substituting the values:
\[ L = 2 \pi (1 \times 10^{-3}) + 2 \pi (2 \times 10^{-3}) = 2 \pi (1 + 2) \times 10^{-3} = 6 \pi \times 10^{-3} \, \text{m} \]

### Step 3: Calculate the Upward Force due to Surface Tension
The upward force due to surface tension is:
\[ F_T = T \cdot L = 7 \times 10^{-2} \cdot 6 \pi \times 10^{-3} \]

### Step 4: Calculate the Volume of Water in the Capillary
The volume of water that will rise in the capillary tube is given by the difference in areas multiplied by the height \( h \):
\[ V = \pi (r_2^2 - r_1^2) h \]
Substituting the values:
\[ V = \pi \left((2 \times 10^{-3})^2 - (1 \times 10^{-3})^2\right) h = \pi \left(4 \times 10^{-6} - 1 \times 10^{-6}\right) h = 3 \pi \times 10^{-6} h \]

### Step 5: Calculate the Weight of the Water Column
The weight of the water column is given by:
\[ W = \rho V g = \rho \cdot 3 \pi \times 10^{-6} h \cdot g \]
Where \( \rho \) (density of water) is approximately \( 1000 \, \text{kg/m}^3 \) and \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).

### Step 6: Set Up the Equation
Setting the upward force equal to the weight of the water gives:
\[ T \cdot L = \rho \cdot 3 \pi \times 10^{-6} h \cdot g \]
Substituting the values we have:
\[ 7 \times 10^{-2} \cdot 6 \pi \times 10^{-3} = 1000 \cdot 3 \pi \times 10^{-6} h \cdot 10 \]

### Step 7: Solve for Height \( h \)
Now we can simplify and solve for \( h \):
\[ 7 \times 10^{-2} \cdot 6 \times 10^{-3} = 30000 \times 10^{-6} h \]
\[ 4.2 \times 10^{-4} = 30000 \times 10^{-6} h \]
\[ h = \frac{4.2 \times 10^{-4}}{30000 \times 10^{-6}} = \frac{4.2}{30} \, \text{m} = 0.014 \, \text{m} = 1.4 \, \text{cm} \]

### Final Answer
The height to which the water will rise in the capillary tube is \( h = 1.4 \, \text{cm} \).

---