The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, up to what height will the water rise in the capillary ?

Explanation

To solve the problem of how high water will rise in a capillary tube when the lower end is immersed in water, given that the same tube shows a capillary depression of 2 cm when immersed in mercury, we can follow these steps:

### Step 1: Understand the given information
- The capillary depression in mercury is given as 2 cm (or 0.02 m).
- The surface tension of mercury (\( \sigma_m \)) is 0.465 N/m.
- The surface tension of water (\( \sigma_w \)) is \( 7.5 \times 10^{-2} \) N/m.
- The contact angle for mercury with glass is 140 degrees, and for water, it is 0 degrees.

### Step 2: Use the formula for capillary action
The formula for capillary rise (or depression) is given by:

\[ h = \frac{2\sigma \cos \theta}{\rho g r} \]

Where:
- \( h \) is the height of the liquid column (capillary rise or depression),
- \( \sigma \) is the surface tension of the liquid,
- \( \theta \) is the contact angle,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( r \) is the radius of the capillary tube.

### Step 3: Set up the ratio of capillary rise in water to capillary depression in mercury
Since we are comparing the capillary rise in water to the capillary depression in mercury, we can set up the equation as follows:

\[ \frac{h_w}{h_m} = \frac{\sigma_w \cos \theta_w}{\sigma_m \cos \theta_m} \cdot \frac{\rho_m}{\rho_w} \]

Where:
- \( h_w \) is the capillary rise in water,
- \( h_m \) is the capillary depression in mercury (which is -0.02 m),
- \( \sigma_w \) and \( \sigma_m \) are the surface tensions of water and mercury respectively,
- \( \theta_w \) and \( \theta_m \) are the contact angles for water and mercury respectively,
- \( \rho_w \) and \( \rho_m \) are the densities of water and mercury respectively.

### Step 4: Substitute the known values
Substituting the known values into the equation:

\[ \frac{h_w}{-0.02} = \frac{(7.5 \times 10^{-2}) \cdot \cos(0)}{(0.465) \cdot \cos(140^\circ)} \cdot \frac{13.6 \times 10^3}{10^3} \]

### Step 5: Calculate the cosine values
- \( \cos(0) = 1 \)
- \( \cos(140^\circ) = -0.766 \) (since it is in the second quadrant)

### Step 6: Substitute and simplify
Now substituting the cosine values:

\[ \frac{h_w}{-0.02} = \frac{(7.5 \times 10^{-2}) \cdot 1}{(0.465) \cdot (-0.766)} \cdot \frac{13.6 \times 10^3}{10^3} \]

### Step 7: Calculate the right-hand side
Calculating the right-hand side:

\[ \frac{h_w}{-0.02} = \frac{7.5 \times 10^{-2}}{0.465 \cdot -0.766} \cdot 13.6 \]

Calculating the denominator:

\[ 0.465 \cdot -0.766 \approx -0.35679 \]

Thus,

\[ \frac{h_w}{-0.02} = \frac{7.5 \times 10^{-2}}{-0.35679} \cdot 13.6 \]

Calculating this gives:

\[ \frac{h_w}{-0.02} = -0.204 \cdot 13.6 \approx -2.77 \]

### Step 8: Solve for \( h_w \)
Now, solving for \( h_w \):

\[ h_w = -2.77 \cdot (-0.02) \approx 0.0554 \text{ m} \approx 5.54 \text{ cm} \]

### Final Answer
The height to which water will rise in the capillary tube is approximately **5.54 cm**.

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