If a 5.0 cm long capillary tube with 0.10 mm internal diameter open at both end is slightly dipped in water having surface tension 75 dyne ` cm ^( -1) ` Find radius of curvature of meniscus at top in mm.

Explanation

To solve the problem of finding the radius of curvature of the meniscus at the top of a capillary tube dipped in water, we can follow these steps:

### Step 1: Identify the given values
- Length of the capillary tube (L) = 5.0 cm
- Internal diameter of the capillary tube (d) = 0.10 mm = 0.01 cm (conversion to cm)
- Surface tension of water (s) = 75 dyne/cm
- Density of water (ρ) = 1 g/cm³
- Acceleration due to gravity (g) = 981 cm/s²

### Step 2: Calculate the radius of the capillary tube
The radius (r) of the capillary tube is half of the diameter:
\[ r = \frac{d}{2} = \frac{0.01 \, \text{cm}}{2} = 0.005 \, \text{cm} \]

### Step 3: Calculate the height of capillary rise (h)
The formula for capillary rise is given by:
\[ h = \frac{2s \cos \theta}{\rho g r} \]
For water-glass interface, the contact angle (θ) is approximately 0°, so cos(θ) = 1. Plugging in the values:
\[ h = \frac{2 \times 75 \, \text{dyne/cm} \times 1}{1 \, \text{g/cm}^3 \times 981 \, \text{cm/s}^2 \times 0.005 \, \text{cm}} \]
Converting dyne to g·cm/s² (1 dyne = 1 g·cm/s²):
\[ h = \frac{150}{0.005 \times 981} = \frac{150}{4.905} \approx 30.58 \, \text{cm} \]

### Step 4: Set up the relationship for radius of curvature
Using the relationship for capillary rise:
\[ h_1 r_1 = h_2 r_2 \]
Where:
- \(h_1 = 30.58 \, \text{cm}\) (calculated height)
- \(r_1 = 0.005 \, \text{cm}\) (radius of the capillary)
- \(h_2 = 5 \, \text{cm}\) (height of the liquid in the tube)
- \(r_2\) is the radius of curvature we need to find.

### Step 5: Solve for \(r_2\)
Substituting the known values into the equation:
\[ 30.58 \times 0.005 = 5 \times r_2 \]
Calculating the left side:
\[ 0.1529 = 5 \times r_2 \]
Now, solving for \(r_2\):
\[ r_2 = \frac{0.1529}{5} \approx 0.03058 \, \text{cm} \]

### Step 6: Convert \(r_2\) to mm
To convert \(r_2\) from cm to mm:
\[ r_2 = 0.03058 \, \text{cm} \times 10 = 0.3058 \, \text{mm} \approx 0.31 \, \text{mm} \]

### Final Answer
The radius of curvature of the meniscus at the top is approximately **0.31 mm**.

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