A capillary tube made of glass of radius 0.15 mm is dipped vertically in a beaker filled with methylene iodide (surface tension = `0.05 "Nm"^(-1) ` density = 667 kg `m^(-3)` ) which rises to height h in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opp. sides of the capillary) make an angle of `60^(@)` with one another. Then h is close to `( g = 10 "ms"^(-2))`

Options

A. `0.172`m
B. `0.049`m
C. `0.087`m
D. `0.137` m

Explanation

To solve this problem, we will use the formula for the height of liquid rise in a capillary tube, which is given by:

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

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

### Step 1: Identify the given values
- Radius of the capillary tube, \( r = 0.15 \, \text{mm} = 0.15 \times 10^{-3} \, \text{m} \)
- Surface tension, \( T = 0.05 \, \text{N/m} \)
- Density of methylene iodide, \( \rho = 667 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- The angle of contact \( \theta = 30^\circ \) (since the tangents make an angle of \( 60^\circ \), the angle of contact is half of that).

### Step 2: Calculate \( \cos \theta \)
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]

### Step 3: Substitute the values into the formula
Now we can substitute the values into the formula for \( h \):

\[ h = \frac{2 \times 0.05 \times \frac{\sqrt{3}}{2}}{0.15 \times 10^{-3} \times 667 \times 10} \]

### Step 4: Simplify the equation
First, simplify the numerator:
\[ 2 \times 0.05 \times \frac{\sqrt{3}}{2} = 0.05 \sqrt{3} \]

Now, simplify the denominator:
\[ 0.15 \times 10^{-3} \times 667 \times 10 = 0.15 \times 667 \times 10^{-2} = 100.05 \times 10^{-2} \]

### Step 5: Calculate \( h \)
Now we can write:
\[ h = \frac{0.05 \sqrt{3}}{100.05 \times 10^{-2}} \]

Calculating \( \sqrt{3} \approx 1.732 \):
\[ h \approx \frac{0.05 \times 1.732}{100.05 \times 10^{-2}} = \frac{0.0866}{1} \approx 0.0866 \, \text{m} = 0.087 \, \text{m} \]

### Final Result
Thus, the height \( h \) is approximately \( 0.087 \, \text{m} \).

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