A glass tube of radius `0.4mm` is dipped vertically in water. Find upto what height the water will rise in the capillary? If the tube in inclined at an angle of `60^(@)` with the vertical, how much length of the capillary is occupied by water? Suface of water`=7.0xx10^(-2)N//m`,density of water `=10^(3)kg//m^(3)`.

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Calculate the height of water rise in the vertical capillary tube

The formula for the height \( h \) to which water rises in a capillary tube is given by:

\[ h = \frac{2T \cos(\theta)}{\rho g R} \]

Where:
- \( T \) = surface tension of water = \( 7.0 \times 10^{-2} \, \text{N/m} \)
- \( \theta \) = angle with respect to vertical = \( 0^\circ \) (since the tube is vertical)
- \( \rho \) = density of water = \( 10^3 \, \text{kg/m}^3 \)
- \( g \) = acceleration due to gravity = \( 9.8 \, \text{m/s}^2 \)
- \( R \) = radius of the tube = \( 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} \)

Substituting the values into the formula:

\[ h = \frac{2 \times (7.0 \times 10^{-2}) \cos(0^\circ)}{(10^3) \times (9.8) \times (0.4 \times 10^{-3})} \]

Since \( \cos(0^\circ) = 1 \):

\[ h = \frac{2 \times (7.0 \times 10^{-2})}{(10^3) \times (9.8) \times (0.4 \times 10^{-3})} \]

Calculating the denominator:

\[ (10^3) \times (9.8) \times (0.4 \times 10^{-3}) = 3.92 \]

Now substituting this value back into the equation for \( h \):

\[ h = \frac{2 \times (7.0 \times 10^{-2})}{3.92} = \frac{0.14}{3.92} \approx 0.0357 \, \text{m} = 3.57 \times 10^{-2} \, \text{m} \]

### Step 2: Calculate the length of the capillary tube occupied by water when inclined at \( 60^\circ \)

When the tube is inclined at an angle of \( 60^\circ \) with the vertical, we can relate the height \( h \) to the length \( L \) of the tube occupied by water using the cosine of the angle:

\[ L \cos(60^\circ) = h \]

Since \( \cos(60^\circ) = \frac{1}{2} \):

\[ L \cdot \frac{1}{2} = h \implies L = 2h \]

Substituting the value of \( h \):

\[ L = 2 \times (3.57 \times 10^{-2}) = 7.14 \times 10^{-2} \, \text{m} \]

### Final Answers

1. The height to which water rises in the vertical capillary tube is \( 3.57 \times 10^{-2} \, \text{m} \).
2. The length of the capillary tube occupied by water when inclined at \( 60^\circ \) is \( 7.14 \times 10^{-2} \, \text{m} \).