A tube of `1 mm` bore is dipped into a vessel containing a liquid of density `0.8 g//cm^(3)`, surface tension `30 "dyne"//"cm"` and angle of contact zero. Calcualte the length which the liquid will occupy in the tube when the tube is held `(a)` vertical `(b)` inclined to the vertical at an angle of `30^(@)`.

Answer

To solve the problem step by step, we will calculate the height of the liquid in the tube when it is held vertically and then when it is inclined at an angle of 30 degrees.

### Given Data:
- Diameter of the tube, \( D = 1 \, \text{mm} = 0.1 \, \text{cm} \)
- Radius of the tube, \( R = \frac{D}{2} = 0.05 \, \text{cm} \)
- Density of the liquid, \( \rho = 0.8 \, \text{g/cm}^3 \)
- Surface tension, \( T = 30 \, \text{dyne/cm} \)
- Angle of contact, \( \theta = 0^\circ \)
- Acceleration due to gravity, \( g = 980 \, \text{cm/s}^2 \)

### (a) When the tube is held vertically:
1. **Use the formula for height of liquid rise in a capillary tube:**
\[ h = \frac{2T}{R \rho g} \]

2. **Substitute the values into the formula:**
\[ h = \frac{2 \times 30}{0.05 \times 0.8 \times 980} \]

3. **Calculate the denominator:**
\[ 0.05 \times 0.8 \times 980 = 39.2 \]

4. **Now calculate \( h \):**
\[ h = \frac{60}{39.2} \approx 1.53 \, \text{cm} \]

### (b) When the tube is inclined at an angle of \( 30^\circ \):
1. **The effective height \( h' \) when inclined is given by:**
\[ h' = \frac{h}{\cos(30^\circ)} \]

2. **Calculate \( \cos(30^\circ) \):**
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \]

3. **Substitute the value of \( h \):**
\[ h' = \frac{1.53}{0.866} \]

4. **Calculate \( h' \):**
\[ h' \approx 1.76 \, \text{cm} \]

### Final Answers:
- (a) The length of the liquid in the tube when held vertically is approximately \( 1.53 \, \text{cm} \).
- (b) The length of the liquid in the tube when inclined at \( 30^\circ \) is approximately \( 1.76 \, \text{cm} \).