The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury. (a) Find the depression of mercury column in the capillary. (b) If the length dipped inside is half the answer of part (a), find the angle made by the mercury surface at the end of the capillary with the vertical. Surface tension of mercury `=0.465Nm^-1` and the contact angle of mercury with glass `=135^@`.

Explanation

To solve the problem step by step, we'll address both parts of the question regarding the capillary tube dipped in mercury.

### Part (a): Find the depression of the mercury column in the capillary.

1. **Identify the given values:**
- Radius of the capillary tube, \( R = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Surface tension of mercury, \( S = 0.465 \, \text{N/m} \)
- Contact angle of mercury with glass, \( \theta = 135^\circ \)
- Density of mercury, \( \rho = 13600 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Use the formula for the depression of the mercury column in the capillary:**
\[ h = \frac{2S \cos \theta}{\rho R g} \]

3. **Calculate \( \cos \theta \):**
\[ \cos(135^\circ) = -\frac{1}{\sqrt{2}} \approx -0.7071 \]

4. **Substitute the values into the formula:**
\[ h = \frac{2 \times 0.465 \times (-0.7071)}{13600 \times (1 \times 10^{-3}) \times 10} \]

5. **Calculate the numerator:**
\[ 2 \times 0.465 \times -0.7071 \approx -0.6571 \]

6. **Calculate the denominator:**
\[ 13600 \times 1 \times 10^{-3} \times 10 = 136 \]

7. **Final calculation for \( h \):**
\[ h = \frac{-0.6571}{136} \approx -0.00484 \, \text{m} \approx -4.84 \, \text{mm} \]

8. **Interpret the result:**
The negative sign indicates a depression, so the depression of the mercury column in the capillary is approximately \( 4.84 \, \text{mm} \).

### Part (b): Find the angle made by the mercury surface at the end of the capillary with the vertical.

1. **Given that the length dipped inside is half the answer from part (a):**
\[ h' = \frac{h}{2} = \frac{-4.84 \, \text{mm}}{2} = -2.42 \, \text{mm} \]

2. **Use the relationship for the capillary rise/depression:**
\[ \frac{h'}{h} = \frac{2S \cos \theta'}{\rho R g} \div \frac{2S \cos \theta}{\rho R g} \]

3. **This simplifies to:**
\[ \frac{h'}{h} = \frac{\cos \theta'}{\cos \theta} \]

4. **Substituting the known values:**
\[ \frac{-2.42}{-4.84} = \frac{\cos \theta'}{\cos 135^\circ} \]

5. **Calculate \( \frac{h'}{h} \):**
\[ \frac{1}{2} = \frac{\cos \theta'}{-0.7071} \]

6. **Rearranging gives:**
\[ \cos \theta' = -0.7071 \times \frac{1}{2} = -0.35355 \]

7. **Calculate \( \theta' \):**
\[ \theta' = \cos^{-1}(-0.35355) \approx 111^\circ \]

### Final Answers:
- (a) The depression of the mercury column in the capillary is approximately \( 4.84 \, \text{mm} \).
- (b) The angle made by the mercury surface at the end of the capillary with the vertical is approximately \( 111^\circ \).