A glass U-tube is such that the diameter of one limb is `3.0mm` and that of the other is 6.0mm. The tube is inverted vertically with the open ends below the surface of water in a beaker. What is the difference between the height to which water rises in the two limbs? Surface tension of water is `0.07Nm^(-1)`. Assume that the angle of contact between water and glass is `0^(@)`.

Explanation

To solve the problem, we need to determine the height difference of water in the two limbs of the U-tube due to the difference in diameters and the effect of surface tension.

### Step-by-Step Solution:

1. **Identify the Radii of the Limbs:**
- The diameter of the first limb (D1) is 3.0 mm, so the radius (r1) is:
\[ r_1 = \frac{D_1}{2} = \frac{3.0 \, \text{mm}}{2} = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \]
- The diameter of the second limb (D2) is 6.0 mm, so the radius (r2) is:
\[ r_2 = \frac{D_2}{2} = \frac{6.0 \, \text{mm}}{2} = 3.0 \, \text{mm} = 3.0 \times 10^{-3} \, \text{m} \]

2. **Apply the Pressure Balance Condition:**
- The pressure at the same height in both limbs must be equal. Therefore, we can write:
\[ P_A = P_B + \frac{2T}{r_1} \quad \text{(for the first limb)} \]
\[ P_C = P_D + \frac{2T}{r_2} \quad \text{(for the second limb)} \]
- Since \(P_A = P_C\), we can equate the two expressions:
\[ P_B + \frac{2T}{r_1} = P_D + \frac{2T}{r_2} \]

3. **Rearranging for Pressure Difference:**
- Rearranging gives:
\[ P_B - P_D = \frac{2T}{r_2} - \frac{2T}{r_1} \]
- This can be simplified to:
\[ P_B - P_D = 2T \left( \frac{1}{r_2} - \frac{1}{r_1} \right) \]

4. **Relate Pressure Difference to Height Difference:**
- The pressure difference can also be expressed in terms of the height difference (h) of the water columns:
\[ P_B - P_D = \rho g h \]
- Therefore, we can set the two expressions equal:
\[ \rho g h = 2T \left( \frac{1}{r_2} - \frac{1}{r_1} \right) \]

5. **Substituting Known Values:**
- Given:
- Surface tension, \(T = 0.07 \, \text{N/m}\)
- Density of water, \(\rho = 1000 \, \text{kg/m}^3\)
- Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\)
- Substitute the values into the equation:
\[ 1000 \cdot 10 \cdot h = 2 \cdot 0.07 \left( \frac{1}{3.0 \times 10^{-3}} - \frac{1}{1.5 \times 10^{-3}} \right) \]

6. **Calculating the Right Side:**
- Calculate \(\frac{1}{r_2} - \frac{1}{r_1}\):
\[ \frac{1}{3.0 \times 10^{-3}} - \frac{1}{1.5 \times 10^{-3}} = \frac{1}{3.0} - \frac{1}{1.5} = \frac{1}{3.0} - \frac{2}{3.0} = -\frac{1}{3.0} \]
- Substitute back:
\[ 10000h = 2 \cdot 0.07 \cdot \left(-\frac{1}{3.0}\right) \]
- Simplifying gives:
\[ 10000h = -\frac{0.14}{3.0} = -0.04667 \]

7. **Solving for h:**
- Rearranging for h:
\[ h = -\frac{0.04667}{10000} = -0.000004667 \, \text{m} = -4.667 \, \text{mm} \]
- The negative sign indicates that the height difference is in the opposite direction, so we take the absolute value:
\[ h = 4.67 \, \text{mm} \]

### Final Answer:
The difference between the height to which water rises in the two limbs is approximately **4.67 mm**.

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