Mercury is poured into a U-tube in which the cross-sectional area of the left-hand limb is three times smaller than that of the right one. The level of the mercury in the narrow limb is a distandce `l=30 cm` from the upper end of the tube. How much will the mercury level rise in the right-hand limb if the left one is filled to the top with water?

Explanation

To solve the problem, we need to analyze the situation in the U-tube when the left limb is filled with water and the mercury levels adjust accordingly.

### Step-by-Step Solution:

1. **Understanding the U-tube Setup**:
- Let the cross-sectional area of the left limb (A1) be \( A \).
- The cross-sectional area of the right limb (A2) is \( 3A \) (since it is three times larger).
- The height of mercury in the left limb is given as \( l = 30 \, \text{cm} \).

2. **Filling the Left Limb with Water**:
- When the left limb is filled to the top with water, the pressure exerted by the water column must be balanced by the pressure exerted by the mercury in the right limb.

3. **Pressure Balance**:
- The pressure exerted by the water column (height \( h \)) in the left limb can be expressed as:
\[ P_{\text{water}} = \rho_{\text{water}} g h \]
- The pressure exerted by the mercury column in the right limb (height \( h_m \)) is:
\[ P_{\text{mercury}} = \rho_{\text{mercury}} g h_m \]
- At equilibrium, the pressures must be equal:
\[ \rho_{\text{water}} g h = \rho_{\text{mercury}} g h_m \]

4. **Relating Heights and Areas**:
- Since the area of the left limb is \( A \) and the area of the right limb is \( 3A \), the volume of water added to the left limb will cause a rise in mercury in the right limb.
- The volume of water added (height \( h \) in left limb) will equal the volume of mercury displaced in the right limb:
\[ A h = 3A h_m \]
- This simplifies to:
\[ h = 3 h_m \quad \Rightarrow \quad h_m = \frac{h}{3} \]

5. **Finding the Height of Water**:
- The height of water \( h \) in the left limb can be expressed in terms of the height of mercury \( h_m \):
\[ h = 30 + 3h_m \]
- Substituting \( h_m = \frac{h}{3} \) into the equation gives:
\[ h = 30 + 3\left(\frac{h}{3}\right) \quad \Rightarrow \quad h = 30 + h \]
- This leads to a contradiction unless we consider the initial height of mercury in the left limb.

6. **Final Calculation**:
- To find the rise in the right limb, we need to solve for \( h_m \) using the pressure balance:
\[ \rho_{\text{water}} g h = \rho_{\text{mercury}} g h_m \]
- Rearranging gives:
\[ h_m = \frac{\rho_{\text{water}}}{\rho_{\text{mercury}}} h \]
- Given that \( \rho_{\text{water}} \approx 1000 \, \text{kg/m}^3 \) and \( \rho_{\text{mercury}} \approx 13600 \, \text{kg/m}^3 \):
\[ h_m = \frac{1000}{13600} \cdot 30 \approx 0.0221 \, \text{m} = 2.21 \, \text{cm} \]

### Conclusion:
The mercury level will rise approximately **2.21 cm** in the right-hand limb.