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 will analyze the situation involving a U-tube filled with mercury and water. The left limb has a smaller cross-sectional area compared to the right limb. Let's break down the steps to find out how much the mercury level will rise in the right-hand limb when the left one is filled to the top with water.

### Step-by-Step Solution:

1. **Understanding the U-tube Setup**:
- Let the cross-sectional area of the left limb be \( A \).
- Then the cross-sectional area of the right limb is \( 3A \) (since it is three times larger).
- The mercury level in the left limb is at a distance \( l = 30 \, \text{cm} \) from the top of the tube.

2. **Setting Up the Pressure Equation**:
- When the left limb is filled with water, the pressure at the same height in both limbs must be equal.
- Let \( h_w \) be the height of the water column in the left limb and \( h_m \) be the height of the mercury column in the right limb.
- The pressure exerted by the water column is given by:
\[ P_{\text{water}} = \rho_{\text{water}} \cdot g \cdot h_w \]
- The pressure exerted by the mercury column is:
\[ P_{\text{mercury}} = \rho_{\text{mercury}} \cdot g \cdot h_m \]

3. **Relating the Heights**:
- The height of the water column in the left limb can be expressed as:
\[ h_w = 30 \, \text{cm} + 3x \]
where \( x \) is the rise in the mercury level in the right limb.
- The height of the mercury column in the right limb will be:
\[ h_m = x \]

4. **Equating the Pressures**:
- Since the pressures must balance out:
\[ \rho_{\text{water}} \cdot g \cdot (30 + 3x) = \rho_{\text{mercury}} \cdot g \cdot x \]
- We can cancel \( g \) from both sides:
\[ \rho_{\text{water}} \cdot (30 + 3x) = \rho_{\text{mercury}} \cdot x \]

5. **Substituting the Densities**:
- The density of water \( \rho_{\text{water}} = 1 \, \text{g/cm}^3 \) and the density of mercury \( \rho_{\text{mercury}} = 13.6 \, \text{g/cm}^3 \):
\[ 1 \cdot (30 + 3x) = 13.6 \cdot x \]

6. **Solving for \( x \)**:
- Rearranging the equation gives:
\[ 30 + 3x = 13.6x \]
\[ 30 = 13.6x - 3x \]
\[ 30 = 10.6x \]
\[ x = \frac{30}{10.6} \approx 2.83 \, \text{cm} \]

### Final Answer:
The mercury level will rise approximately \( 2.83 \, \text{cm} \) in the right-hand limb.