A cylindrical vessel is filled with mercury and water of equal weights. The total height of the two liquid layers is 29.2 cm. If the specific gravity of mercury is 13.6, then the height of mercury column will be

Options

A. 1 cm
B. 1.5 cm
C. 2 cm
D. 3 cm

Explanation

To solve the problem, we need to find the height of the mercury column in a cylindrical vessel filled with equal weights of mercury and water, given that the total height of the two liquid layers is 29.2 cm and the specific gravity of mercury is 13.6.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a cylindrical vessel with two liquid layers: mercury and water.
- The total height of both layers combined is 29.2 cm.
- The specific gravity of mercury is given as 13.6, which means the density of mercury is 13.6 times the density of water.

2. **Setting Up the Variables:**
- Let \( H_m \) be the height of the mercury column.
- Let \( H_w \) be the height of the water column.
- According to the problem, we know:
\[ H_m + H_w = 29.2 \text{ cm} \quad \text{(Equation 1)} \]

3. **Weight Equivalence:**
- Since the weights of the mercury and water are equal, we can express this mathematically:
\[ \text{Weight of mercury} = \text{Weight of water} \]
- The weight can be expressed as:
\[ \text{Weight} = \text{Volume} \times \text{Density} \]
- For mercury:
\[ \text{Weight of mercury} = H_m \times \rho_m \]
- For water:
\[ \text{Weight of water} = H_w \times \rho_w \]
- Since the specific gravity of mercury is 13.6, we have:
\[ \rho_m = 13.6 \times \rho_w \]

4. **Substituting Densities:**
- From the weight equivalence, we can write:
\[ H_m \times (13.6 \times \rho_w) = H_w \times \rho_w \]
- Dividing both sides by \( \rho_w \) (assuming \( \rho_w \neq 0 \)):
\[ 13.6 H_m = H_w \quad \text{(Equation 2)} \]

5. **Substituting Equation 2 into Equation 1:**
- Now, substitute \( H_w \) from Equation 2 into Equation 1:
\[ H_m + 13.6 H_m = 29.2 \]
- This simplifies to:
\[ 14.6 H_m = 29.2 \]

6. **Solving for \( H_m \):**
- Now, divide both sides by 14.6:
\[ H_m = \frac{29.2}{14.6} \approx 2 \text{ cm} \]

### Final Answer:
The height of the mercury column \( H_m \) is approximately **2 cm**.