A piece of steel floats in mercury. The specific gravity of mercury and steel are 13.6 and 7.8 respectively. For covering the whole piece some water is filled above the mercury. What part of the piece is inside the mercury

Options

A. 0.57
B. 0.5
C. 0.47
D. 0.62

Answer

To solve the problem of determining what part of a steel piece is submerged in mercury, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a piece of steel floating in mercury. The specific gravity (SG) of mercury is 13.6, and that of steel is 7.8. We need to find the fraction of the steel piece that is submerged in mercury.

2. **Convert Specific Gravity to Density**:
- The specific gravity (SG) is the ratio of the density of a substance to the density of water.
- Therefore, the density of mercury (ρ_Hg) can be calculated as:
\[ \rho_{Hg} = SG_{Hg} \times \rho_{water} = 13.6 \times 1000 \, \text{kg/m}^3 = 13600 \, \text{kg/m}^3 \]
- The density of steel (ρ_S) is:
\[ \rho_{S} = SG_{S} \times \rho_{water} = 7.8 \times 1000 \, \text{kg/m}^3 = 7800 \, \text{kg/m}^3 \]

3. **Apply the Principle of Floatation**:
- According to Archimedes' principle, the weight of the steel piece must equal the weight of the mercury displaced by the submerged part of the steel.
- Let \( V \) be the total volume of the steel piece and \( V_1 \) be the volume of the steel piece submerged in mercury.

4. **Set Up the Equation**:
- The weight of the steel piece is given by:
\[ W_{steel} = \rho_{S} \cdot V \cdot g \]
- The buoyant force (weight of the mercury displaced) is given by:
\[ F_{buoyant} = \rho_{Hg} \cdot V_1 \cdot g \]
- Setting these equal gives:
\[ \rho_{S} \cdot V = \rho_{Hg} \cdot V_1 \]

5. **Express Volume in Terms of Height**:
- If we assume the cross-sectional area of the steel piece is \( A \) and its height is \( H \), then:
\[ V = A \cdot H \quad \text{and} \quad V_1 = A \cdot h_1 \]
- Substituting these into the equation gives:
\[ \rho_{S} \cdot (A \cdot H) = \rho_{Hg} \cdot (A \cdot h_1) \]
- The area \( A \) cancels out:
\[ \rho_{S} \cdot H = \rho_{Hg} \cdot h_1 \]

6. **Solve for the Height Submerged**:
- Rearranging gives:
\[ h_1 = \frac{\rho_{S}}{\rho_{Hg}} \cdot H \]
- Substituting the densities:
\[ h_1 = \frac{7800}{13600} \cdot H = 0.5735 \cdot H \]

7. **Determine the Fraction Submerged**:
- The fraction of the steel piece submerged in mercury is:
\[ \frac{h_1}{H} = \frac{7800}{13600} \approx 0.5735 \]
- This means approximately 57.35% of the steel piece is submerged in mercury.

### Final Answer:
Approximately 57.35% of the piece of steel is inside the mercury.