A cubical metal block of edge 12 cm floats in mercry with one fifth of the height inside the mercury. Water poured till the surface of the block is just immersed in it. Find the height of the block is just immersed in it. Find the height of the water column to be poured. Specific gravity of mercury =13.6.

Explanation

To solve the problem, let's break it down step by step.

### Step 1: Understand the problem
We have a cubical metal block with an edge length of 12 cm floating in mercury. The block is floating such that one-fifth of its height is submerged in mercury. We need to find the height of the water column that needs to be poured until the block is completely immersed.

### Step 2: Calculate the submerged height of the block in mercury
The total height of the block is 12 cm. Since one-fifth of the height is submerged in mercury:
\[ \text{Submerged height in mercury} = \frac{1}{5} \times 12 \text{ cm} = 2.4 \text{ cm} \]

### Step 3: Calculate the volume of the block
The volume \( V \) of the cubical block can be calculated using the formula for the volume of a cube:
\[ V = \text{edge}^3 = 12^3 = 1728 \text{ cm}^3 \]

### Step 4: Calculate the weight of the block
The weight of the block \( W \) can be expressed in terms of its density \( \rho_b \) and volume \( V \):
\[ W = \rho_b \cdot V \cdot g \]
where \( g \) is the acceleration due to gravity. However, since we are looking for a ratio, we can ignore \( g \) for now.

### Step 5: Calculate the buoyant force acting on the block
The buoyant force \( F_b \) acting on the block is equal to the weight of the mercury displaced by the submerged part of the block:
\[ F_b = \rho_{Hg} \cdot V_{Hg} \cdot g \]
Where \( V_{Hg} \) is the volume of mercury displaced. The volume of mercury displaced is equal to the submerged volume of the block:
\[ V_{Hg} = \text{Submerged height} \times \text{Base area} = 2.4 \text{ cm} \times (12 \text{ cm})^2 = 2.4 \times 144 = 345.6 \text{ cm}^3 \]

### Step 6: Calculate the density of mercury
Given that the specific gravity of mercury is 13.6, the density of mercury \( \rho_{Hg} \) can be calculated as:
\[ \rho_{Hg} = 13.6 \times \rho_{water} = 13.6 \times 1 \text{ g/cm}^3 = 13.6 \text{ g/cm}^3 \]

### Step 7: Set up the equilibrium condition
At equilibrium, the weight of the block is equal to the buoyant force:
\[ \rho_b \cdot V = \rho_{Hg} \cdot V_{Hg} \]
Substituting the values we have:
\[ \rho_b \cdot 1728 = 13.6 \cdot 345.6 \]

### Step 8: Solve for the density of the block
Calculating the right side:
\[ 13.6 \cdot 345.6 = 4693.76 \]
Now we can find \( \rho_b \):
\[ \rho_b = \frac{4693.76}{1728} \approx 2.72 \text{ g/cm}^3 \]

### Step 9: Calculate the height of the water column
When water is poured, the total volume of the block will be equal to the volume of mercury plus the volume of water:
\[ V = V_{Hg} + V_{water} \]
Let \( x \) be the height of the water column. The volume of water is:
\[ V_{water} = x \cdot (12^2) = 144x \]
The volume of mercury will be:
\[ V_{Hg} = (12 - x) \cdot (12^2) = 144(12 - x) \]
Setting up the equation:
\[ 1728 = 345.6 + 144x \]
Solving for \( x \):
\[ 1728 - 345.6 = 144x \\ 1382.4 = 144x \\ x = \frac{1382.4}{144} \approx 9.6 \text{ cm} \]

### Final Answer
The height of the water column to be poured is approximately **9.6 cm**.