A cubical block of steel of each side equal to 1 is floating on mercury in vessel. The densities of steel and mercury are `rho_(s) " and " rho_(m)`. The height of the block above the mercury level is given by

Options

A. `l(1+(rho_(s))/(rho_(m)))`
B. `l(1-(rho_(s))/(rho_(m)))`
C. `l(1+(rho_(m))/(rho_(s)))`
D. `l(1-(rho_(m))/(rho_(s)))`

Explanation

To solve the problem of finding the height of a cubical block of steel floating on mercury, we can follow these steps:

### Step 1: Understand the Problem
We have a cubical block of steel with each side equal to 1 meter, floating in mercury. We need to find the height \( h \) of the block that is above the mercury level. The density of steel is \( \rho_s \) and the density of mercury is \( \rho_m \).

### Step 2: Define the Variables
Let:
- The total height of the block = 1 meter
- The height of the block submerged in mercury = \( 1 - h \)
- The height of the block above the mercury level = \( h \)

### Step 3: Apply the Principle of Buoyancy
According to Archimedes' principle, the buoyant force acting on the block is equal to the weight of the liquid displaced by the submerged part of the block.

The buoyant force \( F_b \) can be expressed as:
\[ F_b = \text{(density of mercury)} \times \text{(volume submerged)} \times g \]
Where:
- Density of mercury = \( \rho_m \)
- Volume submerged = \( (1 - h) \times 1^2 = 1 - h \) (since the area of the base is \( 1 \, \text{m}^2 \))
- \( g \) is the acceleration due to gravity (which will cancel out later)

Thus, the buoyant force becomes:
\[ F_b = \rho_m (1 - h) g \]

### Step 4: Calculate the Weight of the Block
The weight \( W \) of the block can be expressed as:
\[ W = \text{(density of steel)} \times \text{(volume of the block)} \times g \]
Where:
- Density of steel = \( \rho_s \)
- Volume of the block = \( 1^3 = 1 \, \text{m}^3 \)

Thus, the weight of the block becomes:
\[ W = \rho_s \cdot 1 \cdot g = \rho_s g \]

### Step 5: Set Up the Equation
For the block to be in equilibrium (floating), the buoyant force must equal the weight of the block:
\[ \rho_m (1 - h) g = \rho_s g \]

### Step 6: Simplify the Equation
We can cancel \( g \) from both sides:
\[ \rho_m (1 - h) = \rho_s \]

### Step 7: Solve for \( h \)
Rearranging the equation gives:
\[ \rho_m - \rho_m h = \rho_s \]
\[ \rho_m h = \rho_m - \rho_s \]
\[ h = \frac{\rho_m - \rho_s}{\rho_m} \]

### Step 8: Final Expression for Height Above Mercury Level
Now we can express \( h \) in a more convenient form:
\[ h = 1 - \frac{\rho_s}{\rho_m} \]

### Conclusion
The height of the block above the mercury level is given by:
\[ h = 1 - \frac{\rho_s}{\rho_m} \]