A cubical block is initially on water such that its `(4)/(5)th` volume is submersed in water. Now oil is poured on water and when block attains equilibrium its half volume is in water and half volume is in oil. The relative density of oil is:

Options

A. `(4)/(5)`
B. `(3)/(5)`
C. `(2)/(5)`
D. `(5)/(3)`

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the Initial Condition
Initially, the cubical block is floating in water with \(\frac{4}{5}\) of its volume submerged. Let the volume of the cube be \(V\) and the density of the cube be \(\rho_c\). The volume submerged in water is \(\frac{4V}{5}\).

### Step 2: Apply the Principle of Buoyancy
According to Archimedes' principle, the buoyant force acting on the cube is equal to the weight of the water displaced. The weight of the cube can be expressed as:
\[ mg = \rho_c \cdot V \cdot g \]
The buoyant force \(F_b\) can be expressed as:
\[ F_b = \rho_w \cdot \left(\frac{4V}{5}\right) \cdot g \]
Setting these two equal (since the block is in equilibrium):
\[ \rho_c \cdot V \cdot g = \rho_w \cdot \left(\frac{4V}{5}\right) \cdot g \]
We can cancel \(g\) and \(V\) from both sides:
\[ \rho_c = \frac{4}{5} \rho_w \]

### Step 3: Analyze the New Condition
Now, oil is poured on the water, and the block reaches a new equilibrium where half of its volume is submerged in water and half in oil. This means:
- Volume submerged in water = \(\frac{V}{2}\)
- Volume submerged in oil = \(\frac{V}{2}\)

### Step 4: Calculate the Buoyant Forces
The total buoyant force acting on the block is the sum of the buoyant forces from water and oil:
\[ F_{b, \text{total}} = F_{b, \text{water}} + F_{b, \text{oil}} \]
Where:
- \(F_{b, \text{water}} = \rho_w \cdot \left(\frac{V}{2}\right) \cdot g\)
- \(F_{b, \text{oil}} = \rho_o \cdot \left(\frac{V}{2}\right) \cdot g\)

Setting the total buoyant force equal to the weight of the block:
\[ \rho_w \cdot \left(\frac{V}{2}\right) \cdot g + \rho_o \cdot \left(\frac{V}{2}\right) \cdot g = \rho_c \cdot V \cdot g \]
Cancelling \(g\) and \(V\):
\[ \frac{1}{2} \rho_w + \frac{1}{2} \rho_o = \rho_c \]

### Step 5: Substitute the Density of the Cube
Substituting \(\rho_c = \frac{4}{5} \rho_w\) into the equation:
\[ \frac{1}{2} \rho_w + \frac{1}{2} \rho_o = \frac{4}{5} \rho_w \]
Multiplying through by 2 to eliminate the fractions:
\[ \rho_w + \rho_o = \frac{8}{5} \rho_w \]
Rearranging gives:
\[ \rho_o = \frac{8}{5} \rho_w - \rho_w = \frac{3}{5} \rho_w \]

### Step 6: Find the Relative Density of Oil
The relative density of the oil is given by:
\[ \text{Relative Density} = \frac{\rho_o}{\rho_w} = \frac{3/5 \cdot \rho_w}{\rho_w} = \frac{3}{5} \]

Thus, the relative density of the oil is \(\frac{3}{5}\).

### Final Answer
The relative density of the oil is \(\frac{3}{5}\).

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