A block of density `rho` floats in a liquid with its one third volume immersed. The density of the liquid is

Options

A. `rho`
B. `(rho)/(3)`
C. `(rho)/(2)`
D. `3 rho`

Answer

To solve the problem, we need to find the density of the liquid (ρ_L) in which a block of density (ρ) is floating with one-third of its volume submerged.

### Step-by-Step Solution:

1. **Understand the Situation**:
- We have a block with density ρ.
- The block is floating, meaning it is in equilibrium.
- One-third of the block's volume (V/3) is submerged in the liquid.

2. **Identify the Forces Acting on the Block**:
- The weight of the block (W) acts downward.
- The buoyant force (F_b) acts upward.

3. **Write the Expression for the Weight of the Block**:
- The weight of the block can be expressed as:
\[ W = \text{Density of block} \times \text{Volume of block} \times g = \rho V g \]

4. **Write the Expression for the Buoyant Force**:
- The buoyant force can be expressed as:
\[ F_b = \text{Density of liquid} \times \text{Volume submerged} \times g = \rho_L \left(\frac{V}{3}\right) g \]

5. **Set Up the Equation for Equilibrium**:
- Since the block is floating in equilibrium, the buoyant force equals the weight of the block:
\[ F_b = W \]
- Substituting the expressions for weight and buoyant force:
\[ \rho_L \left(\frac{V}{3}\right) g = \rho V g \]

6. **Simplify the Equation**:
- Cancel out \(g\) and \(V\) from both sides (assuming \(V \neq 0\)):
\[ \rho_L \left(\frac{1}{3}\right) = \rho \]

7. **Solve for the Density of the Liquid**:
- Multiply both sides by 3 to isolate ρ_L:
\[ \rho_L = 3\rho \]

### Final Answer:
The density of the liquid (ρ_L) is \(3\rho\).

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