A body floats in water with its one-third volume above the surface. The same body floats in a liquid with one-third volume immersed. The density of the liquid is

Options

A. 9 times more than that of water
B. 2 times more than that of water
C. 3 times more than that of water
D. 1.5 times more than that of water

Explanation

To solve the problem step by step, we will apply Archimedes' principle and analyze the conditions under which the body floats in both water and the liquid.

### Step 1: Understand the problem
We have a body that floats in water with one-third of its volume above the surface. This means that two-thirds of the volume is submerged. In another liquid, the same body floats with one-third of its volume submerged, meaning two-thirds is above the surface.

### Step 2: Define variables
Let:
- \( V \) = total volume of the body
- \( \rho \) = density of water
- \( \rho_L \) = density of the liquid
- \( g \) = acceleration due to gravity
- \( m \) = mass of the body

### Step 3: Apply Archimedes' Principle for water
For the first situation (in water):
- Volume submerged = \( \frac{2V}{3} \)
- Weight of displaced water = \( \rho \cdot g \cdot \frac{2V}{3} \)
- Weight of the body = \( m \cdot g \)

According to Archimedes' principle:
\[ \rho \cdot g \cdot \frac{2V}{3} = m \cdot g \]
Cancelling \( g \) from both sides:
\[ \rho \cdot \frac{2V}{3} = m \quad \text{(Equation 1)} \]

### Step 4: Apply Archimedes' Principle for the liquid
For the second situation (in the liquid):
- Volume submerged = \( \frac{V}{3} \)
- Weight of displaced liquid = \( \rho_L \cdot g \cdot \frac{V}{3} \)

According to Archimedes' principle:
\[ \rho_L \cdot g \cdot \frac{V}{3} = m \cdot g \]
Cancelling \( g \) from both sides:
\[ \rho_L \cdot \frac{V}{3} = m \quad \text{(Equation 2)} \]

### Step 5: Equate the two equations
From Equation 1 and Equation 2, we can set them equal to each other since both equal \( m \):
\[ \rho \cdot \frac{2V}{3} = \rho_L \cdot \frac{V}{3} \]

### Step 6: Simplify the equation
Cancelling \( \frac{V}{3} \) from both sides (assuming \( V \neq 0 \)):
\[ \rho \cdot 2 = \rho_L \]
Thus, we find:
\[ \rho_L = 2\rho \]

### Step 7: Conclusion
The density of the liquid is twice the density of water.

### Final Answer
The density of the liquid is \( 2 \times \rho \), where \( \rho \) is the density of water.

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