An object is put in turn, in three liquids having different densities . The object floats with `3//5, 2//9` and `8//11` parts of its volume inside the liquid surface in liquids of densities `rho_(1), rho_(2)` and `rho_(3)` respectively . Which of the following is correct ?

Options

A. `rho_(1)gtrho_(2)gtrho_(3)`
B. `rho_(3)ltrho_(1)ltrho_(2)`
C. `rho_(3)gtrho_(1)gtrho_(2)`
D. `rho_(1)ltrho_(3)ltrho_(2)`

Explanation

To solve the problem, we need to analyze the relationship between the volume of the object submerged in each liquid and the densities of those liquids. Here's a step-by-step solution:

### Step 1: Understand the concept of buoyancy
When an object floats in a liquid, the weight of the liquid displaced by the submerged part of the object is equal to the weight of the object. This is known as Archimedes' principle.

### Step 2: Set up the relationship
Let:
- \( V_0 \) = volume of the object submerged in the liquid
- \( V \) = total volume of the object
- \( \rho \) = density of the liquid
- \( g \) = acceleration due to gravity (constant)

According to Archimedes' principle, the thrust (upward force) is equal to the weight of the displaced liquid:
\[ \text{Thrust} = V_0 \cdot \rho \cdot g \]
Since the object is in equilibrium, this thrust equals the weight of the object:
\[ V_0 \cdot \rho \cdot g = m \cdot g \]
Where \( m \) is the mass of the object. Since \( g \) is constant, we can simplify this to:
\[ V_0 \cdot \rho = m \]

### Step 3: Express \( V_0 \) in terms of \( \rho \)
From the equation above, we can express the volume submerged \( V_0 \) as:
\[ V_0 = \frac{m}{\rho} \]
This shows that the volume submerged is inversely proportional to the density of the liquid.

### Step 4: Analyze the given fractions
For the three liquids, we have:
- For liquid 1: \( V_0 = \frac{3}{5}V \) → \( \rho_1 \)
- For liquid 2: \( V_0 = \frac{2}{9}V \) → \( \rho_2 \)
- For liquid 3: \( V_0 = \frac{8}{11}V \) → \( \rho_3 \)

### Step 5: Write the relationships
Using the inverse relationship, we can write:
\[ \rho_1 \propto \frac{1}{\frac{3}{5}} = \frac{5}{3} \]
\[ \rho_2 \propto \frac{1}{\frac{2}{9}} = \frac{9}{2} \]
\[ \rho_3 \propto \frac{1}{\frac{8}{11}} = \frac{11}{8} \]

### Step 6: Compare the densities
Now we need to compare these values:
- \( \rho_1 \) is proportional to \( \frac{5}{3} \) (approximately 1.67)
- \( \rho_2 \) is proportional to \( \frac{9}{2} \) (approximately 4.5)
- \( \rho_3 \) is proportional to \( \frac{11}{8} \) (approximately 1.375)

### Step 7: Determine the order of densities
From the proportional values, we can see:
- \( \rho_3 < \rho_1 < \rho_2 \)

### Conclusion
Thus, the correct relationship between the densities is:
\[ \rho_3 < \rho_1 < \rho_2 \]

### Final Answer
The correct option is: \( \rho_3 < \rho_1 < \rho_2 \).

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