A hollow cylindrical container floats in water with half its length immersed. A liquid of specific gravity `sigma` is slowly poured into the container until the levels of the liquid inside the container and of water outside are the same. It is observed that the container is submerged to two-thirds its length. Then the value of `sigma` is

Options

A. `1/2`
B. `1/3`
C. `1/4`
D. `1/3`

Explanation

To solve the problem step by step, we will analyze the situation of the hollow cylindrical container floating in water and the effects of pouring a liquid with specific gravity \( \sigma \) into it.

### Step 1: Understand the initial condition
Initially, the hollow cylindrical container is floating with half of its length immersed in water. This means that the buoyant force acting on the container equals the weight of the container.

### Step 2: Set up the equations for the initial condition
Let:
- \( V \) = total volume of the container
- \( \rho_l \) = density of water
- \( m \) = mass of the container

The buoyant force \( F_B \) can be expressed as:
\[ F_B = \rho_l \cdot \text{Volume submerged} \cdot g = \rho_l \cdot \left(\frac{V}{2}\right) \cdot g \]
Since the container is floating, the weight of the container \( mg \) is equal to the buoyant force:
\[ mg = \rho_l \cdot \left(\frac{V}{2}\right) \cdot g \]
Dividing both sides by \( g \):
\[ m = \frac{\rho_l V}{2} \]

### Step 3: Analyze the condition after pouring the liquid
After pouring the liquid, the container is submerged to two-thirds of its length. This means the new volume submerged is:
\[ \text{Volume submerged} = \frac{2V}{3} \]

### Step 4: Set up the equations for the new condition
Now, the total weight of the system (container + liquid) must equal the new buoyant force:
\[ F_{B}' = \rho_l \cdot \left(\frac{2V}{3}\right) \cdot g \]
The mass of the liquid poured into the container can be expressed as:
\[ \text{Mass of liquid} = \sigma \cdot \left(\frac{2V}{3}\right) \]
Thus, the total weight of the system becomes:
\[ mg + \sigma \cdot \left(\frac{2V}{3}\right) \cdot g \]

### Step 5: Equate the forces
Setting the total weight equal to the buoyant force:
\[ \frac{\rho_l V}{2} g + \sigma \cdot \left(\frac{2V}{3}\right) g = \rho_l \cdot \left(\frac{2V}{3}\right) g \]
Dividing through by \( g \) and \( V \):
\[ \frac{\rho_l}{2} + \sigma \cdot \frac{2}{3} = \rho_l \cdot \frac{2}{3} \]

### Step 6: Solve for \( \sigma \)
Rearranging the equation gives:
\[ \sigma \cdot \frac{2}{3} = \rho_l \cdot \frac{2}{3} - \frac{\rho_l}{2} \]
Finding a common denominator (which is 6):
\[ \sigma \cdot \frac{2}{3} = \frac{4\rho_l}{6} - \frac{3\rho_l}{6} \]
\[ \sigma \cdot \frac{2}{3} = \frac{\rho_l}{6} \]
Multiplying both sides by \( \frac{3}{2} \):
\[ \sigma = \frac{\rho_l}{6} \cdot \frac{3}{2} = \frac{3\rho_l}{12} = \frac{\rho_l}{4} \]

### Step 7: Find the specific gravity
Since the specific gravity \( \sigma \) is defined as the ratio of the density of the liquid to the density of water:
\[ \sigma = \frac{\rho_l/4}{\rho_l} = \frac{1}{4} \]

### Final Answer
The value of \( \sigma \) is \( \frac{1}{4} \).

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