A block of volume V and density `rho` is floating in a liquid of density `2 rho` filled in a vessel. Now the vesset starts falling freely with acceleration `g`. Then the volume of block inside the liquid in the falling condition is

Options

A. V
B. `V//2`
C. arbitray
D. zero

Answer

To solve the problem step-by-step, we will analyze the situation of the block floating in the liquid both when the vessel is at rest and when it is falling freely.

### Step 1: Understand the Initial Condition
- The block has a volume \( V \) and density \( \rho \).
- The liquid has a density \( 2\rho \).
- When the block is floating in the liquid at rest, the weight of the block is balanced by the buoyant force acting on it.

### Step 2: Calculate the Weight of the Block
- The weight of the block \( W \) can be calculated as:
\[ W = \text{mass} \times g = \rho V g \]

### Step 3: Calculate the Buoyant Force
- The buoyant force \( F_b \) acting on the block is given by Archimedes' principle:
\[ F_b = \text{density of liquid} \times g \times \text{volume displaced} \]
- Let the volume of the block submerged in the liquid be \( V_d \). Since the density of the liquid is \( 2\rho \), we have:
\[ F_b = 2\rho g V_d \]

### Step 4: Set Up the Equation for Equilibrium
- At equilibrium (when the vessel is not falling), the weight of the block equals the buoyant force:
\[ \rho V g = 2\rho g V_d \]
- Dividing both sides by \( g \) and \( \rho \) (assuming \( g \neq 0 \) and \( \rho \neq 0 \)):
\[ V = 2 V_d \]
- Thus, the volume of the block submerged is:
\[ V_d = \frac{V}{2} \]

### Step 5: Analyze the Condition When the Vessel is Falling Freely
- When the vessel falls freely with acceleration \( g \), the effective gravitational force acting on the block is zero because both the block and the liquid are in free fall.
- In this condition, the buoyant force becomes zero because the pressure difference that creates buoyancy is eliminated.

### Step 6: Conclusion on the Volume of the Block Inside the Liquid
- Since the buoyant force is zero, the block will not be forced down into the liquid. Therefore, the volume of the block that is submerged can be arbitrary.
- The block can float at any position within the liquid, meaning the volume submerged can be anywhere from \( 0 \) to \( V \).

### Final Answer
The volume of the block inside the liquid in the falling condition is arbitrary.

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