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, we need to analyze the situation step by step:

### Step 1: Understand the Scenario
We have a block of volume \( V \) and density \( \rho \) floating in a liquid of density \( 2\rho \). The vessel containing the liquid is falling freely with an acceleration equal to \( g \).

### Step 2: Identify Forces Acting on the Block
In normal conditions (when the vessel is not falling), the block experiences two main forces:
1. The gravitational force acting downwards: \( F_g = mg = \rho V g \)
2. The buoyant force acting upwards: \( F_b = \text{density of liquid} \times \text{volume of liquid displaced} \times g = 2\rho V' g \), where \( V' \) is the volume of the block submerged in the liquid.

### Step 3: Analyze the Condition of Free Fall
When the vessel is in free fall, both the block and the liquid are accelerating downwards with acceleration \( g \). This means that the effective weight of the block and the buoyant force are both negated by the free fall.

### Step 4: Pressure Difference in Free Fall
In free fall, the pressure difference between any two points in the liquid becomes zero because both points are accelerating downwards at the same rate. Therefore, the buoyant force acting on the block becomes zero.

### Step 5: Conclusion About the Block's Volume in Liquid
Since the buoyant force is zero, there is no upward force acting on the block. This means that the block can occupy any volume inside the liquid without being pushed upwards. Therefore, the volume of the block submerged in the liquid can be arbitrary.

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

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