A wooden block is floating in a liquid. About `50%` of its volume is inside the liquid when the vessel is stationary, Percentage volume immersed when the vessel moves upwards with acceleration `a=g.2` is

Options

A. `75%`
B. `25%`
C. `50%`
D. `33.33%`

Explanation

To solve the problem of determining the percentage volume of a wooden block that is immersed in a liquid when the vessel moves upwards with an acceleration of \( a = g/2 \), we can follow these steps:

### Step 1: Understand the Initial Condition
When the vessel is stationary, we know that 50% of the volume of the wooden block is submerged in the liquid. This means that the buoyant force (Fb) acting on the block is equal to the weight of the block (mg).

**Equation:**
\[ Fb = mg \]
Where:
- \( Fb = \rho_{liquid} \cdot V_{submerged} \cdot g \)
- \( V_{submerged} = \frac{V}{2} \) (where \( V \) is the total volume of the block)

### Step 2: Set Up the Forces When the Vessel is Accelerating
When the vessel accelerates upwards with an acceleration \( a = g/2 \), we need to consider the effective gravitational force acting on the block. The effective acceleration due to gravity becomes \( g_{effective} = g + a \).

**Effective Gravity:**
\[ g_{effective} = g + \frac{g}{2} = \frac{3g}{2} \]

### Step 3: Write the Equation for Buoyant Force in Accelerating Frame
In the accelerating frame, the buoyant force (Fb') can be expressed as:
\[ Fb' = \rho_{liquid} \cdot V_{submerged} \cdot g_{effective} \]

### Step 4: Set Up the Equation for the Block in the Accelerating Frame
The weight of the block remains the same, \( mg \), but we need to account for the effective gravity:
\[ mg = \rho_{liquid} \cdot V_{submerged} \cdot g_{effective} \]

### Step 5: Substitute the Effective Gravity into the Equation
Substituting \( g_{effective} \) into the equation gives:
\[ mg = \rho_{liquid} \cdot V_{submerged} \cdot \frac{3g}{2} \]

### Step 6: Solve for the Submerged Volume
From the initial condition, we know:
\[ m = \rho_{block} \cdot V \]
Substituting this into the buoyant force equation:
\[ \rho_{block} \cdot V \cdot g = \rho_{liquid} \cdot V_{submerged} \cdot \frac{3g}{2} \]

### Step 7: Cancel out g and Rearrange
Canceling \( g \) from both sides gives:
\[ \rho_{block} \cdot V = \rho_{liquid} \cdot V_{submerged} \cdot \frac{3}{2} \]
Rearranging for \( V_{submerged} \):
\[ V_{submerged} = \frac{2 \rho_{block}}{3 \rho_{liquid}} \cdot V \]

### Step 8: Calculate the Percentage Volume Immersed
To find the percentage of volume submerged:
\[ \text{Percentage submerged} = \left( \frac{V_{submerged}}{V} \right) \times 100 = \left( \frac{2 \rho_{block}}{3 \rho_{liquid}} \right) \times 100 \]

### Conclusion
Given that the wooden block is less dense than the liquid, we can conclude that the percentage volume submerged when the vessel is accelerating upwards with \( a = g/2 \) will be less than 50%.

Thus, the percentage volume immersed in the liquid when the vessel moves upwards with acceleration \( a = g/2 \) is approximately **33.33%**.