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

### Step 1: Understand the Initial Condition
When the vessel is stationary, 50% of the volume of the wooden block is submerged in the liquid. This means that the weight of the block is balanced by the buoyant force acting on it.

### Step 2: Establish the Relationship at Rest
Let:
- \( V_B \) = total volume of the block
- \( V_I \) = volume of the block immersed in the liquid
- \( \rho_B \) = density of the block
- \( \rho_L \) = density of the liquid

From the problem, we know:
\[ V_I = 0.5 V_B \]

Using the principle of buoyancy:
\[ \text{Weight of block} = \text{Buoyant force} \]
\[ V_B \cdot \rho_B \cdot g = V_I \cdot \rho_L \cdot g \]
Substituting \( V_I = 0.5 V_B \):
\[ V_B \cdot \rho_B \cdot g = 0.5 V_B \cdot \rho_L \cdot g \]
Cancelling \( g \) and \( V_B \) (assuming \( V_B \neq 0 \)):
\[ \rho_B = 0.5 \rho_L \]
This implies:
\[ \frac{\rho_L}{\rho_B} = 2 \]

### Step 3: Analyze the Condition When Vessel Moves Upwards
Now, when the vessel accelerates upwards with \( a = g/2 \), we need to consider the forces acting on the block.

### Step 4: Set Up the Equation of Motion
The forces acting on the block are:
- Buoyant force \( F_B = V_I \cdot \rho_L \cdot g \) (upwards)
- Weight of the block \( W = V_B \cdot \rho_B \cdot g \) (downwards)

Using Newton's second law:
\[ F_B - W = M \cdot a \]
Where \( M \) is the mass of the block:
\[ V_B \cdot \rho_B \cdot g \]
Thus:
\[ V_I \cdot \rho_L \cdot g - V_B \cdot \rho_B \cdot g = V_B \cdot \rho_B \cdot \frac{g}{2} \]

### Step 5: Rearrange the Equation
Rearranging gives:
\[ V_I \cdot \rho_L \cdot g = V_B \cdot \rho_B \cdot g + V_B \cdot \rho_B \cdot \frac{g}{2} \]
\[ V_I \cdot \rho_L = V_B \cdot \rho_B \left(1 + \frac{1}{2}\right) \]
\[ V_I \cdot \rho_L = V_B \cdot \rho_B \cdot \frac{3}{2} \]

### Step 6: Solve for the Immersed Volume
Now, substituting \( \rho_B = 0.5 \rho_L \):
\[ V_I \cdot \rho_L = V_B \cdot (0.5 \rho_L) \cdot \frac{3}{2} \]
Cancelling \( \rho_L \) (assuming \( \rho_L \neq 0 \)):
\[ V_I = V_B \cdot 0.5 \cdot \frac{3}{2} \]
\[ V_I = \frac{3}{4} V_B \]

### Step 7: Calculate the Percentage Immersed
The percentage of the volume immersed is:
\[ \text{Percentage immersed} = \left( \frac{V_I}{V_B} \right) \times 100 = \left( \frac{3/4 V_B}{V_B} \right) \times 100 = 75\% \]

### Final Answer
The percentage volume immersed when the vessel moves upwards with acceleration \( a = g/2 \) is **75%**.

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