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%`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- When the vessel is stationary, 50% of the wooden block's volume is submerged in the liquid.
- Let the volume of the wooden block be \( V \).
- The submerged volume \( V_s \) is given by:
\[ V_s = \frac{V}{2} \]

2. **Applying the Principle of Buoyancy**:
- The buoyant force \( F_b \) acting on the block is equal to the weight of the liquid displaced:
\[ F_b = \rho_l \cdot g \cdot V_s = \rho_l \cdot g \cdot \frac{V}{2} \]
- The weight of the wooden block \( W \) is given by:
\[ W = \rho_w \cdot g \cdot V \]
- For the block to float, the buoyant force must equal the weight of the block:
\[ \rho_l \cdot g \cdot \frac{V}{2} = \rho_w \cdot g \cdot V \]
- Simplifying this gives:
\[ \rho_l = 2 \cdot \rho_w \]

3. **Analyzing the Condition When the Vessel is Accelerating**:
- When the vessel accelerates upwards with \( a = \frac{g}{2} \), we need to consider the pseudo force acting on the block.
- The effective weight acting downwards on the block becomes:
\[ W' = \rho_w \cdot g \cdot V + \rho_w \cdot \frac{g}{2} \cdot V = \rho_w \cdot g \cdot V \left(1 + \frac{1}{2}\right) = \frac{3}{2} \rho_w \cdot g \cdot V \]

4. **Setting Up the New Equilibrium Condition**:
- The new buoyant force must balance this effective weight:
\[ F_b = \rho_l \cdot g \cdot V_d \]
- Where \( V_d \) is the new displaced volume. Setting the forces equal gives:
\[ \rho_l \cdot g \cdot V_d = \frac{3}{2} \rho_w \cdot g \cdot V \]
- Canceling \( g \) from both sides, we have:
\[ \rho_l \cdot V_d = \frac{3}{2} \rho_w \cdot V \]

5. **Substituting the Density Relationship**:
- We know \( \rho_l = 2 \cdot \rho_w \):
\[ 2 \cdot \rho_w \cdot V_d = \frac{3}{2} \rho_w \cdot V \]
- Dividing both sides by \( \rho_w \) (assuming \( \rho_w \neq 0 \)):
\[ 2 \cdot V_d = \frac{3}{2} V \]
- Solving for \( V_d \):
\[ V_d = \frac{3}{4} V \]

6. **Calculating the Percentage Volume Immersed**:
- The percentage volume immersed is given by:
\[ \text{Percentage immersed} = \left(\frac{V_d}{V}\right) \times 100 = \left(\frac{3/4 \cdot V}{V}\right) \times 100 = 75\% \]

### Final Answer:
The percentage volume immersed when the vessel moves upwards with an acceleration of \( a = \frac{g}{2} \) is **75%**.