A cubical block is floating in a liquid with half of its volume immersed in the liquid. When the whole system acelerates upwards with a net acceleration of g/3. The fraction of volume immersed in the liquid will be `(3)/(x)` Find x.

Answer

To solve the problem, we will analyze the forces acting on the cubical block when it is floating in a liquid and when the entire system accelerates upwards.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- The cubical block is floating in a liquid with half of its volume immersed.
- Let the total volume of the block be \( V \).
- Therefore, the volume submerged in the liquid is \( V_{\text{submerged}} = \frac{V}{2} \).

2. **Applying the Equilibrium Condition**:
- In equilibrium (when the block is floating), the buoyant force equals the weight of the block.
- The buoyant force \( F_b \) can be expressed as:
\[ F_b = \rho_f \cdot V_{\text{submerged}} \cdot g = \rho_f \cdot \frac{V}{2} \cdot g \]
- The weight of the block \( W \) is given by:
\[ W = \rho_b \cdot V \cdot g \]
- Setting the buoyant force equal to the weight:
\[ \rho_f \cdot \frac{V}{2} \cdot g = \rho_b \cdot V \cdot g \]
- Canceling \( g \) and \( V \) from both sides:
\[ \frac{\rho_f}{2} = \rho_b \implies \rho_f = 2 \rho_b \quad (1) \]

3. **Considering the Accelerated Motion**:
- Now, the entire system accelerates upwards with an acceleration of \( \frac{g}{3} \).
- The effective gravitational acceleration \( g' \) acting on the block becomes:
\[ g' = g + \frac{g}{3} = \frac{4g}{3} \]

4. **Finding the New Buoyant Force**:
- Let the new volume submerged when the system accelerates be \( V'_{\text{submerged}} \).
- The buoyant force in this case is:
\[ F_b' = \rho_f \cdot V'_{\text{submerged}} \cdot g' = \rho_f \cdot V'_{\text{submerged}} \cdot \frac{4g}{3} \]

5. **Setting Up the Equation for the Accelerated Case**:
- The weight of the block remains the same:
\[ W = \rho_b \cdot V \cdot g \]
- Setting the new buoyant force equal to the weight of the block:
\[ \rho_f \cdot V'_{\text{submerged}} \cdot \frac{4g}{3} = \rho_b \cdot V \cdot g \]
- Canceling \( g \) from both sides:
\[ \rho_f \cdot V'_{\text{submerged}} \cdot \frac{4}{3} = \rho_b \cdot V \]

6. **Substituting the Relation from Equation (1)**:
- Substitute \( \rho_f = 2 \rho_b \) into the equation:
\[ 2 \rho_b \cdot V'_{\text{submerged}} \cdot \frac{4}{3} = \rho_b \cdot V \]
- Dividing both sides by \( \rho_b \):
\[ 2 V'_{\text{submerged}} \cdot \frac{4}{3} = V \]
- Rearranging gives:
\[ V'_{\text{submerged}} = \frac{3V}{8} \]

7. **Finding the Fraction of Volume Immersed**:
- The fraction of the volume immersed is:
\[ \frac{V'_{\text{submerged}}}{V} = \frac{3V/8}{V} = \frac{3}{8} \]
- According to the question, this fraction is given as \( \frac{3}{x} \):
\[ \frac{3}{8} = \frac{3}{x} \]
- Cross-multiplying gives:
\[ 3x = 24 \implies x = 8 \]

### Final Answer:
The value of \( x \) is \( 8 \).