If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid (`a_(y)`) for calculation of pressure, effective `g` is used. A closed box horizontal base `6 m` by `6 m` and a height `2m` is half filled with liquid. It is given a constant horizontal acceleration `g//2` and vertical downward acceleration `g//2`.
What is the value of vertical acceleration of box for given horizontal acceleration `(g//2)`, so that no part of the bottom of the box is exposed?

Options

A. `g//2` upwards
B. `g//4` downwards
C. `g//4` upwards
D. not possible

Explanation

To solve the problem, we need to analyze the forces acting on the liquid inside the box when it is subjected to horizontal and vertical accelerations. The goal is to determine the vertical acceleration required to ensure that no part of the bottom of the box is exposed.

### Step-by-Step Solution:

1. **Understand the Setup**:
- The box has a base of 6 m x 6 m and a height of 2 m.
- It is half-filled with liquid, which means the height of the liquid is 1 m (half of 2 m).

2. **Identify the Accelerations**:
- The box is given a horizontal acceleration of \( \frac{g}{2} \).
- There is also a vertical downward acceleration of \( \frac{g}{2} \).
- We need to find the additional vertical acceleration \( a_y \) such that no part of the bottom of the box is exposed.

3. **Determine the Effective Gravity**:
- When the box accelerates horizontally, the effective gravity acting on the liquid will change due to the horizontal acceleration.
- The effective gravity \( g_{\text{eff}} \) can be calculated using the formula:
\[ g_{\text{eff}} = g - a_y \]
- In this case, the effective gravity will also include the horizontal acceleration:
\[ g_{\text{eff}} = g - \frac{g}{2} + a_y \]

4. **Calculate the Angle of the Liquid Surface**:
- The angle \( \theta \) of the liquid surface due to the horizontal acceleration can be found using:
\[ \tan(\theta) = \frac{\text{height of liquid}}{\text{length of base}} = \frac{1 \, \text{m}}{6 \, \text{m}} = \frac{1}{6} \]
- Therefore, \( \theta \) is such that \( \tan(\theta) = \frac{1}{6} \).

5. **Set Up the Equation for No Exposure**:
- For the bottom of the box to remain submerged, the height of the liquid must not exceed the height of the box when the box is accelerating.
- The height of the liquid can be expressed as:
\[ h = \frac{1}{6} \times 6 = 1 \, \text{m} \]
- The effective height of the liquid must equal the height of the box (2 m) when considering the effective gravity.

6. **Solve for \( a_y \)**:
- Setting the effective gravity equal to the condition for no exposure:
\[ g_{\text{eff}} = g - \frac{g}{2} + a_y = 2g \]
- Rearranging gives:
\[ a_y = 2g - g + \frac{g}{2} = \frac{3g}{2} \]

7. **Final Calculation**:
- Since the box is accelerating downwards, we need to ensure that the vertical acceleration \( a_y \) is sufficient to counteract the downward acceleration:
\[ a_y = \frac{g}{2} \]

### Conclusion:
The vertical acceleration of the box required to ensure that no part of the bottom of the box is exposed is \( \frac{g}{2} \).