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`.
The maximum value of water pressure in the box is equal to

Options

A. `1.4MPa`
B. `0.14MPa`
C. `0.104MPa`
D. `0.014MPa`

Explanation

To find the maximum value of water pressure in the box that is half-filled with liquid and subjected to constant horizontal and vertical accelerations, we can follow these steps:

### Step 1: Understand the problem
The box has dimensions of 6 m x 6 m x 2 m and is half-filled with liquid. The box is experiencing a horizontal acceleration of \( \frac{g}{2} \) and a vertical downward acceleration of \( \frac{g}{2} \). We need to find the maximum pressure at the bottom of the box.

### Step 2: Determine the effective gravitational acceleration
When the box is accelerated, the effective gravitational acceleration \( g' \) can be calculated using the formula:
\[ g' = g + a \]
where \( a \) is the acceleration due to the box's motion.

In this case, since the box is accelerating horizontally and vertically downward, we need to consider both components. The effective acceleration will be:
\[ g' = g + \frac{g}{2} = g + \frac{g}{2} = \frac{3g}{2} \]

### Step 3: Calculate the height of the liquid column
Since the box is half-filled with liquid, the height of the liquid column \( h \) is:
\[ h = \frac{2 \, \text{m}}{2} = 1 \, \text{m} \]

### Step 4: Calculate the maximum pressure
The pressure at a depth in a fluid is given by the formula:
\[ P = P_{\text{atm}} + \rho g' h \]
where:
- \( P_{\text{atm}} \) is the atmospheric pressure (approximately \( 10^5 \, \text{Pa} \)),
- \( \rho \) is the density of the liquid (for water, \( \rho \approx 10^3 \, \text{kg/m}^3 \)),
- \( g' \) is the effective gravitational acceleration,
- \( h \) is the height of the liquid column.

Substituting the values:
\[ P = 10^5 + (10^3) \left(\frac{3g}{2}\right) (1) \]
Using \( g \approx 10 \, \text{m/s}^2 \):
\[ P = 10^5 + (10^3) \left(\frac{3 \times 10}{2}\right) (1) \]
\[ P = 10^5 + (10^3) \left(15\right) \]
\[ P = 10^5 + 15000 \]
\[ P = 115000 \, \text{Pa} \]

### Step 5: Convert pressure to MegaPascal
To convert the pressure from Pascal to MegaPascal:
\[ P = \frac{115000}{10^6} = 0.115 \, \text{MPa} \]

### Final Answer
The maximum value of water pressure in the box is:
\[ \boxed{0.115 \, \text{MPa}} \]