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`.
Water pressure at the bottomof centre of the box is equal to (atmospheric pressure `=10^(5)N//m^(2)` density of water `=1000 kg//m^(3),g=10m//s^(2)`)

Options

A. `1.1MPa`
B. `0.11MPa`
C. `0.101MPa`
D. `0.011MPa`

Explanation

To solve the problem, we need to determine the water pressure at the bottom of the container when it is subjected to both horizontal and vertical accelerations.

### Step-by-Step Solution:

1. **Understand the Effective Gravity**:
The effective gravity (`g_eff`) in a liquid when the container is accelerated both horizontally and vertically can be calculated using vector addition. The horizontal acceleration is `g/2` and the vertical acceleration is also `g/2`.

\[ g_{eff} = \sqrt{(g/2)^2 + (g/2)^2} = \sqrt{(g^2/4) + (g^2/4)} = \sqrt{(g^2/2)} = \frac{g}{\sqrt{2}} \]

2. **Calculate the Depth of the Liquid**:
The container is half-filled with liquid. Given the height of the container is `2 m`, the depth of the liquid (`h`) is:

\[ h = \frac{2}{2} = 1 \text{ m} \]

3. **Calculate the Pressure at the Bottom**:
The pressure at the bottom of the container due to the liquid is given by the hydrostatic pressure formula:

\[ P = P_{atm} + \rho g_{eff} h \]

where:
- \( P_{atm} = 10^5 \, \text{N/m}^2 \) (atmospheric pressure)
- \( \rho = 1000 \, \text{kg/m}^3 \) (density of water)
- \( g_{eff} = \frac{g}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, \text{m/s}^2 \)
- \( h = 1 \, \text{m} \)

Substituting the values:

\[ P = 10^5 + 1000 \cdot \left(\frac{10}{\sqrt{2}}\right) \cdot 1 \]

\[ P = 10^5 + 1000 \cdot \frac{10}{\sqrt{2}} = 10^5 + \frac{10000}{\sqrt{2}} \approx 10^5 + 7071.07 \]

\[ P \approx 10^5 + 7071.07 \approx 107071.07 \, \text{N/m}^2 \]

4. **Final Answer**:
The water pressure at the bottom of the center of the box is approximately:

\[ P \approx 107071.07 \, \text{N/m}^2 \]