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 angle of the free surface with the horizons is equal to

Options

A. `30^@`
B. `tan^(-1) 2/3`
C. `tan^(-1)` `1/3`
D. `45^@`

Explanation

To find the angle of the free surface of the liquid in the container that is being accelerated both horizontally and vertically, we can follow these steps:

### Step 1: Identify the accelerations
The container is experiencing:
- A horizontal acceleration \( a_x = \frac{g}{2} \) (to the right)
- A vertical downward acceleration \( a_y = \frac{g}{2} \)

### Step 2: Determine the effective gravitational acceleration
When the container is accelerated, the effective gravitational acceleration \( g_{\text{effective}} \) can be calculated by considering both the gravitational force and the pseudo forces due to the acceleration of the container.

The effective gravitational acceleration can be represented as a vector:
- The downward component is \( g - a_y = g - \frac{g}{2} = \frac{g}{2} \)
- The horizontal component is \( a_x = \frac{g}{2} \)

Thus, we have:
- \( g_{\text{effective}} = \sqrt{(g - a_y)^2 + a_x^2} = \sqrt{\left(\frac{g}{2}\right)^2 + \left(\frac{g}{2}\right)^2} \)

### Step 3: Calculate \( g_{\text{effective}} \)
Calculating \( g_{\text{effective}} \):
\[ g_{\text{effective}} = \sqrt{\left(\frac{g}{2}\right)^2 + \left(\frac{g}{2}\right)^2} = \sqrt{\frac{g^2}{4} + \frac{g^2}{4}} = \sqrt{\frac{g^2}{2}} = \frac{g}{\sqrt{2}} \]

### Step 4: Determine the angle of the free surface
The angle \( \theta \) that the free surface makes with the horizontal can be found using the tangent function:
\[ \tan(\theta) = \frac{\text{vertical component}}{\text{horizontal component}} = \frac{g - a_y}{a_x} = \frac{\frac{g}{2}}{\frac{g}{2}} = 1 \]

### Step 5: Calculate \( \theta \)
Since \( \tan(\theta) = 1 \), we can find \( \theta \):
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Conclusion
The angle of the free surface with the horizontal is \( 45^\circ \).

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