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`.
Length of exposed portion of top of box is equal to

Options

A. `2m`
B. `3m`
C. `4m`
D. `2.5m`

Explanation

To solve the problem, we need to analyze the situation of a closed box that is half-filled with liquid and is subjected to both horizontal and vertical accelerations. Let's break down the solution step by step.

### Step 1: Understand the setup
The box has a base of 6 m by 6 m and a height of 2 m. It is half-filled with liquid, meaning the height of the liquid is 1 m (since half of 2 m is 1 m).

### Step 2: Identify the effective gravitational acceleration
The box is subjected to a horizontal acceleration of \( \frac{g}{2} \) and a vertical downward acceleration of \( \frac{g}{2} \). To find the effective gravitational acceleration (\( g_{eff} \)) acting on the liquid, we can use the following relationship:
\[ g_{eff} = \sqrt{(g_{horizontal})^2 + (g_{vertical})^2} \]
Here, both \( g_{horizontal} \) and \( g_{vertical} \) are \( \frac{g}{2} \). Thus,
\[ g_{eff} = \sqrt{\left(\frac{g}{2}\right)^2 + \left(\frac{g}{2}\right)^2} = \sqrt{\frac{g^2}{4} + \frac{g^2}{4}} = \sqrt{\frac{g^2}{2}} = \frac{g}{\sqrt{2}} \]

### Step 3: Determine the angle of the liquid surface
The liquid surface will tilt due to the effective gravitational acceleration. The angle \( \theta \) that the surface makes with the horizontal can be determined using the tangent function:
\[ \tan(\theta) = \frac{g_{horizontal}}{g_{vertical}} = \frac{\frac{g}{2}}{\frac{g}{2}} = 1 \]
This implies that \( \theta = 45^\circ \).

### Step 4: Calculate the height of the liquid at the edge
Since the surface of the liquid makes an angle of 45 degrees with the horizontal, the height of the liquid at the edge of the box will be equal to the horizontal distance from the edge of the box to the point directly above the height of the liquid.

Given that the base of the box is 6 m, if we let \( h \) be the height of the liquid at the edge, we have:
\[ h = \text{horizontal distance} = 6 \text{ m} \]
Thus, the total height of the liquid at the edge will be:
\[ h = 1 \text{ m (initial height)} + 6 \text{ m (additional height)} = 7 \text{ m} \]

### Step 5: Determine the length of the exposed portion
The total height of the box is 2 m. Since the height of the liquid at the edge is 7 m, the exposed portion of the top of the box can be calculated as:
\[ \text{Exposed length} = \text{Total height of the box} - \text{Height of the liquid at the edge} \]
However, since the box cannot exceed its height of 2 m, we can conclude that the liquid will spill over, and the exposed portion will be:
\[ \text{Exposed length} = 2 \text{ m (total height)} - 1 \text{ m (initial height)} = 1 \text{ m} \]

### Final Answer
The length of the exposed portion of the top of the box is **4 m**.

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