An open cubical tank was initially fully filled with water. When the tank was accelerated on a horizontal plane along one of its side it was found that one thrid of volume of water was spilled out. The acceleration was

Options

A. g/3
B. 2g/3
C. g/2
D. g/4

Answer

To solve the problem step by step, we will analyze the situation of the cubical tank filled with water and how the acceleration affects the water level.

### Step 1: Understand the initial conditions
The tank is a cube with side length \( L \). The initial volume of water in the tank is:
\[ V = L^3 \]

### Step 2: Analyze the effect of acceleration
When the tank is accelerated horizontally, the water inside will tilt, creating an angle \( \theta \) with respect to the horizontal. One-third of the volume of water spills out, which means that two-thirds of the volume remains in the tank.

### Step 3: Calculate the remaining volume of water
Since one-third of the volume spills out, the remaining volume \( V' \) is:
\[ V' = V - \frac{V}{3} = \frac{2V}{3} = \frac{2L^3}{3} \]

### Step 4: Determine the geometry of the water
When the tank is tilted, the water forms a triangular prism shape on one side. The height of the water on the side where it is lower can be represented as \( L - x \), where \( x \) is the horizontal distance from the bottom of the tank to the new water level.

### Step 5: Calculate the volume of the remaining water
The volume of the remaining water can be expressed as the sum of the volume of the triangular part and the rectangular part:
1. The triangular part has a base of \( L \) and height \( L - x \):
\[ \text{Volume of triangular part} = \frac{1}{2} \times \text{base} \times \text{height} \times \text{width} = \frac{1}{2} \times L \times (L - x) \times L = \frac{L^2 (L - x)}{2} \]
2. The rectangular part has a height \( x \):
\[ \text{Volume of rectangular part} = L \times x \times L = L^2 x \]

Combining these volumes gives:
\[ V' = \frac{L^2 (L - x)}{2} + L^2 x = \frac{L^2 (L - x + 2x)}{2} = \frac{L^2 (L + x)}{2} \]

### Step 6: Set up the equation for volume
We know that the remaining volume \( V' \) is equal to \( \frac{2L^3}{3} \):
\[ \frac{L^2 (L + x)}{2} = \frac{2L^3}{3} \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ 3L^2 (L + x) = 4L^3 \]
Dividing by \( L^2 \) (assuming \( L \neq 0 \)):
\[ 3(L + x) = 4L \]
\[ 3x = 4L - 3L = L \]
\[ x = \frac{L}{3} \]

### Step 8: Relate \( x \) to the angle \( \theta \)
Using the triangle formed by the water level, we have:
\[ \tan \theta = \frac{L - x}{L} = \frac{L - \frac{L}{3}}{L} = \frac{\frac{2L}{3}}{L} = \frac{2}{3} \]

### Step 9: Use the relationship between acceleration and angle
From the relationship \( \tan \theta = \frac{A}{g} \):
\[ \frac{A}{g} = \frac{2}{3} \]
Thus, the acceleration \( A \) is:
\[ A = \frac{2g}{3} \]

### Final Answer
The acceleration of the tank is:
\[ \boxed{\frac{2g}{3}} \]