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 of finding the acceleration of the open cubical tank that spills one-third of its water when accelerated, we can follow these steps:

### Step 1: Understand the Problem
We have an open cubical tank filled with water. When the tank is accelerated horizontally, it spills out one-third of its volume of water. We need to find the acceleration of the tank.

### Step 2: Define the Variables
Let:
- \( H \) = side length of the cube
- \( V \) = total volume of water in the tank = \( H^3 \)
- The volume of water spilled = \( \frac{1}{3} V = \frac{1}{3} H^3 \)

### Step 3: Determine the Height of the Water Surface
When the tank is accelerated, the water surface will tilt. Let \( h \) be the height to which the water surface dips due to acceleration. The volume of the spilled water can be expressed in terms of the height \( h \) and the area of the triangle formed by the tilted water surface.

### Step 4: Calculate the Volume of Water Spilled
The volume of the spilled water can be calculated as:
\[ \text{Volume of spilled water} = \text{Area of triangle} \times H \]
The area of the triangle formed by the tilted water surface is:
\[ \text{Area} = \frac{1}{2} \times h \times H \]
Thus, the volume of spilled water becomes:
\[ \text{Volume} = \frac{1}{2} \times h \times H^2 \]

### Step 5: Set Up the Equation
From the previous steps, we can set up the equation:
\[ \frac{1}{3} H^3 = \frac{1}{2} h H^2 \]
Dividing both sides by \( H^2 \):
\[ \frac{1}{3} H = \frac{1}{2} h \]
Thus, we can express \( h \) in terms of \( H \):
\[ h = \frac{2}{3} H \]

### Step 6: Relate the Angle and Acceleration
The angle \( \theta \) that the water surface makes with the horizontal can be related to the acceleration \( a \) and gravitational acceleration \( g \) using:
\[ \tan \theta = \frac{h}{H} \]
Substituting \( h \):
\[ \tan \theta = \frac{\frac{2}{3} H}{H} = \frac{2}{3} \]

### Step 7: Find the Acceleration
Using the relationship \( \tan \theta = \frac{a}{g} \):
\[ \frac{a}{g} = \frac{2}{3} \]
Thus, the acceleration \( a \) can be expressed as:
\[ a = \frac{2}{3} g \]

### Step 8: Conclusion
The acceleration of the tank is:
\[ \boxed{\frac{2g}{3}} \]