Question

A cubical container is completely filled with a liquid as shown in figure. Now if it starts moving with an acceleration a = 5 m/s², find out how much water will spill out from the tank.

• A. 31.25 m³
• B. 62.5 m³
• C. 625 m³
• D. None of these

Answer 1

Answer: (A)

31.25 m³

Answer 2

When a liquid in a container accelerates horizontally, its free surface becomes inclined. The angle of inclination θ with the horizontal is given by the formula:

$\tan(\theta) = \frac{a}{g}$

Given: Acceleration a = 5 m/s² The side length of the cubical container L = 5 m. We assume the acceleration due to gravity g = 10 m/s².

1. Calculate the angle of inclination: $\tan(\theta) = \frac{5 \text{ m/s}^2}{10 \text{ m/s}^2} = 0.5$

2. Determine the shape of the spilled liquid: Since the tank is initially completely filled, when it accelerates to the right, the liquid will pile up at the back wall (left side) and spill out. The liquid level at the front wall (right side) will drop. The new free surface of the liquid will pass through the top edge of the back wall (where it spills) and extend downwards towards the front wall. Let h be the vertical drop in the liquid level from the top of the container at the front wall. The horizontal distance across which this drop occurs is the length of the container L = 5 m. From the geometry of the tilted surface: $\tan(\theta) = \frac{h}{L}$ Substituting the values: $0.5 = \frac{h}{5 \text{ m}}$ $h = 0.5 \times 5 \text{ m} = 2.5 \text{ m}$ This h is the height of the triangular wedge of liquid that spills out.

3. Calculate the volume of the spilled liquid: The spilled liquid forms a wedge (a triangular prism). The base of this prism is a right-angled triangle with base L = 5 m and height h = 2.5 m. The depth of this prism (perpendicular to the base triangle) is also the side length of the cube, L = 5 m. The volume of the spilled liquid V_spilled is the volume of this triangular prism: $V_{\text{spilled}} = (\text{Area of base triangle}) \times (\text{depth})$ $V_{\text{spilled}} = \left( \frac{1}{2} \times \text{base} \times \text{height} \right) \times \text{depth}$ $V_{\text{spilled}} = \left( \frac{1}{2} \times L \times h \right) \times L$ $V_{\text{spilled}} = \frac{1}{2} \times L^2 \times h$ Substitute the values L = 5 m and h = 2.5 m: $V_{\text{spilled}} = \frac{1}{2} \times (5 \text{ m})^2 \times (2.5 \text{ m})$ $V_{\text{spilled}} = \frac{1}{2} \times 25 \text{ m}^2 \times 2.5 \text{ m}$ $V_{\text{spilled}} = \frac{1}{2} \times 62.5 \text{ m}^3$ $V_{\text{spilled}} = 31.25 \text{ m}^3$

The volume of water that will spill out from the tank is 31.25 m³.