A solid cube of side 8 cm is dropped into a rectangular container of length 16 cm, breadth 8 cm and height 15 cm which is partly filled with water. If the cube is completely submerged, then the rise of water level (in cm) is:
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Calculating Water Level Rise When a Cube is Submerged
This problem involves understanding the concept of volume displacement. When an object is submerged in a liquid, it pushes aside, or displaces, a volume of liquid equal to the volume of the submerged part of the object. In this case, a solid cube is completely submerged in water within a rectangular container. The volume of water displaced will cause the water level in the container to rise.
Given Dimensions
- Side of the solid cube = 8 cm
- Length of the rectangular container = 16 cm
- Breadth of the rectangular container = 8 cm
- Height of the rectangular container = 15 cm
Volume of the Solid Cube
First, we need to calculate the volume of the solid cube. The formula for the volume of a cube is side × side × side.
Volume of cube $= \text{side}^3$
Volume of cube $= (8 \text{ cm})^3$
Volume of cube $= 8 \times 8 \times 8 \text{ cm}^3$
Volume of cube $= 512 \text{ cm}^3$
Volume of Displaced Water
When the cube is completely submerged, the volume of water displaced is equal to the volume of the cube.
Volume of displaced water = Volume of cube
Volume of displaced water $= 512 \text{ cm}^3$
Relating Displaced Volume to Water Level Rise
The displaced water occupies a volume within the rectangular container. This volume forms a rectangular prism with the length and breadth of the container as its base dimensions and the rise in water level as its height. Let 'h' be the rise in water level in centimeters.
The volume of this displaced water can also be calculated using the dimensions of the container and the rise in water level:
Volume of displaced water $= \text{Length of container} \times \text{Breadth of container} \times \text{Rise in water level (h)}$
We know the volume of displaced water is 512 cm<sup>3</sup>, the length of the container is 16 cm, and the breadth is 8 cm. We can set up an equation to find 'h'.
$512 \text{ cm}^3 = 16 \text{ cm} \times 8 \text{ cm} \times h$
$512 \text{ cm}^3 = (16 \times 8) \text{ cm}^2 \times h$
$512 \text{ cm}^3 = 128 \text{ cm}^2 \times h$
Solving for the Rise in Water Level
Now, we can solve the equation for 'h':
$h = \frac{512 \text{ cm}^3}{128 \text{ cm}^2}$
$h = 4 \text{ cm}$
Therefore, the rise of the water level in the container is 4 cm.
Summary of Calculation
| Item | Dimension/Value | Calculation |
|---|---|---|
| Cube Side | 8 cm | - |
| Container Length | 16 cm | - |
| Container Breadth | 8 cm | - |
| Volume of Cube | $8^3 \text{ cm}^3$ | 512 cm<sup>3</sup> |
| Volume of Displaced Water | 512 cm<sup>3</sup> | Equals Volume of Cube |
| Displaced Water Volume Formula | Length × Breadth × h | $16 \times 8 \times h = 128h$ |
| Equation for h | $128h = 512$ | - |
| Rise in Water Level (h) | $\frac{512}{128} \text{ cm}$ | 4 cm |
Revision Table: Solid Cube and Water Displacement
| Concept | Description | Formula/Principle |
|---|---|---|
| Volume of a Cube | Space occupied by a cube. | Side × Side × Side ($s^3$) |
| Volume Displacement | When an object is submerged in a fluid, it pushes aside a volume of fluid equal to its own volume (or the submerged part's volume). | Volume of Object = Volume of Displaced Fluid (for fully submerged object) |
| Volume of a Rectangular Prism | Space occupied by a rectangular box shape. | Length × Breadth × Height ($l \times b \times h$) |
| Water Level Rise in Container | The increase in height of the water level due to a submerged object. The volume of this increased level section is the volume of the displaced water. | Volume Displaced = Base Area of Container × Rise in Height |
Additional Information: Archimedes' Principle
The concept of volume displacement is closely related to Archimedes' Principle. While the principle primarily focuses on buoyant force (the upward force exerted by the fluid), it stems from the fact that a submerged or partially submerged object displaces a volume of fluid. The weight of the displaced fluid is equal to the buoyant force.
In this specific problem, we focused purely on the volume aspect: the volume of the submerged cube is equal to the volume of the displaced water. This displaced water has to go somewhere, and in a container with a fixed base area, it results in a rise in the liquid level.
The height of the container (15 cm) is important only to ensure that the final water level plus the initial water level (which isn't given but must be < 15 cm - 4 cm) does not exceed the container's height, meaning the cube is indeed fully submerged within the container's limits and doesn't cause overflow or partial submersion due to hitting the top.
Important Questions from Solid Figures
A cylindrical tube, open at both ends, is made of a metal sheet which is 0.5 cm thick. Its outer radius is 4 cm and length is 2 m. How much metal (in cm 3) has been used in making the tube?
The volume of a right circular cone is 308 cm 3 and the radius of its base is 7 cm. What is the curved surface area (in cm 2) of the cone? (Take π = \(\frac{22}{7} \) )
The slant height and radius of a right circular cone are in the ratio 29 ∶ 20. If its volume is 4838.4 π cm 3, then its radius is:
Six cubes, each of edge 2 cm, are joined end to end. What is the total surface area of the resulting cuboid in cm 2?
The radii of two cylinders are in the ratio 3 : 4 and their heights are in the ratio 8 : 5. The ratio of their volumes is equal to:
