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How Big How Heavy - Archimedes Principle Concept (Displacement)

Grade 5CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Volume is defined as the amount of space an object occupies. You can visualize the volume of a solid box by imagining how many unit cubes, each measuring 1 cm×1 cm×1 cm1 \text{ cm} \times 1 \text{ cm} \times 1 \text{ cm}, are needed to fill it completely.

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The Principle of Displacement explains that when an object is immersed in a liquid, it pushes the liquid out of its way to occupy that space. Visually, this is seen as the water level rising in a container when an object like a stone is dropped into it.

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For irregular objects like stones or marbles, we cannot use a ruler to measure volume. Instead, we use a measuring cylinder. The volume of the object is exactly equal to the volume of water it displaces (the 'rise' in the water level).

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Archimedes' Principle (Simplified for Grade 5): When an object is fully submerged in water, the Volume of the Object is equal to the Volume of the water that overflows or rises. Imagine a tub filled to the brim; if you step in, the water that spills out is equal to the space your body takes up.

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There is a direct relationship between solid volume and liquid capacity: 1 cubic centimeter1 \text{ cubic centimeter} (1 cm31 \text{ cm}^{3}) of solid space is equal to 1 milliliter1 \text{ milliliter} (1 ml1 \text{ ml}) of liquid.

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We can calculate the volume of displaced water using the formula: Final Level−Initial LevelFinal \text{ Level} - Initial \text{ Level}. If you see a cylinder where the water was at 30 ml30 \text{ ml} and it moves to 50 ml50 \text{ ml} after adding a ball, the ball's volume is the 20 ml20 \text{ ml} difference.

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The 'Eureka Can' is a visual demonstration tool. It is a container with a spout at the top. When an object is lowered into a Eureka can filled to the spout, the displaced water flows out into a separate measuring beaker, making it easy to see exactly how much space the object occupies.

📐Formulae

Volume of a Cuboid=Length×Breadth×HeightVolume \text{ of a Cuboid} = Length \times Breadth \times Height

Volume of a Cube=side×side×sideVolume \text{ of a Cube} = side \times side \times side

Volume of Displaced Water=Final Water Level−Initial Water LevelVolume \text{ of Displaced Water} = Final \text{ Water Level} - Initial \text{ Water Level}

1 cm3=1 ml1 \text{ cm}^{3} = 1 \text{ ml}

1 Liter=1000 ml=1000 cm31 \text{ Liter} = 1000 \text{ ml} = 1000 \text{ cm}^{3}

💡Examples

Problem 1:

Ravi has a measuring cylinder filled with 150 ml150 \text{ ml} of water. He drops 55 identical glass marbles into the cylinder, and the water level rises to 175 ml175 \text{ ml}. What is the volume of a single marble?

Solution:

  1. Total volume of water displaced = Final Level−Initial LevelFinal \text{ Level} - Initial \text{ Level} \ 2. Total displacement = 175 ml−150 ml=25 ml175 \text{ ml} - 150 \text{ ml} = 25 \text{ ml} \ 3. Volume of 55 marbles = 25 ml25 \text{ ml} \ 4. Volume of 11 marble = 255=5 ml\frac{25}{5} = 5 \text{ ml} \ 5. Since 1 ml=1 cm31 \text{ ml} = 1 \text{ cm}^{3}, the volume of one marble is 5 cm35 \text{ cm}^{3}.

Explanation:

First, find the total volume displaced by all marbles by subtracting the initial water level from the final level. Then, divide by the number of marbles to find the volume of just one.

Problem 2:

A metal box has a length of 4 cm4 \text{ cm}, a breadth of 3 cm3 \text{ cm}, and a height of 2 cm2 \text{ cm}. If this box is dropped into a large bucket of water, how much water (in milliliters) will it displace?

Solution:

  1. Volume of the metal box = l×b×hl \times b \times h \ 2. Volume = 4 cm×3 cm×2 cm4 \text{ cm} \times 3 \text{ cm} \times 2 \text{ cm} \ 3. Volume = 24 cm324 \text{ cm}^{3} \ 4. Using the conversion 1 cm3=1 ml1 \text{ cm}^{3} = 1 \text{ ml}, the box will displace 24 ml24 \text{ ml} of water.

Explanation:

To find the displacement, we first calculate the volume of the solid object using its dimensions. Since the object is solid, it will displace a volume of water exactly equal to its own calculated volume.