9.6. Volume and Capacity

There is not much difference between these two words.

(a) Volume refers to the amount of space occupied by an object.

(b) Capacity refers to the quantity that a container holds.

In this chunk, we differentiate between two related concepts: volume and capacity. Volume is about the space that an object takes up, whereas capacity refers to how much liquid or substance a container can hold. For example, if you have a box filled with toys, the volume is the space the box occupies, while the capacity is how many toys can fit inside the box.

Think of a big balloon filled with air. The volume of the balloon is how much space the air takes inside it, while the capacity would be how much air the balloon could hold if it were empty.

Note: If a water tin holds 100 cm3 of water then the capacity of the water tin is 100 cm3.

Capacity is also measured in terms of litres. The relation between litre and cm3 is, 1 mL = 1 cm3, 1 L = 1000 cm3. Thus, 1 m3 = 1000000 cm3 = 1000 L.

This chunk provides important relationships between different units of measurement. For example, a small amount of liquid measured in millilitres (mL) is equivalent to cubic centimeters (cm3). Also, 1 litre is equal to 1000 cubic centimeters, and this gives us a straightforward path to convert larger volumes of liquid into more manageable units.

Imagine filling a bottle with water. If you know that 1 litre of water is the same as 1000 cm3, then if you fill your water bottle with 500 cm3 of water, you can say that you have filled it with half a litre.

Example 8: Find the height of a cuboid whose volume is 275 cm3 and base area is 25 cm2.

Solution: Volume of a cuboid = Base area × Height

Hence height of the cuboid = Volume/Base area = 275/25 = 11 cm

Height of the cuboid is 11 cm.

This example illustrates how you can find the height of a cuboid if you know its volume and the area of its base. The formula used here states that the volume is the product of the base area and the height. By rearranging this formula, we can solve for height, making it easy to find what we're looking for.

Imagine a shoebox filled with shoes. If you know how much space all the shoes take together (the volume), and you already know the area of the base of the shoebox, you can figure out how tall the shoebox is. In this case, finding the height is just like figuring out how deep the shoes are packed into the box.

Example 9: A godown is in the form of a cuboid of measures 60 m × 40 m × 30 m. How many cuboidal boxes can be stored in it if the volume of one box is 0.8 m3?

Solution: Volume of one box = 0.8 m3 Volume of godown = 60 × 40 × 30 = 72000 m3

Volume of the godown / Volume of one box = 72000 / 0.8 = 90000

Hence, the number of cuboidal boxes that can be stored in the godown is 90,000.

In this example, we calculate how many smaller boxes can fit into a larger space, or godown, which is a cuboidal structure. First, we find the total volume of the godown by multiplying its dimensions. Then, we divide this volume by the volume of one smaller box to get the total number of boxes that can fit inside.

Picture a large warehouse storing small boxes of products. If the warehouse can store a total of 72,000 m3 of product and each box takes up 0.8 m3, you can easily calculate how many boxes can fit by simply dividing the total warehouse capacity by the capacity of one box. It's like figuring out how many jars of jam can fit in a pantry!

Example 10: A rectangular paper of width 14 cm is rolled along its width and a cylinder of radius 20 cm is formed. Find the volume of the cylinder.

Solution: Height of the cylinder = h = 14 cm Radius = r = 20 cm

Volume of the cylinder = V = π r2 h = (22/7) × 20 × 20 × 14 = 17600 cm3.

Hence, the volume of the cylinder is 17600 cm3.

This example explores how to compute the volume of a cylinder formed by rolling a piece of rectangular paper. The width of the paper becomes the cylinder's height, and the radius is given. By using the formula for the volume of a cylinder, we substitute the known values to find the volume.

Think of wrapping a gift with a piece of wrapping paper. When you roll the wrapping paper into a cylindrical shape around the gift, the height of the cylinder is how wide the paper is, and the radius is how round the cylinder is. You can calculate how much space the cylindrical gift takes up just as you would for any other object!

Example 11: A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.

Solution: Length of the paper becomes the perimeter of the base of the cylinder and width becomes height. Let radius of the cylinder = r and height = h

Perimeter of the base of the cylinder = 2πr = 11 or 2πr = 11 Therefore, r = 11/(2π) m.

Volume of the cylinder = V = πr2h. Hence, the volume of the cylinder is 38.5 cm3.

In this example, a rectangular paper is utilized to create a cylinder. The perimeter of the base of the cylinder is formulated using the length of the paper. Knowing the height allows the calculation of the volume by substituting the radius found from the perimeter back into the cylinder volume formula.

Imagine rolling up a piece of parchment paper to create a scroll. The way the parchment is folded gives it both height and circularity, similar to how you can calculate its volume based on how tightly it's rolled.

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