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Mensuration - Volume and Capacity

Grade 8CBSE

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

🔑Concepts

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Volume refers to the amount of space occupied by a three-dimensional object. For a cuboid, it is the product of its length, breadth, and height (V=l×b×hV = l \times b \times h). Capacity is the volume of substance that a container can hold.

A 3D cuboid showing length l, breadth b, and height h.
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The volume of a cube is calculated by cubing the length of its side (V=a3V = a^3). Since all edges are equal, it is the simplest three-dimensional regular shape for volume calculation.

A cube with all sides marked as a.
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The volume of a cylinder is found by multiplying the area of its circular base (πr2\pi r^2) by its height (hh). Thus, V=πr2hV = \pi r^2 h.

A cylinder with radius r and height h indicated.
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Units of capacity are interlinked with units of volume. Commonly, 1000cm3=1L1000 cm^3 = 1 L and 1m3=1000L1 m^3 = 1000 L (which is 1kL1 kL). 1cm31 cm^3 is exactly equivalent to 1mL1 mL.

📐Formulae

Volume of a Cuboid = l×b×hl \times b \times h

Volume of a Cube = a3a^3 (where aa is the side length)

Volume of a Cylinder = πr2h\pi r^2 h (where rr is radius and hh is height)

Area of base of a Cuboid = l×bl \times b

Area of base of a Cylinder = πr2\pi r^2

1cm3=1mL1 cm^3 = 1 mL

1000cm3=1L1000 cm^3 = 1 L

1m3=1000L=1,000,000cm31 m^3 = 1000 L = 1,000,000 cm^3

💡Examples

Problem 1:

Find the volume of a cuboidal stone slab that is 2m2 m long, 1.5m1.5 m wide, and 0.2m0.2 m thick.

Solution:

Given: Length l=2ml = 2 m, Breadth b=1.5mb = 1.5 m, Height (thickness) h=0.2mh = 0.2 m. Using the formula for Volume of a Cuboid: V=l×b×hV = l \times b \times h V=2×1.5×0.2V = 2 \times 1.5 \times 0.2 V=3×0.2=0.6m3V = 3 \times 0.2 = 0.6 m^3. The volume of the stone slab is 0.6m30.6 m^3.

Explanation:

To find the volume, we identify the three dimensions of the cuboid and multiply them. Since all units are already in meters, the resulting volume is in cubic meters.

Problem 2:

A cylindrical tank has a base radius of 70cm70 cm and a height of 2m2 m. Find the capacity of the tank in liters.

Solution:

Given: Radius r=70cm=0.7mr = 70 cm = 0.7 m, Height h=2mh = 2 m. Step 1: Calculate Volume in m3m^3: V=πr2hV = \pi r^2 h V=227×(0.7)2×2V = \frac{22}{7} \times (0.7)^2 \times 2 V=227×0.49×2V = \frac{22}{7} \times 0.49 \times 2 V=22×0.07×2=3.08m3V = 22 \times 0.07 \times 2 = 3.08 m^3. Step 2: Convert to Liters: Since 1m3=1000L1 m^3 = 1000 L, Capacity =3.08×1000=3080L= 3.08 \times 1000 = 3080 L.

Explanation:

First, ensure units are consistent by converting the radius to meters. Use the cylinder volume formula to find the space in cubic meters, then multiply by 10001000 to find the capacity in liters.

Problem 3:

Find the volume of a cube whose total surface area is 600cm2600 cm^2.

A cube with side length 'a' to be determined.

Solution:

Total Surface Area of a cube = 6a26 a^2 Given, 6a2=6006 a^2 = 600 a2=6006=100a^2 = \frac{600}{6} = 100 a=100=10cma = \sqrt{100} = 10 cm Volume of the cube = a3a^3 V=10×10×10=1000cm3V = 10 \times 10 \times 10 = 1000 cm^3

Explanation:

First, we use the surface area formula to find the side length of the cube. Once the side length is known, we cube it to find the total volume.

Problem 4:

A cylindrical well is 20m20 m deep and has a diameter of 7m7 m. Find the volume of earth dug out to make the well.

A cylindrical well showing diameter 7m and depth 20m.

Solution:

Diameter = 7m7 m, so Radius (rr) = 72=3.5m\frac{7}{2} = 3.5 m Depth (hh) = 20m20 m Volume of earth = Volume of cylinder = πr2h\pi r^2 h V=227×3.5×3.5×20V = \frac{22}{7} \times 3.5 \times 3.5 \times 20 V=227×72×72×20V = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 V=11×7×10=770m3V = 11 \times 7 \times 10 = 770 m^3

Explanation:

The 'earth dug out' represents the volume of the cylindrical hole. We calculate this using the radius (half of diameter) and the depth as the height.