Mathematics Volume of a Sphere

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`color{red} ♦` Volume of a Sphere

Volume of a Sphere

Now, let us see how to go about measuring the volume of a sphere.
First, take two or three spheres of different radii, and a container big enough to be able to put each of the spheres into it, one at a time. Also, take a large trough in which you can place the container.

Then, fill the container up to the brim with water [see Fig. 13.30(a)].

Now, carefully place one of the spheres in the container. Some of the water from the container will over flow into the trough in which it is kept [see Fig. 13.30(b)].

Carefully pour out the water from the trough into a measuring cylinder (i.e., a graduated cylindrical jar) and measure the water over flowed [see Fig. 13.30(c)].

Suppose the radius of the immersed sphere is r (you can find the radius by measuring the diameter of the sphere). Then evaluate ` 4/3 π r^3` . Do you find this value almost equal to the measure of the volume over flowed?

Once again repeat the procedure done just now, with a different size of sphere.

Find the radius `R` of this sphere and then calculate the value of ` 4/3 π r^3` . Once again this value is nearly equal to the measure of the volume of the water displaced (over flowed) by the sphere.

What does this tell us? We know that the volume of the sphere is the same as the measure of the volume of the water displaced by it.

By doing this experiment repeatedly with spheres of varying radii, we are getting the same result, namely, the volume of a sphere is equal to ` 4/3 π` times the cube of its radius. This gives us the idea that

Volume of a Sphere ` = 4/3 π r^3`

where `r` is the radius of the sphere.

Later, in higher classes it can be proved also. But at this stage, we will just take it as true.

Since a hemisphere is half of a sphere, can you guess what the volume of a hemisphere will be? Yes, it is `1/2` of ` 4/3 π r^3 = 2/3 π r^3`

So, Volume of a Hemisphere ` = 2/3 π r^3`
where `r` is the radius of the hemisphere.
Q 3245734663

Find the volume of a sphere of radius `11.2 cm`.
Class 9 Chapter 13 Example 17

Required volume `= 4/3 π r^3`

` = 4/3 xx (22)/7 xx 11.2 xx 11.2 xx 11.2 cm^3 = 5887.32 cm^3`
Q 3255734664

A shot-putt is a metallic sphere of radius `4.9 cm`. If the density of the metal is `7.8 g` per `c m^3`, find the mass of the shot-putt.

Class 9 Chapter 13 Example 18

Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere.
Now, volume of the sphere `= 4/3 pi r^3`

` = 4/3 xx (22)/7 xx 4.9 xx 4.9 xx 4.9 cm^3`

`= 493 cm^3` (nearly)

Further, mass of `1 cm^3` of metal is `7.8 g`.

Therefore, mass of the shot-putt `= 7.8 xx 493 g`
`= 3845.44 g = 3.85 k g` (nearly)
Q 3265734665

A hemispherical bowl has a radius of `3.5 cm`. What would be the volume of water it would contain?
Class 9 Chapter 13 Example 19

The volume of water the bowl can contain

` = 2/3 pi r^3`

` = 2/3 xx (22)/7 xx 3.5 xx 3.5 xx 3.5 cm^3 = 89.8 cm^3`