Class 10 Volume of a Combination of Solids

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class-10 chapter-13

Volume of a Combination of Solids

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♦ Volume of a Combination of Solids

Volume of a Combination of Solids

Here, we shall see how to calculate their volumes. It may be noted that in calculating the surface area, we have not added the surface areas of the two constituents, because some part of the surface area disappeared in the process of joining them.

However, this will not be the case when we calculate the volume. The volume of the solid formed by joining two basic solids will actually be the sum of the volumes of the constituents, as we see in the examples below.

Q 3200701618

Shanta runs an industry in a shed which is in the shape of a cuboid
surmounted by a half cylinder (see Fig. 13.12). If the base of the shed is of
dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume
of air that the shed can hold. Further, suppose the machinery in the shed
occupies a total space of `300 m^3`, and there are 20 workers, each of whom
occupy about 0.08 m3 space on an average. Then, how much air is in the
shed? (Take `π = 22/7` )
Class 10 Chapter 13 Example 5

Solution:

The volume of air inside the shed (when there are no people or machinery)
is given by the volume of air inside the cuboid and inside the half cylinder, taken
together.

Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively.
Also, the diameter of the half cylinder is 7 m and its height is `15 m`.

So, the required volume = volume of the cuboid `+1/2` volume of the cylinder

` = [15 xx 7 xx 8 xx 1/2 xx 22/7 xx 7/2 xx 7/2 xx 15] m^3 = 1128.75 m^3`

Next, the total space occupied by the machinery `= 300 m^3`
And the total space occupied by the workers `= 20 × 0.08 m^3 = 1.6 m^3`
Therefore, the volume of the air, when there are machinery and workers

`= 1128.75 – (300.00 + 1.60) = 827.15 m^3`

Q 3210801710

A juice seller was serving his customers using glasses as shown in Fig. 13.13.
The inner diameter of the cylindrical glass was `5 cm`, but the bottom of the glass had a
hemispherical raised portion which reduced the capacity of the glass. If the height of a glass
was `10 cm`, find the apparent capacity of the glass and its actual capacity. (Use `π = 3.14`.)
Class 10 Chapter 13 Example 6

Solution:

Since the inner diameter of the glass` = 5 cm` and height `= 10 cm`,

the apparent capacity of the glass `= πr^2 h`

`= 3.14 × 2.5 × 2.5 × 10 cm^3 = 196.25 cm^3`

But the actual capacity of the glass is less by the volume of the hemisphere at the
base of the glass.

i.e., it is less by `2/3 pi r^3 =2/3 xx 3.14 xx 2.5 xx 2.5 xx 2.5 cm^3 = 32.71 cm^3`

So, the actual capacity of the glass = apparent capacity of glass – volume of the
hemisphere

`= (196.25 – 32.71) cm^3`

`= 163.54 cm^3`

Q 3230801712

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The
height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a
right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy.
(Take `π = 3.14`)
Class 10 Chapter 13 Example 7

Solution:

Let BPC be the hemisphere and ABC be the cone standing on the base
of the hemisphere (see Fig. 13.14). The radius BO of the hemisphere (as well as

of the cone) `= 1/2 xx 4 cm = 2 cm` .

So, volume of the toy ` = 2/3 pi r^3 + 1/3 pi r^2 h`

`= [ 2/3 xx 3.14 xx (2)^3 +1/3 xx 3.14 xx (2)^2 xx 2 ] cm^3 = 25.12 cm^3`

Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of
the base of the right circular cylinder `= HP = BO = 2 cm`, and its height is

`EH = AO + OP = (2 + 2) cm = 4 cm`

So, the volume required = volume of the right circular cylinder – volume of the toy

`= (3.14 × 2^2 × 4 – 25.12) cm^3`

`= 25.12 cm^3`

Hence, the required difference of the two volumes `= 25.12 cm^3`.