EXAMPLES ON VOLUME OF SPHERE AND HEMISPHERE

Example 1 :

Find the mass of 200 steel spherical ball bearings, each of which has radius 0.7 cm, given that the density of steel is 7.95 g/cm³. (Mass = Volume x Density)

Solution :

radius of spherical ball = 0.7 cm 

Volume of one spherical ball  =  (4/3) Π r³

=  (4/3) (22/7)  ⋅  0.7 ⋅  0.7 ⋅  0.7

=  4.312/3

=  1.437

Volume of 200 steel spherical ball  =  200 ⋅ 1.437

=  287.46 cm³

1 cm³ = 7.95 g

Therefore mass of 200 spherical ball bearings

=  287.46 (7.95)

=  2285.307 gram

1000 gram  =  1 kg

=  2285.307/1000

=  2.29 kg

Volume of 200 spherical balls = 2.29 kg

Example 2 :

The outer and inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume.

Solution :

From this information we have to find the volume

Outer radius (R)  =  12 cm

Inner radius (r)  =  10 cm

Volume of hollow sphere  =  (4/3) Π (R³-r³)

=  (4/3)(22/7) (12³-10³)

=  (88/21) (1728-1000)

=  (88/21) (728)

=  (64064/21)

=  3050.67 cm³

Volume of hollow sphere is 3050.67 cm³.

Example 3 :

The volume of a solid hemisphere is 1152 Π cu.cm. Find its curved surface area.

Solution :

Volume of hollow sphere  =  1152 Π

(2/3) Π r³  =  1152 Π

r³  =  1152 Π (3/2Π)

r³  =  (576 x 3)/2

r³  =  1728

r  =  ∛1728

r  =  12 cm

Curved surface area  =  2Πr²

=  2Π(12)²

=  2Π(144)

=  288Π cm³

Curved surface area  =  288Π cm³

Example 4 :

Find the volume of the largest right circular cone that can be cut of a cube whose edge is 14 cm.

Solution :

Since it is cube length of all sides will be equal that is 14 cm. Diameter and height of cone are 14 cm.

r  =  14/2  ==>  7

h  =  14 cm

Volume of cone  =  (1/3) Π r² h 

=  (1/3) ⋅ (22/7)  ⋅ 7² ⋅ 14

=  (1/3) ⋅ 22 ⋅ 49 ⋅ 2  

=  (49 ⋅ 44)/3

=  2156/3

=  718.67 cm³

Volume of cone is 718.67 cm³.

Example 5 :

The radius of a spherical balloon increase from 7 cm to 14 cm as air is being pumped into it. Find the ratio of volumes of the balloon in the two cases.

Solution :

Let r₁ and r₂ are the radii of two spherical balloon

r₁ : r₂  =  7 : 14

Volume of one spherical balloon  =  (4/3) Π r³

(4/3) Π (7)³ :  (4/3) Π 14³

7³ : 14³

7 ⋅ 7 ⋅ 7 : 14 ⋅ 14 ⋅ 14

1 : 8

So, the required ratio is 1 : 8.

Example 6 :

A sphere has radius 12.6 cm. Find the volume and surface area. Give your answers to 3 significant figures. Take π = 3.142

Solution :

Radius = 12.6 cm

Volume of sphere = (4/3) Π r³

= (4/3) Π (12.6)³

= (4/3) x 3.14 x 2000.37

= 8374.90

Approximately 8375 cm³

Example 7 :

Find the radius of an opened hemisphere whose surface area is 1762 cm² Give your answers to 3 significant figures. Take π = 3.142

Solution :

Surface area of hemisphere = 2Π r²

Given that, 2 Π r² = 1762 cm²

2 x 3.14 x r² = 1762

r² = 1762/(2 x 3.14)

r² = 280.57

r = 16.75 cm

So, the radius of the hemisphere is 17 cm approximately.

Example 8 :

Find the radius of a sphere whose volume is 288π cm³ Give your answers to 3 significant figures.

Solution :

Volume of sphere = (4/3) Π r³

(4/3) Π r³ = 288π

(4/3) r³ = 288

r³ = 288 x (3/4)

r³ = 72 x 3

r³ = 216

r = 6 cm

So, the radius of the sphere is 6 cm.

Example 9 :

A bowl has the form of a hollow hemisphere of radius 8.4 cm. Find the external surface area and the volume of the bowl. Give your answers to 3 significant figures. Take π = 3.142

Solution :

Radius = 8.4 cm

External surface area = 3Π r²

= 3 x 3.14 x (8.4)²

= 664.67 cm²

Example 10 :

A hemispherical bowl has a radius of 15 cm. If it is filled completely with water and covered with a lid,

(a) find the volume of the water

(b) find the surface area of the bowl (including the lid). Give your answers to 3 significant figures. Take π = 3.142

Solution :

a)  Radius = 15 cm

Volume of water in the hemispherical bowl = (2/3) Π r³

= (2/3) x 3.14 x 15³

= (2/3) x 3.14 x 3375

= 7065 cm³

b) Surface area of bowl = 2Π r²

= 2 x 3.14 x 15²

= 1413 cm²

Example 11 :

A metal sphere has a radius of 8 cm. Find its volume and surface area. Give your answers to 3 significant figures. Take π = 3.142

Solution :

Radius = 8 cm

a)  Volume of water in the hemispherical bowl = (4/3) Π r³

= (4/3) x 3.14 x 8³

= (2/3) x 3.14 x 512

= 1071.78 cm³

b) Surface area of bowl = 4Π r²

= 4 x 3.14 x 8²

= 803.84 cm²

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