1. 10.3 Volume of Combination of Solids
Suresh runs an industry in a shed which is in the shape of a cuboied surmounted by a half cylinder . The base of the shed is of dimensions 7 m X 15 m . and height of the cuboid 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 m3 and there are 20 workers each of whom occupies about 0.08 m3 space on an average. Then how much air is in the shed ?
The volume of air inside the shed ( When there are no people or machinary ) is given by the volume of air inside the cuboid and inside the half cylinder taken together.
15 m
7 m
8 m
The length,breadth and height of the cuboid are 15 m,7 m and 8 m respectively .

2. The total space occupied by the machinery = 300 m3
Also the diameter of the half cylinder is d = 7 m , radius = r = 3.5 m and its height is h = 15 m.
The required Volume = Volume of the cuboid
Volume of the cylinder
The total space occupied by the machinery = 300 m3
The total space occupied by 20 workers
The volume of the air, when there are machinery and workers =

3. Note : In calculating the surface area of combination of solids , We can not add the surface areas of the two solids
Because some part of the surface areas disappears 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.

4. If the diameter of the cross section of a wire is decreased by 5%, by what percentage should the length be increased so that the volume remains the same ?
Try this
Solution : Diameter of the wire be d and radius of the wire be r1 = x
If the diameter of the a wire is decreased by 5% then radius =
Let the volume of the wire ( cylinder ) before decreasing the diameter of the cross section V1 , radius r1 and its length ( height ) h1
Let the volume of the wire ( cylinder ) after decreasing the diameter of the cross section V2 , radius r2 and its length ( height ) h2
Volume remains same after decreasing the diameter of the cross section of wire =
If the diameter of the cross section of a wire is decreased by 5%, by
percentage should the length be increased so that the volume remains the same

5. If surface area of a sphere and cube are equal then
Surface area of a sphere and cube are equal . Then find the ratio of their volumes.
Try this
Solution : Let “a” be the side of cube and surface area of the cube
Let “r” be the radius of the sphere and surface area of the sphere
If surface area of a sphere and cube are equal then
Volume of the cube
Volume of the sphere
Ratio of their volumes

6. Example – 10 Volume of the Solid toy = Volume of the cone
A Solid toy is in the form of a right circular cylinder with hemispherical shape at one end and a cone at the other end . Their common diameter is 4.2 cm and the height of the cylinderical and conical portions are 12 cm and 7 cm respectively. Find the volume of the solid toy.
Example – 10
Solution : height of the cone
Height of the cylinder
Common Diameter of Cone, cylinder , hemisphere = d = 4.2 cm
Common Radius of Cone,cylinder , hemisphere = r = d / 2
Volume of the Solid toy =
Volume of the cone
Volume of the cylinder
Volume of the hemisphere

7. 7 cm
12 cm
2.1 cm

8. Voume of the ice cream in cylinderical container
A cylinderical container is filled with ice cream whose diameter is 12 cm and height is 15 cm . The whole ice cream is distributed to 10 children in equal cones having hemispherical tops. If the height of the conical portion is twice the diameter of its base, find the diameter of the ice cream cone .
Example - 11
h =15 cm
Voume of the ice cream in cylinderical container
d =- 12 cm
Solution : diamete fo the cylinderical container d= 12 cm
Radius of the cylinderical container
Height of the cylinderical container = h =15 cm
Volume of the ice cream in cylinderical container =

9. Volume of the ice cream in a cone
Let radius of the ice cream in cone be
Diameter of cone
the height of the conical portion is twice the diameter of its base
Height of the cone = 2 (diameter of the base ) =
Volume of the ice cream in conical portion
Volume of the ice cream in a cone =
+ Volume of ice cream in hemispherical portion
4x cm
The whole is cream is distributed to No. of children = 10
Volume of the ice cream in cylinderical container
Volume of the ice cream in a cone
Diameter of the cone

10. Solution : ABCD is a cylinder OLM is a cone LMN is a Hemisphere
Example - 12
A solid consisting of a right circular cone standing on a hemisphere , is placed upright in a right circular cylinder full of water and touches the bottom. Find the volume of water left in the cylinder , given that the radius of the cylinder is 3 cm and its height is 6 cm . Thus radius of the hemisphere is 2 cm and the height of the cone is 4 cm.
Solution : ABCD is a cylinder
OLM is a cone
LMN is a Hemisphere
A Solid consisting of a right circular cone standing on hemisphere , is placed upright in a right circular cylinder full of water and touches the bottom
When a solid is immersed in the cylinder full of water , then volume of water displaced from cylinder is equal to the volume of the solid.
Radius of the cylinder = r = 3 cm and height of the cylinder = h = 6 cm
Radius of the hemisphere = r = 2 cm and height of the cone = h = 4 cm
Volume of the cylinder
Volume of the hemisphere

11. The Volume of water displaced from cylinder =
Radius of the Cone = r = 2 cm and height of the cone = h = 4 cm
Volume of the cone
Volume of hemisphere and cone
The Volume of water displaced from cylinder =
Volume of the cylinder
Volume of hemisphere and cone
Volume of the water left in the cylinder =
Volume of the Water in the cylinder
The Volume of water displaced from cylinder

12. Example - 13
A cylinderical pencil is sharpened to produce a perfect cone at one end with no over all loss of its length. The diameter of the pencil is 1 cm and the length of the conical portion is 2 cm . Calculate the volume of the shavings . Give your answer correct to two places if it is in decimal use = 355 / 113
2 cm
2 cm
Solution : diameter of the cylinderical pencil = d = 1 cm
Radius of the cylinderical pencil = r = d / 2 = ½ = 0.5 cm
0.5 cm
0.5 cm
height of the cylinderical pencil sharpened = h = 2 cm
Radius of the sharpened conical shape pencil = r = 0.5 cm
Lenth of the sharpened conical shape pencil = h = 2 cm
Volume of the cylinderical Pencil
Volume of the conical shape of Pencil
Volume of the shavings =
Volume of the cylinderical shape of the pencil
Volume of the conical shape of the pencil
1 cm

13. Exercise – 10.3
1. An iron pillar consists of a cylinderical portion of 2.8 m height and 20 cm diameter and a cone of 42 cm height surmounting it. Find the weight of the pillar if 1 cm3 of iron weighs 7.5 g.
Solution : diameter of cylinderical iron pillar = d = 20 cm
Radius of the cylinderical iron pillar
42cm
Height of the cylinderical Iron pillar
10cm
Radius of cone = Radius of cylinderical iron pillar = 10 cm
Height of the cone = h2 = 42 cm
2.8 m
Volume of the cylinder
Volume of the Iron pillar =
Volume of the cylinder
20 cm
gms
Weight of the pillar if 1 cm3 of iron weighs 7.5 g.

14. Surface area of the toy = Curved surface area of the cone
Exercise 10.3
2. Toy is made in the form of hemisphere surmounted by a right cone whose circular base is joined with the plane surface fo the hemisphere . The radius of the base of the cone is 7 cm . and its volume is 3/2 of the hemisphere. Calculate the height of the cone and the suface area of the toy correct to 2 places of decimal
Solution : Radius of the base of the cone = r = 7 cm
Let height of the cone = h cm
Volume of the cone = 3/2 X Volume of the hemisphere
Slant height of the cone =
Surface area of the toy =
Curved surface area of the cone
Curved surface area of the hemisphere

15. Exercise – 10.3
3. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 7 cm .
Solution : Side of the cube = diameter of the cone = 7 cm.
Radius of the cone =
The volume of the largest right circular cone that can be cut out of a cube =

16. Exercise – 10.3
4. Cylindrical tub of radius 5 cm. and length 9.8 cm is full of water . A Solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. The radius of the hemisphere is 3.5 cm and height of cone outside the hemisphere is 5 cm. Find the volume of water left in the tub.
Solution : Radius of cylindrical tub = 5cm
Height of the cylindrical tub = 9.8 cm
Volume of the water in cylindrical tub =
cm3
A Solid in the form of right circular cone mounted on a hemisphere is immersed into the tub.
h = 9.8 òÜ….Ò$
Radius of the hemisphere = r1 = 3.5 cm
Volume of the hemisphere =
r = 5 òÜ….Ò$
= cm3

17. cm3 A¿êÅçÜ… : 10.3 Volume of the cone outside the hemisphere cm3
Height of the cone outside the hemisphere = h1 = 5 cm radius of cone = r1 = 3.5 cm
Volume of the cone outside the hemisphere
cm3
Volume of the water left in the tub =
Volume of the water in the cylindrical tub
Volume of the a solid in the form of right circular cone mounted on a hemisphere
h = 9.8 òÜ….Ò$
cm3
r = 5 òÜ….Ò$

18. Solution: height of the solid cylinder = 10 cm
Exerices 10.3
5. In the Adjacent figure , the height of a solid cylinder is 10 cm and diameter is 7 cm. Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown the figure. Find the volume of the remaining solid.
Solution: height of the solid cylinder = 10 cm
Diameter of the solid cylinder = 7 cm
Radius of the solid cylinder = r = d/2 =7/2 = 3.5 cm
Volume of the Solid cylinder =
h = 10cm
Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown the figure
Radius of conical hole = r1 = 3 cm and height of the conical hole = h1= 4 cm
Volume of the cone
r = 3.5 cm

19. Volume of the Solid cylinder Volume fo the conical hole
If two equal conical holes of radius 3 cm and height 4 cm are cut off then the volume of the remaining solid. =
Volume of the Solid cylinder
Volume fo the conical hole
h = 10cm
r = 3.5 cm

20. Exercise 10.3
6. Spherical marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm which contains some water. Find the number of marbles that should be dropped in to the beaker , So that water level rises by 5.6 cm .
Solution : diameter of cylindrical beaker = d = 7 cm
Radius of the cylindrical beaker = r = d / 2 = 7 / 2 = 3.5 cm
d = 7cm
Diameter of the spherical marble dropped into cylindrical beaker = d = 1.4 cm ; r =d/ = 1.4/2= 0.7 cm
r = 0.7cm
h = 5.6cm
When spherical marbles are dropped into a cylindrical beaker , the water level rises = 5.6 cm
Number of marbles should be dropped into beaker so that water level rises by 5.6 cm =
r = 3.5cm
Volume of the water in cylinderical beaker
Volume of spherical marble

21. cm3 cm3 cm3 Execise – 10.3 Solution : Volume of the cuboid =
7. A pen stand is made of wood in the shape of cuboid with three conical depressions to hold the pens . The dimensions of the cuboid are 15cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm . Find the volume of wood in the entire stand.
Solution : Volume of the cuboid =
cm3
Radius of the depression = r = 0.5 cm and depth = h = 1.4 cm
Volume of the three conical depressions =
cm3
Volume of wood in the entire pen stand =
Volume of the cuboid – Volume of Three conical depressions
cm3