8 m 10 3 Volume of Combination of

8 m 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 m 3 and there are 20 workers each of whom occupies about 0. 08 m 3 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 m given by the volume of air 5 1 inside the cuboid and inside 7 m the half cylinder taken together. 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 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 m 3 The total space occupied by 20 workers The volume of the air, when there are machinery and workers =

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.

If the diameter of the cross section of a wire is decreased by 5%, by Try this what percentage should the length be increased so that the volume remains the same ? Solution : Diameter of the wire be d and radius of the wire be r 1 =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 V 1 , radius r 1 and its length ( height ) h 1 Let the volume of the wire ( cylinder ) after decreasing the diameter of the cross section V 2 , radius r 2 and its length ( height ) h 2 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

Try this Surface area of a sphere and cube are equal. Then find the ratio of their volumes. 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

Example – 10 A Solid toy is in the form of a right circular cylinder with hemispherical shape at one end 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. 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 cm 12 cm 2. 1 cm

Example - 11 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. h =15 cm 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 = Voume of the ice cream in cylinderical container

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 a cone = Volume of the ice cream in conical portion + Volume of ice cream in hemispherical portion 4 x 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

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 = 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

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

Solution : diameter of the cylinderical pencil = d = 1 cm Radius of the cylinderical pencil = r = d / 2 = ½ = 0. 5 cm height of the cylinderical pencil sharpened = h = 2 cm 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 0. 5 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

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 cm 3 of iron weighs 7. 5 g. Solution : diameter of cylinderical iron pillar = d = 20 cm Height of the cylinderical Iron pillar 42 cm Radius of the cylinderical iron pillar 10 cm Radius of cone = Radius of cylinderical iron pillar = 10 cm Height of the cone = h 2 = 42 cm 2. 8 m Volume of the Iron pillar = Volume of the cylinder 20 cm Weight of the pillar if 1 cm 3 of iron weighs 7. 5 g. gms

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 th cone is 7 cm. and its volume is 3/2 of the hemisphere. Calculate the height of the con 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

Exercise – 10. 3 3. Find the volume of the largest right circular cone that can be cut 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 of a cube =

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 = 5 cm Height of the cylindrical tub = 9. 8 cm Volume of the water in cylindrical tub = cm 3 h = 9. 8 òÜ…. Ò$ A Solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. Radius of the hemisphere = r 1 = 3. 5 cm r=5 Volume of the hemisphere = òÜ…. Ò$ = 89. 83 cm 3

10. 3 Height of the cone outside the hemisphere = h 1 = 5 cm radius of cone = r 1 = 3. 5 cm Volume of the cone outside the hemisphere cm 3 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$ cm 3 r=5 òÜ…. Ò$

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 = d/2 =7/2 = 3. 5 cm Volume of the Solid cylinder = h = 10 cm Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown the figure Radius of conical hole = r 1 = 3 cm and height of the conical hole = h 1= 4 cm Volume of the cone r = 3. 5 cm

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 = 10 cm r = 3. 5 cm

Exercise 6. Spherical marbles of diameter 1. 4 cm are dropped into a cylindrical beaker 10. 3 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 Diameter of the spherical marble dropped into cylindrical beaker = d = 1. 4 cm ; r =d/2 = 1. 4/2= 0. 7 cm d = 7 cm r = 0. 7 cm Radius of the cylindrical beaker = d / 2 = 7 / 2 = 3. 5 cm 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. 5 cm Volume of the water in cylinderical beaker Volume of spherical marble

Execise – 10. 3 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 15 cm 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 = cm 3 Radius of the depression = r = 0. 5 cm and depth = 1. 4 cm Volume of the three conical depressions = cm 3 Volume of wood in the entire pen stand = Volume of the cuboid – Volume of Three conical depressions cm 3