Let R be the radius of the given cone and r be radius of small cone.
Let h be the height of small cone.
19. A semi-circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find
(i) the radius of the cone.
(ii) the (lateral) surface area of the cone.
Solution:
We have given that:
(i)Given radius of the semi circular lamina, r = 35 cm
A cone is formed by folding it.
So the slant height of the cone, l = 35 cm
Let r1 be radius of cone.
Semicircular perimeter of lamina becomes the base of the cone.
r = 2 r1
r = 2 r1
35 = 2 r1
r1 = 35/2= 17.5 cm
Hence the radius of the cone is 17.5 cm.
(ii) Curved surface area of the cone = r1 l
= (22/7)×17.5×35
= 22×17.5×5
= 1925 cm2
Hence the lateral surface area of the cone is 1925 cm2.
Exercise 17.3
1. Find the surface area of a sphere of radius :
(i) 14 cm
(ii) 10.5 cm
Solution:
(i) Given radius of the sphere, r = 14 cm
Surface area of the sphere = 4r2
= 4×(22/7)×142
= 4×22×14×2
= 2464 cm3
Hence the surface area of the sphere is 2464 cm2.
(ii) Given radius of the sphere, r = 10.5 cm
Surface area of the sphere = 4r2
= 4×(22/7)×10.52
= 1386 cm3
Hence the surface area of the sphere is 1386 cm2.
2. Find the volume of a sphere of radius :
(i) 0.63 m
(ii) 11.2 cm
Solution:
(i) Given radius of the sphere, r = 0.63 m
Volume of the sphere, V = (4/3)r3
= (4/3)×(22/7)×0.633
= 1.047 m3
= 1.05 m3 (approx)
Hence the volume of the sphere is 1.05 m3.
(ii) Given radius of the sphere, r = 11.2 cm
Volume of the sphere, V = (4/3)r3
= (4/3)×(22/7)×11.23
= 5887.317 cm3
= 5887.32 cm3 (approx)
Hence the volume of the sphere is 5887.32 cm3.
3. Find the surface area of a sphere of diameter:
(i) 21 cm
(ii) 3.5 cm
Solution:
We have given that:
(i) Given diameter of the sphere, d = 21 cm
Radius, r = d/2 = 21/2 = 10.5
Surface area of the sphere = 4r2
= 4×(22/7)×10.52
= 1386 cm2
Hence the surface area of the sphere is 1386 cm2.
(ii) Given diameter of the sphere, d = 3.5 cm
Radius, r = d/2 = 3.5/2 = 1.75
Surface area of the sphere = 4r2
= 4×(22/7)×1.752
= 38.5 cm2
Hence the surface area of the sphere is 38.5 cm2.
4. A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm3, find the mass of the shot-put.
Solution:
We have given that:
Given radius of the metallic sphere,r = 4.9 cm
Volume of the sphere, V = (4/3)r3
V = (4/3)×(22/7)×4.93
V = 493.005
V = 493 cm^3 (approx)
Given Density = 7.8 g per cm3
Density = Mass/ Volume
Mass = Density × Volume
= 7.8×493
= 3845.4 g
Hence the mass of the shot put is 3845.4 g.
5. Find the diameter of a sphere whose surface area is 154 cm2 .
Solution:
We have given that:
Given surface area of the sphere = 154 cm2
Surface area of the sphere = 4r2
4×(22/7)×r2 = 154
r2 = 154 ×7/(22×4)
= 49/4
6. Find:
(i) the curved surface area.
(ii) the total surface area of a hemisphere of radius 21 cm.
Solution:
(i) Given radius of the hemisphere, r = 21 cm
Curved surface area of the hemisphere = 2r2
= 2×(22/7)×212
= 2×22×3×21
= 2772 cm2
Therefore, the curved surface area of the hemisphere is 2772 cm2.
(ii)Total surface area of the hemisphere = 3r2
= 3×(22/7)×212
= 3×22×3×21
= 4158 cm2
Hence the total surface area of the hemisphere is 4158 cm2.
7. A hemispherical brass bowl has inner- diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2.
Solution:
We have given that:
Given inner diameter of the brass bowl, d = 10.5 cm
Radius, r = d/2 = 10.5/2 = 5.25 cm
Curved surface area of the bowl = 2r2
= 2×(22/7)×5.252
= 173.25 cm2
Rate of tin plating = Rs.16 per 100 cm2
So total cost = 173.25×16/100 = 27.72
Hence the cost of tin plating the bowl on the inside is Rs.27.72.
8. The radius of a spherical balloon increases from 7 cm to 14 cm as air is jumped into it. Find the ratio of the surface areas of the balloon in two cases.
Solution:
We have given that:
Given radius of the spherical balloon, r = 7 cm
Radius of the spherical balloon after air is pumped, R = 14 cm
Surface area of the sphere = 4r2
Ratio of surface areas of the balloons = 4r2/4R2
= r2/R2
= 72/142
= 1/4
Hence the ratio of the surface areas of the spheres is 1:4.
9. A sphere and a cube have the same surface area. Show that the ratio of the volume of the sphere to that of the cube is √6 ∶√π
Solution:
We have given that:
Let r be the radius of the sphere and a be the side of the cube.
Surface area of sphere = 4r2
Surface area of cube = 6a2
Given sphere and cube has same surface area.
17. Write whether the following statements are true or false. Justify your answer :
(i) The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.
(ii) The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals the volume of a hemisphere of radius r.
(iii) A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is 1∶ 2∶ 3.
Solution:
We have given that:
(i)Let the radius of sphere be r.
Then height of the cylinder, h = 2r
Radius of cylinder = r
Volume of cylinder = r2 h
= ×r2×2r
= 2r3
Volume of sphere = (4/3)r3
= (2/3)× 2r3
= (2/3)× Volume of cylinder
Hence the given statement is true.
(ii)Let the edge of the cube is 2r.
So radius of cone = r
Height of cone, h = 2r
Volume of cone = (1/3)r2 h
= (1/3)r2×2r
= (2/3)r3
= Volume of a hemisphere of radius r
Hence the given statement is true.
(iii)Let r be radius of cone, hemisphere and cylinder.
So the height of the cone = r
Height of cylinder = r
Volume of cone = (1/3)r2 h
= (1/3)r3
Volume of hemisphere = (2/3)r3
Volume of cylinder = r2 h
= r3
Ratio of volume of cone , hemisphere and cylinder = (1/3)r3 : (2/3)r3 : r3
= 1/3∶ 2/3∶ 1
= 1:2:3
Hence the given statement is true.
Exercise 17.4
Take π = 22/7 unless stated otherwise.
1. The adjoining figure shows a cuboidal block of wood through which a circular cylindrical hole of the biggest size is drilled. Find the volume of the wood left in the block.
Solution:
We have given that:
Given diameter of the hole, d = 30 cm
radius of the hole, r = d/2 = 30/2 = 15 cm
Height of the cylindrical hole, h = 70 cm
Volume of the cuboidal block = lbh
= 70×30×30
= 63000 cm3
Volume of cylindrical hole = r2 h
= (22/7)×152×70
= 22×225×10
= 49500 cm3
Volume of the wood left in the block = 63000-49500 = 13500 cm3
Hence the volume of the wood left in the block is 13500 cm3.
2. The given figure shows a solid trophy made of shining glass. If one cubic centimetre of glass costs Rs 0.75, find the cost of the glass for making the trophy.
Solution:
We have given that:
Given side of the cube, a = 28 cm
Radius of the cylinder, r = 28/2 = 14 cm
Height of the cylinder, h = 28 cm
Volume of the cube = a3
= 283
= 28×28×28
= 21952 cm3
Volume of the cylinder = r2 h
= (22/7)×142×28
= 22×2×14×28
= 17248 cm3
Total volume of the solid = Volume of the cube + volume of the cylinder
= 21952+17248
= 39200 cm3
Cost of 1 cm3 glass = Rs.0.75
Total cost of glass = 39200×0.75
= Rs.29400
Hence the cost of making the trophy is Rs.29400.
3. From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.
Solution:
We have given that:
Given edge of the cube, a = 14 cm
Radius of the cone, r = 14/2 = 7 cm
Height of the cone, h = 14 cm
Volume of the cube = a3
= 143
= 14×14×14
= 2744 cm3
Volume of the cone = (1/3)r2 h
= (1/3)×(22/7)×72×14
= 22×7×14/3
= 2156/3 cm3
Volume of the remaining material = Volume of the cube- Volume of the cone
= 2744-2156/3
= (3×2744-2156)/3
= (8232 – 2156)/3
= 6076/3
4. A cone of maximum volume is curved out of a block of wood of size 20 cm x 10 cm x 10 cm. Find the volume of the remaining wood.
Solution:
We have given that:
Given dimensions of the block of wood = 20 cm × 10 cm× 10 cm
Volume of the block of wood = 20×10×10 = 2000 cm3 [Volume = lbh]
Diameter of the cone, d = 10 cm
Radius of the cone , r = d/2 = 10/2 = 5 cm
Height of the cone, h = 20 cm
Volume of the cone = (1/3)r2 h
= (1/3)×(22/7)×52×20
= (22×25×20)/(3×7)
= 11000/21 cm3
Volume of the remaining wood = Volume of block of wood – Volume of cone
= 2000-11000/21
= (21×2000-11000)/21
= (42000-11000)/21
= 31000/21
= 1476.19 cm3
Hence the volume of the remaining wood is 1476.19 cm3.
5. 16 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the box is filled with water. Find the volume of the water filled in the box.
Solution:
We have given that:
Given dimensions of the box = 16 cm ×8 cm ×8 cm
6. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold 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 the wood in the entire stand, correct to 2 decimal places.
Solution:
Dimensions of the cuboid = 15 cm× 10 cm × 3.5 cm
Volume of the cuboid = 15×10×3.5 = 525 cm3
Radius of each depression, r = 0.5 cm
Depth, h = 1.4 cm
Volume of conical depression = (1/3)r2 h
= (1/3)×(22/7)×0.52×1.4
= 22×0.25×1.4/21
= 7.7/21 cm3
Volume of 4 such conical depressions = 4×7.7/21
= 1.467 cm3
Volume of wood in the stand = Volume of the cuboid- Volume of 4 conical depressions
= 525-1.467
= 523.533
= 523.53 cm3
Hence the volume of the wood in the stand is 523.53 cm3.
7. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter that the hemisphere can have? Also, find the surface area of the solid.
Solution:
We have given that:
Given edge of the cube, a = 7 cm
Diameter of the hemisphere, d = 7 cm
Radius, r = d/2 = 7/2 = 3.5 cm
Surface area of the hemisphere = 2r2
= 2×(22/7)×3.52
= 44×12.25/7
= 539/7
= 77 cm2
Surface area of the cube = 6a2
= 6×72
= 6×49
= 294 cm2
Surface area of base of hemisphere = r2
= (22/7)×3.52
= 22×12.25/7
= 38.5 cm2
Surface area of the solid = surface area of the cube + surface area of hemisphere – surface area of the base of hemisphere
= 294+77-38.5
= 332.5 cm2
Hence the surface area of the solid is 332.5 cm2.
8. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder (as shown in the given figure). If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.
Solution:
We have given that:
Given height of the cylinder, h = 10 cm
Radius of the cylinder, r = 3.5 cm
Radius of the hemisphere = 3.5 cm
Total surface area of the article = curved surface area of the cylinder + curved surface area of 2 hemispheres
= 2rh + 2× 2r2
= 2r(h+2r)
= 2×(22/7)×3.5×(10+2×3.5)
= (154/7)×(10+7)
= 22×17
= 374 cm2
Hence the total surface area of the article is 374 cm2.
9. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. If the total height of the toy is 15.5 cm, find the total surface area of the toy.
Solution:
We have given that:
Given radius of the cone, r = 3.5 cm
Radius of hemisphere, r = 3.5 cm
Total height of the toy = 15.5 cm
Height of the cone = 15.5 – 3.5 = 12 cm
10. A circus tent is in the shape of a cylinder surmounted by a cone. The diameter of the cylindrical portion is 24 m and its height is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas used to make the tent.
Solution:
Given diameter of the cylindrical part of tent, d = 24 m
Radius, r = d/2 = 24/2 = 12 m
Height of the cylindrical part, H = 11 m
Since vertex of cone is 16 m above the ground, height of cone, h = 16-11
h = 5 m
Solution:
We have given that:
Given height of the tent above the ground = 85 m
Height of the cylindrical part, H = 50 m
height of the cone, h = 85-50
h = 35 m
Diameter of the base, d = 168 m
Radius of the base of cylindrical part, r = d/2 = 168/2 = 84 m
Radius of cone, r = 84 m
12. From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and the total surface of the remaining solid.
Solution:
We have given that:
Volume of the remaining solid = Volume of the cylinder – Volume of the cone
= r2 H – (1/3)r2h
= r2 (H- h/3)
= (22/7)×72×(30-24/3)
= (22×7)×(30-8)
= (154)×(22)
= 3388 cm3
Volume of the remaining solid is 3388 cm3.
Total surface area of the remaining solid = Curved surface area of cylinder + surface area of top of the cylinder+ curved surface area of the cone
Total surface area of the remaining solid = 2rH + r2 + rl
= r(2H + r + l)
=(22/7)×7(2×30 + 7+ 25)
= 22×(60+32)
= 22×92
= 2024 cm2
Hence the total surface area of the remaining solid is 2024 cm2.
13. The adjoining figure shows a wooden toy rocket which is in the shape of a circular cone mounted on a circular cylinder. The total height of the rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted green and the cylindrical portion red, find the area of the rocket painted with each of these colours.
Also, find the volume of the wood in the rocket. Use π = 3.14 and give answers correct to 2 decimal places.
(i) Given height of the rocket = 26 cm
Height of the cone, H = 6 cm
Height of the cylinder, h = 26-6 = 20 cm
Diameter of the cone = 5 cm
Radius of the cone, R = 5/2 = 2.5 cm
Diameter of the cylinder = 3 cm
Radius of the cylinder, r = 3/2 = 1.5 cm
Curved surface area of the cone = Rl
= 3.14×2.5×6.5
= 51.025 cm2
Base area of cone = R2
= 3.14×2.52
= 3.14×6.25
= 19.625 cm2
Curved surface area of the cylinder = 2rh
= 2×3.14×1.5×20
= 188.4 cm2
Base area of cylinder = r2
= 3.14×1.52
= 3.14×2.25
= 7.065 cm2
Total surface area of conical portion to be painted green = Curved surface area of the cone+ Base area of cone- Base area of cylinder
= 51.025 + 19.625 – 7.065
= 63.585 cm2
= 63.59 cm2
Hence the area of the rocket painted with green colour is 63.59 cm2 .
Total surface area of the cylindrical portion to be painted red = Curved surface area of the cylinder + Base area of cylinder
= 188.4+7.065
= 195.465 cm2
= 195.47 cm2
Hence the area of the rocket painted with red colour is 195.47 cm2 .
(ii) Volume of wood in the rocket = Volume of cone + Volume of cylinder
= (1/3)R2 H + r2 h
= ((R2 H/3) + r2 h))
= 3.14×((2.52×6/3) + 1.52×20)
= 3.14×((6.25×2) + 2.25×20)
= 3.14×(12.5+ 45)
= 3.14×57.5
= 180.55 cm3
Hence the volume of the wood in the rocket is 180.55 cm3
14. The adjoining figure shows a hemisphere of radius 5 cm surmounted by a right circular cone of base radius 5 cm. Find the volume of the solid if the height of the cone is 7 cm. Give your answer correct to two places of decimal.
Solution:
We have given that:
Given radius of the hemisphere,r = 5 cm
Radius of cone, r = 5 cm
Height of the cone, h = 7 cm
Volume of the solid = Volume of the hemisphere + Volume of the cone
= (2/3)r3 + (1/3)r2 h
= (1/3)r2 (2r+h)
= (1/3)×(22/7)×52 (2×5+7)
= (22/21)×25(10+7)
= (22/21)×25×17
= 445.238 cm3
= 445.24 cm3
Hence the volume of the solid is 445.24 cm3 .
15. A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is 2/3 of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimal.
Solution:
We have given that:
Given radius of the cone,r = 3.5 cm
Radius of hemisphere, r = 3.5 cm = 7/2 cm
Volume of hemisphere = (2/3)r3
= (2/3)×(22/7)×(7/2)3
= (2/3)×(22/7)×(7/2)×(7/2)×(7/2)
= (22/3)×(7/2)×(7/2)
= 11×49/6
= 539/6 m3
Volume of cone = 2/3 of volume of hemisphere
= (2/3)× 539/6
= 539/9 m3
Volume of cone = (1/3)r2 h
(1/3)r2 h = 539/9
(1/3)×(22/7)×(7/2)2×h = 539/9
h = 539×3×2/9×11×7
h = 14/3
h = 4.667
h = 4.67 m
Hence the height of the cone is 4.67 m.
16. A circular hall (big room) has a hemispherical roof. The greatest height is equal to the inner diameter, find the area of the floor, given that the capacity of the hall is 48510 m3.
Solution:
We have given that:
Let the radius of the hemisphere be r.
Inner diameter = 2r
Given greatest height equal to inner diameter.
So total height of the hall = 2r
Height of the hemispherical part = r
Height of cylindrical area, h = 2r-r = r
Capacity of the hall = Volume of cylindrical area + volume of hemispherical area
= r^2 h + (2/3)r3
= r3 + (2/3)r3 [∵h = r]
= r3 (1+2/3)
= r3 (3+2)/3
= (5/3)r3
Given capacity of hall = 48510 m3
(5/3)r3 = 48510
(5/3)×(22/7)r3 = 48510
r3 = 48510×3×7/(22×5)
r3 = 9261
Taking cube root on both sides,
r = 21
Area of the floor = r2
= (22/7)×212
= (22×21×3
= 1386 m2
Hence the Area of the floor is 1386 m2 .
17. A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains 41 19/21 m3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building.
Solution:
We have given that:
Let the radius of the dome be r.
Internal diameter = 2r
Given internal diameter is equal to total height.
Total height of the building = 2r
Height of the hemispherical area = r
So height of cylindrical area, h = 2r-r = r
Volume of the building = Volume of cylindrical area + volume of hemispherical area
= r2 h + (2/3)r3
= r3+ (2/3)r3 [∵h = r]
= r3 (1+2/3)
= r3 (3+2)/3
= (5/3)r3
Given Volume of the building =
= 880/21
(5/3)r3= 880/21
(5/3)×(22/7)×r3 = 880/21
r3= 880×3×7/(5×22×21)
r3= 880/110
r3= 8
Taking cube root
r = 2 m
Height of the building = 2r = 2×2 = 4m
Hence the height of the building is 4m.
18. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and the height of the cylinder are 6 cm and 12 cm respectively. If the slant height of the conical portion is 5 cm, find the total surface area and the volume of the rocket. (Use π = 3.14).
Solution:
We have given that:
Given diameter of the cylinder = 6 cm
Radius of the cylinder, r = 6/2 = 3 cm
Height of the cylinder, H = 12 cm
Slant height of the cone, l = 5 cm
Radius of the cone, r = 3 cm
= r2 ((h/3)+H)
= 3.14×32×((4/3)+12)
= 3.14×9×((4+36)/3)
= 3.14×9×(40/3)
= 3.14×3×40
= 376.8 cm3
Hence the volume of the rocket is 376.8 cm3.
19. The adjoining figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and the cone are each of 4 cm. Find the volume of the solid.
Solution:
We have given that:
Given common radius, r = 7 cm
Height of the cone, h = 4 cm
Height of the cylinder, H = 4 cm
Volume of the solid = Volume of the cone + Volume of the cylinder + Volume of the hemisphere
= (1/3)r2 h +r2 H + (2/3) r3
= r2((h/3) + H + (2r/3))
= (22/7)×72×((4/3)+4+(2×7)/3)
= (22×7×((4+12+14)/3)
= 22×7×30/3
= 1540 cm3
Hence the volume of the solid is 1540 cm3.
20. A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. Their common diameter is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the volume of the solid. (Take π = 3.14)
Solution:
We have given that:
Given height of the cylinder, H = 10 cm
Height of the cone, h = 6 cm
Common diameter = 3.5 cm
Common radius, r = 3.5/2 = 1.75 cm
Volume of the solid = Volume of the cone + Volume of the cylinder + Volume of the hemisphere
= (1/3)r2 h +r2 H + (2/3) r3
= r2 ((h/3) + H + (2r/3))
= 3.14×1.752×((6/3)+10+(2×1.75)/3)
= 3.14×3.0625×(2+10+1.167)
= 3.14×3.0625×13.167
= 9.61625×13.167
= 126.617 cm3
= 126.62 cm3
Hence the volume of the solid is 126.62 cm3.
21. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.
Solution:
We have given that:
Given height of the cylinder, H = 13 cm
Radius of the cylinder, r = 5 cm
Radius of the hemisphere, r = 5 cm
Height of the cone, h = 12 cm
Radius of the cone, r = 5 cm
Surface area of the toy = curved surface area of cylinder +curved surface area of hemisphere+ curved surface area of cone
= 2rH+2r2+rl
= r(2H+2r+l)
= (22/7)×5(2×13+2×5+13)
= (110/7)×(26+10+13)
= (110/7)×49
= 110×7
= 770 cm2
Hence the surface area of the toy is 7770 cm2.
22.The adjoining figure shows a model of a solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1∶ 200, find
(i) the total surface area of the solid in π m².
(ii) the volume of the solid in π litres.
Solution:
We have given that:
Given height of the cylinder, h = 8 cm
Radius of the cylinder, r = 3 cm
Radius of hemisphere , r = 3 cm
Scale = 1:200
Hence actual radius, r = 200×3 = 600
Actual height, h = 200×8 = 1600
(i)Total surface area of the solid = Base area of the cylinder + Curved surface area of the cylinder + curved surface area of the hemisphere
= r2+2rh + 2r2
= r(r+2h+2r)
= ×600(600+2×1600+2×600)
= 600 ×(600+3200+1200)
= 600 ×(5000)
= 3000000 cm2
= 300 m2
Hence the total surface area of the solid is 300 m2.
(ii)Volume of the solid = Volume of the cylinder + Volume of the hemisphere
= r2 h + (2/3) r3
= r2 (h+ (2/3)r)
= × 6002 (1600+ (2/3)×600)
= 360000 (1600+400)
= 360000 ×2000
= 720000000 cm3
= 720 m3
= 720000 litres [1 m3 = 1000 litres]
Hence the volume of the solid is 720000 litres.
Exercise 17.5
1. The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36 m, find its radius.
Solution:
We have given that:
Given diameter of the metallic sphere = 6 cm
Radius of the sphere, r = 6/2 = 3 cm
Volume of the sphere, V = (4/3)r3
= (4/3)××33
= 4××9
= 36 cm3
Length of the wire, h = 36 m = 3600 cm
Since the sphere is melted and drawn into a wire, volume remains the same.
Volume of the wire, r2 h = 36
r2×3600 = 36
r2 = 1/100
r = 1/10 = 0.1 cm = 1 mm
Hence the radius of the wire is 1 mm.
2. The radius of a sphere is 9 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire in metres.
Solution:
We have given that:
Radius of the sphere, r = 9 cm
Volume of the sphere, V = (4/3)r3
= (4/3)××93
= 12××81
= 972 cm3
Diameter of the wire = 2 mm
So radius of the wire = 2/2 = 1 mm = 0.1 cm
Since the sphere is melted and drawn into a wire, volume remains the same.
Volume of the wire, r2 h = 972
×0.12×h =972
h = 972/0.12
h = 972/0.01
h = 97200 cm
h = 972 m
Hence the length of the wire is 972 m.
3. A solid metallic hemisphere of radius 8 cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.
Solution:
We have given that:
Given radius of the hemisphere, r = 8 cm
Volume of the hemisphere, V = (2/3)r3
= (2/3)×83
= (1024/3) cm3
Radius of cone, R = 6 cm
Since hemisphere is melted and recasted into a cone, the volume remains the same.
Volume of the cone, (1/3)R2 h = (1024/3)
(1/3)×62 ×h = (1024/3)
36h = 1024
h = 1024/36
= 28.44 cm
Hence the height of the cone is 28.44 cm.
4. A rectangular water tank of base 11 m × 6 m contains water upto a height of 5 m. if the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.
Solution:
We have given that:
Given dimensions of the rectangular water tank = 11 m × 6 m
Height of water in tank = 5 m
Volume of water in tank = 11×6×5 = 330 m3
Radius of the cylindrical tank, r = 3.5 m
Volume of cylindrical tank = r2 h
r2 h = 330
(22/7)×3.52×h = 330
(22/7)×12.25×h = 330
h = 330×7/22×12.25
= 8.57 m
Hence the height of the water level in the tank is 8.57m.
5. The rain water from a roof of dimensions 22 m × 20 m drains into a cylindrical vessel having diameter of base 2 m and height; 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.
Solution:
We have given that:
Given dimensions of the cylindrical vessel = 22 m × 20 m
Let the rainfall be × cm.
Volume of water = (22×20×X)m3
Diameter of the cylindrical base = 2 m
So radius of the cylindrical base = 2/2 = 1 m
Height of the cylindrical base, h = 3.5 m
Volume of cylindrical vessel = r2h
= (22/7)×12×3.5
= 11 m3
Since the rain water collected from the roof just fill the cylindrical vessel, the volumes are equal.
22×20×X = 11
X= 11/22×20
= 1/40 m
= (1/40)×100 cm
= 2.5 cm
Hence the rainfall is 2.5 cm.
6. The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.
Solution:
We have given that:
Given height of the cylinder, h = 9 cm
Diameter of the cylinder = 40 cm
Radius of the cylinder, r = 40/2= 20 cm
Volume of the cylinder = r2h
= × 202×9
= ×400×9
= 3600 cm3
Height of the cone, H = 108 cm
Volume of cone = (1/3)r2h
= (1/3)r2×108
= 36r2
Since volume of cone is equal to the volume of the cylinder, we get
36r2 = 3600
r2 = 3600/36
r2 = 100
Taking square root on both sides,
r = 10 cm
Hence the radius of the cone is 10 cm.
7. Eight metallic spheres, each of radius 2 cm, are melted and cast into a single sphere. Calculate the radius of the new (single) sphere.
Solution:
We have given that:
Given radius of each sphere,r = 2 cm
Volume of a sphere = (4/3)r3
= (4/3)×23
= (4/3)×8
= (32/3) cm3
Volume of 8 spheres = 8×(32/3)
= (256/3) cm3
Let R be radius of new sphere.
Volume of the new sphere = (4/3)R3
Since 8 spheres are melted and casted into a single sphere, volume remains same.
(4/3)R3= (256/3)
4R3 = 256
R3= 256/4 = 64
Taking cube root
R = 4 cm
Hence the radius of the new sphere is 4 cm.
8. A metallic disc, in the shape of a right circular cylinder, is of height 2.5 mm and base radius 12 cm. Metallic disc is melted and made into a sphere. Calculate the radius of the sphere.
Solution:
We have given that:
Given height of the cylinder, h = 2.5 mm = 0.25 cm
Radius of the cylinder, r = 12 cm
Volume of the cylinder = r2h
= × 122×0.25
= ×144×0.25
= 36 cm3
Let R be the radius of the sphere.
Volume of sphere = (4/3)R3
Since metallic disc is melted and made into a sphere, their volumes remains same.
(4/3)R3 = 36
R3= 36×3/4
R3= 27
Taking cube root
R = 3 cm
Hence the radius of the sphere is 3 cm.
9. Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.
Solution:
We have given that:
For same material, density will be same.
Density = mass/Volume
Mass of the smaller sphere, m1 = 1 kg
Mass of the bigger sphere, m2 = 7 kg
The spheres are melted to form a new sphere.
So the mass of new sphere, m3 = 1+7 = 8 kg
Density of smaller sphere = density of new sphere
Let V1 be volume of smaller sphere and V3 be volume of bigger sphere.
m1/V1 = m3/V3
1/V1= 8/V3
V1/ V3= 1/8 …(i)
Given radius of the smaller sphere, r = 3 cm
V1 = (4/3)r3
V1= (4/3) ××33
V1= 36
Let R be radius of new sphere.
V3 = (4/3)R^3
V1/ V_3= 36÷(4/3)R3
V1/ V_3= 27/R3 …(ii)
From (i) and (ii)
1/8 = 27/R3
R3= 27×8 = 216
Taking cube root on both sides,
R = 6 cm
So diameter of the new sphere = 2R = 2×6 = 12 cm
Hence diameter of the new sphere is 12 cm.
10. A hollow copper pipe of inner diameter 6 cm and outer diameter 10 cm is melted and changed into a solid circular cylinder of the same height as that of the pipe. Find the diameter of the solid cylinder.
Solution:
We have given that:
Given inner diameter of the pipe = 6 cm
So inner radius, r = 6/2 = 3 cm
Outer diameter = 10 cm
Outer radius, R = 10/2 = 5 cm
Let h be the height of the pipe.
Volume of pipe = (R2-r2)h
= ×(52-32)×h
= h(25-9)
= 16h cm3
Let r be the radius of solid cylinder.
Volume of solid cylinder = r2h
Since pipe is melted and changed into a cylinder, their volumes remains same.
r2h = 16h
r2 = 16
r = 4 cm
Diameter = 2r = 2×4 = 8 cm
Hence the diameter of the cylinder is 8 cm.
11. A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm and height is 72 cm, find the uniform thickness of the cylinder.
Solution:
We have given that:
Given radius of the sphere, r = 6 cm
Volume of the sphere = (4/3)r3
= (4/3)×63
= 288 cm3
Let r be the internal radius of the hollow cylinder.
External radius of the hollow cylinder, R = 4 cm
Height of hollow cylinder, h = 72 cm
Volume of hollow cylinder = (R2-r2)h
Since sphere is melted and changed into a hollow cylinder, their volumes remain same.
(R2-r2)h = 288
(42-r2)×72 = 288
(42-r2) = 288/72
(42-r2) = 4
16-r2= 4
r2= 16-4
r2= 12
r = 2√3 cm
So thickness = R-r = 4-2√3
= 4- 3.464
= 0.536 cm
= 0.54 cm (approx)
Hence the thickness of the cylinder is 0.54 cm.
12. A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal of the tube is ½ cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone correct to one decimal place.
Solution:
We have given that:
Given internal radius of the tube, r = 3 cm
Thickness of the tube = ½ cm = 0.5 cm
External radius of tube = 3+0.5 = 3.5 cm
Height of the tube, h = 21 cm
Volume of the tube = (R2-r2)h
= (3.52-32)×21
= (12.25-9)×21
= (3.25)×21
= 68.25 cm3
Height of the cone, h = 7 cm
Let r be radius of cone.
Volume of cone = (1/3)r2 h
= (1/3)r2×7
= (7/3)r2
Since tube is melted and changed into a cone, their volumes remain same.
(7/3)r2= 68.25
r2= 68.25×3/7 = 29.25
Taking square root on both sides
r = 5.4 cm
Hence the radius of the cone is 5.4 cm.
13. A hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted into a cone of base diameter 8 cm. Find the height of the cone. (2002)
Solution:
We have given that:
Given internal diameter of hollow sphere = 4 cm
Internal radius, r = 4/2 = 2 cm
External diameter = 8 cm
External radius, R = 8/2 = 4 cm
Volume of the hollow sphere, V = (4/3)(R3-r3)
V = (4/3)×(43-23)
V = (4/3)×(64-8)
V = (4/3)×56
Base diameter of the cone = 8 cm
Radius, r = 8/2 = 4 cm
Volume of cone = (1/3)r2h
= (1/3)×42 ×h
= (16/3)×h
Since sphere is melted and changed into a cone, their volumes remain same.
(4/3)×56 = (16/3)×h
h = 4×56/16
h = 14 cm
Hence the height of the cone is 14 cm.
14. A well with inner diameter 6 m is dug 22 m deep. Soil taken out of it has been spread evenly all round it to a width of 5 m to form an embankment. Find the height of the embankment.
Solution:
We have given that:
Given inner diameter of the well = 6 m
Radius of the well, r = 6/2 = 3 m
Depth of the well, H = 22 m
Volume of the soil dug out of well = r2H
= ×32×22
= 198 m3
Width of the embankment = 5 m
Inner radius of embankment, r = 3 m
Outer radius of embankment, R = 3+5 = 8 m
Let h be height of the soil embankment.
Volume of the soil embankment = (R2– r2)h
= (82– 32)h
= (64-9)h
= 55h
Volume of the soil dug out = volume of the soil embankment
198 = 55h
h = 198/55
h = 3.6 m
Hence the height of the soil embankment is 3.6 m.
15. A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.
Solution:
We have given that:
Given internal diameter of cylindrical can = 21 cm
Radius of the cylindrical can, R = 21/2 cm
Diameter of sphere = 10.5 cm
Radius of the sphere, r = 10.5/2 = 21/4 cm
Let the rise in water level be h.
Rise in volume of water = Volume of sphere immersed
R2 h = (4/3)r3
×(21/2)2 h = (4/3)××(21/4)3
(21/2) ×(21/2)×h = (4/3)× (21/4)×(21/4)× (21/4)
h = 21/12
h = 7/4
h = 1.75 cm
Hence the rise in water level is 1.75 cm.
16. There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water go down when the sphere is removed?
Solution:
We have given that:
Given radius of the glass jar, R = 8 cm
Diameter of the sphere = 12 cm
Radius of the sphere, r = 12/2 = 6 cm
When the sphere is removed from the jar, volume of water decreases.
Let h be the height by which water level will decrease.
Volume of water decreased = Volume of the sphere
R2 h = (4/3)r3
82 h = (4/3)6^3
h = (4/3)×6×6×6/(8×8)
= 18/4 = 9/2 = 4.5 cm
Hence the height by which water level decreased is 4.5 cm.
17. A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows. What is the volume of each of the solid cone submerged?
Solution:
We have given that:
Given height of the cone, h = 20 cm
Diameter of the cone = 16.8 cm
Radius of the cone, r = 16.8/2 = 8.4 cm
Volume of water in the vessel = (1/3)r2 h
= (1/3)×8.42 ×20
= (1/3)×(22/7)×8.4 ×8.4 ×20
= 1478.4 cm3
One third of volume of water in the vessel = (1/3)× 1478.4
= 492.8 cm3
One third of volume of water in the vessel = Volume of water over flown
Volume of water over flown = volume of two equal solid cones dropped into the vessel.
volume of two equal solid cones dropped into the vessel = 492.8 cm3
Volume of one solid cone dropped into the vessel = 492.8/2 = 246.4 cm3
Hence the volume of each of the solid cone submerged is 246.4 cm3.
18. A solid metallic circular cylinder of radius 14 cm and height 12 cm is melted and recast into small cubes of edge 2 cm. How many such cubes can be made from the solid cylinder?
Solution:
Radius of the solid circular cylinder, r = 14 cm
Height, h = 12 cm
Volume of the cylinder = r2 h
= ×142×12
= ×196×12
= 2352
= 2352×22/7
= 7392 cm3
Edge of the cube, a = 2 cm
Volume of cube = a3
= 23 = 8 cm3
Number of cubes made from solid cylinder = 7392/8 = 924.
Hence the number of cubes made from solid cylinder is 924.
19. How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm?
Solution:
We have given that:
Given dimensions of the cuboidal solid = 9 cm× 11 cm× 12 cm
Volume of the cuboidal solid = 9×11×12 = 1188 cm3
Diameter of shot = 3 cm
So radius of shot, r = 3/2 = 1.5 cm
Volume of shot = (4/3)r3
= (4/3)×1.53
= (4/3)×(22/7)×1.53
= 297/21 cm3
Number of shots made from cuboidal lead of solid = 1188÷(297/21)
= 1188×21/297
= 84
Hence the number of shots made from cuboidal lead of solid is 84.
20. How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm?
Solution:
We have given that:
Edge of the cube, a = 44 cm
Volume of cube = a3
= 443
= 85184 cm3
Diameter of shot = 4 cm
So radius of shot, r = 4/2 = 2 cm
Volume of a shot = (4/3)r3
= (4/3)×(22/7)×23
= 704/21 cm3
Number of lead shots made from cube = 85184÷(704/21)
= 85184×21/704
= 2541
Hence the number of shots made from cube is 2541.
21. Find the number of metallic circular discs with 1.5 cm base diameter and height 0.2 cm to be melted to form a circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
We have given that:
Given height of the circular cylinder, h = 10 cm
Diameter of circular cylinder = 4.5 cm
So radius, r = 4.5/2 = 2.25 cm
Volume of circular cylinder = r2h
= ×2.252×10 = 50.625 cm3
Base diameter of circular disc = 1.5 cm
So radius, r = 1.5/2 = 0.75 cm
Height of the circular disc, h = 0.2 cm
Volume of circular disc = r2h
= ×0.752 ×0.2
= 0.1125 cm3
Number of circular discs made from cylinder = 50.625/0.1125 = 450
Hence the number of circular discs made from cylinder is 450.
22. A solid metal cylinder of radius 14 cm and height 21 cm is melted down and recast into spheres of radius 3.5 cm. Calculate the number of spheres that can be made.
Solution:
We have given that:
Given radius of the metal cylinder,r = 14 cm
Height of the metal cylinder, h = 21 cm
Radius of the sphere, R = 3.5 cm
Volume of the metal cylinder = r2h
= (22/7)×142×21
= 22×2×14×21
= 12936 cm3
Volume of sphere = (4/3)R3
= (4/3)×(22/7)×3.53
= 11×49/3
= 539/3 cm3
Number of spheres that can be made = Volume of the metal cylinder/ Volume of sphere
= 12936 ÷539/3
= 12936 ×3/539
= 72
Hence the number of spheres that can be made is 72.
23. A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained.
Solution:
We have given that:
Given radius of the metallic sphere, R = 10.5 cm
Volume of the sphere = (4/3)R3
= (4/3)×10.53
= 1543.5 cm3
Radius of cone, r = 3.5 cm
Height of the cone, h = 3 cm
Volume of the cone = (1/3)r2h
= (1/3)×3.52×3
= 12.25 cm3
Number of cones made from sphere = Volume of sphere / volume of cone
= 1543.5 /12.25
= 126
Hence the number of cones that can be made is 126.
24. A certain number of metallic cones each of radius 2 cm and height 3 cm are melted and recast in a solid sphere of radius 6 cm. Find the number of cones.
Solution:
We have given that:
Given radius of metallic cones,r = 2 cm
Height of cone, h = 3 cm
Volume of cone = (1/3)r2h
= (1/3)×22×3
= 4 cm3
Radius of the solid sphere, R = 6 cm
Volume of the solid sphere = (4/3)R3
= (4/3)63
= 4××2×6×6
= 288 cm3
Number of cones made from sphere = Volume of solid sphere / volume of the cone
= 288/4
= 72
Hence the number of cones that can be made is 72.
25. A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, 2/5 of the water flows out. Find the number of lead shots dropped into the vessel.
Solution:
We have given that:
Given height of the cone, h = 11 cm
Radius of the cone, r = 2.5 cm
Volume of the cone = (1/3)r2h
= (1/3)×2.52×11
= (11/3)×6.25 cm3
When lead shots are dropped into vessel, (2/5) of water flows out.
Volume of water flown out = (2/5)×(11/3)×6.25
= (22/15)×6.25
= (137.5/15) cm3
Radius of sphere, R = 0.25 cm = ¼ cm
Volume of sphere = (4/3)R3
= (4/3)×(1/4)3
= /48 cm3
Number of lead shots dropped = Volume of water flown out/ Volume of sphere
= (137.5/15) ÷48
= (137.5/15) ×48
= 137.5×48/15
= 440
Hence the number of lead shots dropped is 440.
26. The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained?
Solution:
We have given that:
Given surface area of the sphere = 616 cm2
4R2 = 616
4×(22/7)R2= 616
R2= 616×7/4×22
R2= 49
R = 7
Volume of the solid metallic sphere = (4/3)R3
= (4/3)×73
= (1372/3) cm3
Diameter of smaller sphere = 3.5 cm
So radius, r = 3.5/2 = 7/4 cm
Volume of the smaller sphere = (4/3)r3
= (4/3)×(7/4)3
= (343/48) cm3
Number of spheres made = Volume of the solid metallic sphere/ Volume of the smaller sphere
= (1372/3) ÷(343/48)
=1372×48/(3×343)
= 64
Hence the number of spheres made is 64.
27. The surface area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate (i) the radius of the solid sphere. (ii) the number of cones recast. (Use π = 3.14).
Solution:
We have given that:
(i)Given surface area of the solid metallic sphere = 1256 cm2
4R2 = 1256
4×3.14×R2= 1256
R2 = 1256/4×3.14
R2 = 100
R = 10
Hence the radius of solid sphere is 10 cm.
(ii)Volume of the solid sphere = (4/3)R3
= (4/3)×103
= (4000/3) cm3
= 12560/3 cm3
Radius of the cone,r = 2.5 cm
Height of the cone, h = 8 cm
Volume of the cone = (1/3)r2 h
= (1/3)×3.14×2.5^2×8
= 157/3 cm^3
Number of cones made = Volume of the solid sphere/ Volume of the cone
= (12560/3)÷( 157/3)
= (12560/3)×( 3/157)
= 12560/157
= 80
Hence the number of cones made is 80.
28. Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboid pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?
Solution:
We have given that:
Given speed of water flow = 15 km/h
Diameter of pipe = 14 cm
So radius of pipe, r = 14/2 = 7 cm = 0.07 m
Dimensions of cuboidal pond = 50 m × 44 m
Height of water in pond = 21 cm = 0.21 m
So volume of water in pond = 50×44 ×0.21
= 462 m3
Volume of water in pipe = r2 h
= ×0.072×h
= 0.0049h
Volume of water in pond = Volume of water in pipe
462 = 0.0049h
h = 462/0.0049
= 462×7/0.0049×22
= 30000 m
= 30 km [1 km = 1000 m]
Speed = distance/ time
Time = Distance/speed = 30/15 = 2 hr
Hence the time taken is 2 hours.
29. A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate : (i) the total surface area of the can in contact with water when the sphere is in it. (ii) the depth of the water in the can before the sphere was put into the can. Given your answer as proper fractions.
Solution:
We have given that:
(i)Given radius of the cylinder, r = 3.5 cm
Diameter of the sphere = height of the cylinder
= 3.5×2
= 7 cm
So radius of sphere, r = 7/2 = 3.5 cm
Height of cylinder, h = 7 cm
Total surface area of can in contact with water = curved surface area of cylinder+base area of cylinder.
= 2rh+r2
= r(2h+r)
= 22/7)×3.5×(2×7+3.5)
= (22/7)×3.5×(14+3.5)
= 11×17.5
= 192.5 cm2
Hence the surface area of can in contact with water is 192.5 cm2.
(ii) Let the depth of the water before the sphere was put be d.
Volume of cylindrical can = volume of sphere+ volume of water
r2 h = (4/3)r3+r2 d
r2 h = r2 {(4/3)r +d)}
h = (4/3)r +d
d = h – (4/3)r
d = 7 – (4/3)×3.5
d = (21-14)/3
d = 7/3
Hence the depth of the water before the sphere was put into the can is 7/3 cm.