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Mensuration
Sphere
Hemisphere
1. Find the volume and surface area of a sphere of radius
(a) 10.5 cm, (b) 28 cm, (c) 7 cm, (d) 14cm.
Solution:
(a) r = 10. 5 cm
Volume of a sphere = r^3
= x x 10. 5 x 10. 5 x 10. 5
= 4851 cu.cm Surface area of a sphere = 4 r^2 = 4 x x 10. 5 x 10. 5 = 1386 sq.cm
(b) r = 28 cm
Volume of a sphere = x x 28 x 28 x 28 = 91989.33 cu.cm.
Surface area of a sphere = 4r^2
= 4 x x 28 x 28 = 9856 sq.cm.
(c) r = 7cm
Volume of sphere = x x 7 x 7 x 7 = 1437. 33 cu.cm.
Surface area = 4 r^2
= 4 x x 7 x 7= 616 sq.cm.
(d) r = 14 cm
Volume of a sphere = r^3 = x x 14 x 14 x 14 = 11498.67 cm^3
Surface area = 4 r^2
= 4 x x 14 x 14 = 2464 sq.cm.
2. The surface area of a sphere is 1386 sq.cm. find its volume.
Solution : Surface area of sphere = 1386cm^2 4 r^2 = 1386
4 x x r^2 = 1386
r^2 = = 110.
r = 10.
Volume of sphere = r^3
= x x 10.5 x10.5 x10.
= 4851cu.cm.
3. The ratio of the diameters of two spheres is 4 : 5. Find the ratio of their volumes.
Solution: Ratio of diameters of two sphere =
ratio of radii =
r 1 4 r 2 5
4 / 3 r 13 r 1
3 4 3 64
4 / 3 r 23 r 2 5 125
1386 x 7 4 x 22
Radio of their volume = = = =
5 x 5 x 8 x 2 x 2 x 2 4 x 4
Number of shots = 100
7. The diameter of a copper sphere is 6 cm. The sphere is melted and is drawn into a long wire of uniform diameter. If the length of the wire is 36 cm, find its radius. Solution: Diameter of copper sphere = 6 cm.
Radius of copper sphere = 3 cm
Volume of sphere = r^3 = x x 3 x 3 x 3 ----- (1)
Volume of long wire = r^2 h
= x r^2 x 36 ----- (2)
where r is the radius of wire. Volume of sphere = Volume of long wire
x x 3 x 3 x 3 = x r^2 x 36 4 x 3 x 3 36
r^2 = 1 cm
Radius of the wire = 1 cm
8. A cylindrical bowl with base diameter 7 m contains water. A solid sphere is dropped into
it. The water level increases by 4 m. Find the radius of the sphere.
Solution: Let R be the radius of the sphere. Volume of sphere = Volume of water designed 4 22 7 7 14 3 7 2 2 3
x 3 x x 5 x 5 x 8 = x x x x x x
x = = 100
r^2 = = 1
R^3 = x x x
(diameter of cylinder = 14 cm
Increase in height = 4 m ) 4 22 22 7 7 14 3 7 7 2 2 3
R^3 = R = = 3.
Radius of the sphere = 3.5 cm
9. An iron cone of diameter 8 cm and height 12 cm is melted and recast into lead shots of radius 2 mm. How many lead shots are obtained?
Solution: diameter of cone = 8 cm radius r = 4 cm height = 12 cm
Volume of the cone = r^2 h
= x 4 x 4 x 12 ----- (1)
Radius of lead shot = 2 mm = cm = cm
Volume of a lead shot = x x x ----- (2)
Volume of curve Volume of lead shot
4 x 4 x 12 x 5 x 5 x 5 4
10. A hemispherical bowl of internal diameter 36 cm contains a liquid. This liquid is to be filled into
cylindrical shaped bottled of radius 3 cm and height 6 cm. How many bottles are required to empty the
bowl?
Solution: Diameter of a hemispherical bowl = 36 cm.
radius r = 18 cm
Volume of hemisphere = r^3 = x 18 x 18 x 18 ----- (1)
Volume of cylinder = r^2 h (Here r = 3 cm; h = 6 cm)
x x R^3 = x x x
7 x 7 x 7 7 2 x 4 2
No. of shots = =
1 3 4 1 1 1 3 5 5 5
x 4 x 4 x 12
x x x
Radius = 3 cm
Hence height of the cylinder = 13 – 3 = 10 cm.
radius of cylinder = 3 cm = radius of hemisphere
Total volume = Volume of cylinder
= r^2 h + r^3
= x 3 x 3 x 10 + x 3 x 3 x 3 = 90 + 18 = 108
= = 339.43 cm^3.
13. A cylindrical boiler 2m high and 3.5m in diameter, has a hemispherical lid, find the volume of the interior of the boiler, including the part covered by the lid.
Solution:
2
Height of cylinder = 2m
Total volume = Volume of cylinder + Volume of hemispherical lid 2 3 2 3 = 6.125 + 3.573 22 7
108 x 22 7
radius of cylinder = = 1.75m
= r^2 h + r^3
= 1.75 x 1.75 x 2 + 1.75 x 1.75 x 1.
= 9.698 = 9.698 x = 30.48 m^3
14. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the total surface area of the solid.
Solution: 7 2 the height of the cylinder = 19 – 7 = 12 cm
Total surface area = Surface area of cylinder + 2 (surface of hemisphere) = 2 rh + 2 (2 r^2 )
= [ 2 x 3.5 x 12 ] + 2 x 2 x x 3.5 x 3.
= 84 + 49 = 133
133 x 22 7
15. A hemispherical shell is made of metal which weights 3 gm per cubic centimeter of metal. Find the weight of the hemisphere if its internal diameter is 10 cm and has thickness 1 cm. Solution: Internal diameter = 10 cm internal radius = 5 cm
Thickness = 1 cm external radius = 6 cm
= 572 gm
Since the radius of cylinder is = 3.5 cm
= = 418 cm^2
3.5 3.
Volume of hemisphere = (R^3 – r^3 )
= x (6^3 – 53 )
= x (216 – 125)
= x x 91 cm^3
Weight of hemisphere = x x 91 x 3 gm
R^3 – r^3 = 61
125 – r^3 = 61 r^3 = 64
r = 4 cm
radius of internal radius = 4 cm
Thickness = R – r = 5 – 4 cm
1 cm
R^3 – r^3 ) =
