Find the Find the surface area of a sphere of radius :
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Surface area of a shere = 4πr2, where r is radius
(i) r is 10.5 cm
⇒ surface area = 4 × (22/7) × (10.5)2 = 1386 cm2
(ii) r is 5.6 cm
⇒ surface area = 4 × (22/7) × 5.62 = 394.24 cm2
(iii) r is 14 cm
⇒ surface area = 4 × (22/7) × 142 = 2464 cm2
Find the surface area of a sphere of diameter :
(i) 14 cm
(ii) 21 cm
(iii) 3.5 cm
Surface area of a sphere of diameter ‘d’ = πd2
(i) d is 14 cm
⇒ surface area = (22/7) × (14)2 = 616 cm2
(ii) d is 21 cm
⇒ surface area = (22/7) × (21)2 = 1386 cm2
(iii) d is 3.5 cm
⇒ surface area = (22/7) × (3.5)2 = 38.5cm2
Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. (Use π = 3.14)
We have,
Radius = 10 cm
Total surface area of a hemisphere = 2πr2
⇒ Total surface area of a hemisphere = 2 × 3.14 × 10 × 10 = 628 cm2
Total surface area of a solid hemisphere = 3πr2
⇒ Total surface area of a solid hemisphere = 3 × 3.14 × 10 × 10 = 942 cm2
The surface area of a sphere is 5544 cm2, find its diameter.
Let the radius of the sphere be r cm.
We know that, surface area of a sphere = 4πr2
Given, surface area of a sphere is 5544 cm2
⇒ 4πr2 = 5544
⇒ 4× × r2 = 5544
⇒ r2 = 441
⇒ r = 21 cm
Thus diameter = 42 cm
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of the plating it on the inside at the rate of Rs. 4 per 100 cm2.
Surface area of a hemisphere = 2πr2
Given, diameter of the hemisphere bowl is 10.5 cm
Surface area of the bowl = 2 × (22/7) × (10.5/2)2 = 173.25 cm2
Given, cost of the plating it on the inside at the rate of Rs. 4 per 100 cm2
Cost of plating the hemisphere bowl
The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of Rs. 2 per sq. m.
Surface area of a hemisphere = 2πr2
Given, dome of a building is in the form of a hemisphere. Its radius is 63 dm.
1 dm = 0.1m
Thus, 63 dm = 6.3 m
Surface area of the dome = 2 × (22/7) × (6.3)2 = 249.48 m2
Cost of painiting it at Rs. 2 per sq. m. = 249.48 × 2 = Rs. 498.96
Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-fourth of the earth’s surface is covered by water?
Surface area of a spere = 4πr2
Given, earth is a sphere of radius 6370 km.
Surface area of earth = 4 × (22/7) × 63702
⇒ Surface area of earth = 510109600 km2
Now, three-fourth of the earth’s surface is covered by water
Area covered by land = 1/4 × 510109600 = 127527400 km2
A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the shape if the length of the shape is 7 cm.
Given, cylinder of same height and radius is placed on the top of a hemisphere.
Also, length of the shape is 7 cm
⇒ r + r = 7
⇒ r = 3.5 cm
Curved surface area of a hemisphere = 2πr2
Curved surface area of a cylinder = 2πr2
Total surface area of the shape = 4πr2
⇒ Total surface area of the shape = 4 × (22/7) × (3.5)2
⇒ Total surface area of the shape = 154 cm2
A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at Rs. 7 per 100 cm2.
From the figure,
Radius of the hemisphere = 8 cm
Height of the cone = 15 cm
Lateral length of the cone = √(h2 + r2) = √(152 + 82) = 17 cm
Curved surface area of hemisphere = 2πr2
Curved surface area of the cone = πrl
Total curved surface area of the toy = 2πr2 + πrl
⇒ Total curved surface area of the toy = 2 × (22/7) × 82 + (22/7) × 8 × 17
⇒ Total curved surface area of the toy = 829.714 cm2
Cost of painting the toy at Rs. 7 per 100 cm2
A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m2.
Curved surface area of a cylinder = 2πrh
Curved surface area of hemisphere = 2πr2
Given, external diameter of the cylinder be 1.4 m and its length be 8 m
Thus r = 0.7m and h = 8m
Total curved surface area = 2πrh + 2πr2
⇒ Total curved surface area = 2 × (22/7) × 0.7 × 8 + 2 × (22/7) × 0.72 = 38.28 m2
Cost of painting it on the outside at the rate of Rs. 10 per m2 = 38.28 × 10 = Rs. 382.80
The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Surface area of a shephe = 4πr2
Ratio of surface areas of spheres = square of ratio of their radius
Given, diameter of the moon is approximately one fourth of the diameter of the earth
∴ rm = 1/4 × re
⇒ rm : re = 1 : 4
Ratio of their surface area = 1 : 16
A hemi-spherical dome of a building needs to be painted. IF the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is Rs. 5 per 100 cm2.
Circumference of a circle = 2πr
Surface area of a hemisphere = 2πr2
Given, base of the dome is 17.16 m
⇒ 2 × (22/7) × r = 17.6
⇒ r = 2.8 m
Surface area of the hemisphere = 2 × (22/7) × 2.82 = 49.28 m2 = 492800 cm2
Cost of painting is Rs. 5 per 100 cm2,
Cost of painting the dome = (492800/100) × 5 = Rs. 24640
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig. 21.11. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of a radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint cost 5 paise per cm2.
Surface area of a sphere = 4πr2
Curved surface area of a cylinder = 2πrh
Base area of a cylinder = πr2
Given, front compound wall of a house is decorated by wooden spheres of diameter 21 cm placed on small supports.
Each support is a cylinder of a radius 1.5 cm and height 7 cm and is to be painted black
Thus, the surface area of the sphere would be reduced by the cylindrical supports base.
Total surface area of the spherers = 8 × (4 × (22/7) × (21/2)2 – (22/7) × 1.52) = 11031.43 cm2
Total curved surface area of the spheres = 8 × 2 × (22/7) × 1.5 × 7 = 528 cm2
Silver paint costs 25 paise per cm2 and black paint cost 5 paise per cm2
Total cost of painting = 11031.43 × 0.25 + 528 × 0.05 = Rs. 2784.26
Find the volume of a sphere whose radius is :
(i) 2 cm
(ii) 3.5 cm
(iii) 10.5 cm
Volume of a sphere = (4/3)πr3
(i) radius is 2 cm
⇒ Volume of the sphere = (4/3) × (22/7) × 23 = 33.52 cm3
(ii) radius is 3.5 cm
⇒ Volume of the sphere = (4/3) × (22/7) × 3.53 = 179.67 cm3
(iii) radius is 10.5 cm
⇒ Volume of the sphere = (4/3) × (22/7) × (10.5)3 = 4851 cm3
Find the volume of a sphere whose diameter is:
(i) 14 cm
(ii) 3.5 dm
(iii) 2.1 m
Volume of sphere = (1/6)πd3
(i) Diamter is 14 cm
⇒ Volume of the sphere = (1/6) × (22/7) × 143 = 1437.33 cm3
(ii) Diamter is 3.5 dm = 35 cm
⇒ Volume of the sphere = (1/6) × (22/7) × 353 = 22.46 dm3
(iii) Diameter is 2.1 m
⇒ Volume of the sphere = (1/6) × (22/7) × 2.13 = 4.851 m3
A hemispherical tank has inner radius of 2.8 m. Find its capacity in litres.
Volume of a hemisphere = (2/3)πr3
Given, hemispherical tank has inner radius of 2.8 m
⇒ Volume of the tank = (2/3) × (22/7) × 2.83
⇒ Volume of the tank = 45.976 m3 = 45976 litres
A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.
Volume of a hemisphere = (2/3)πr3
Given, hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm.
Outer radius = 5 + 0.25 cm = 5.25 cm
Volume of steel used in the making of the bowl
⇒ Volume of steel used in the making of the bowl = 41.28 cm3
How many bullet can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?
Volume of a cube = side3
Volume of a sphere = (4/3)πr3
Given, edge of a cube is 22 cm and the bullet are of 2 cm diameter
Thus, volume of the cube = 22 × 22 × 22 = 10648 cm3
Volume of each bullet = (4/3) × (22/7) × 13 = 4.19 cm3
No. of bullet that can be made out of the cube of lead = 10648/4.19 = 2541
A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made.
Volume of a sphere = (4/3)πr3
Given, shopkeeper has one laddoo of radius 5 cm
Volume of the laddoo = (4/3) × (22/7) × 53
Now, from this one laddoo, laddoos of radius 2.5 cm are to be made.
Volume of laddoo of radius 2.5cm = (4/3) × (22/7) × 2.53
∴ No. of laddoos of radius 2.5 cm that can be made
A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be cm and 2 cm, find the diameter of the third ball.
Volume of a sphere = (4/3)πr3
Total volume remains same during recasting.
Given, spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls and the diameters of two balls are cm and 2 cm.
Let the diameter of the third ball be ‘a’ cm.
⇒ 27/8 = (27/64) + 1 + (a3/8)
⇒ a3/8 = 125/64
⇒ a3 = 125/8
⇒ a = 5/2 cm
A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises cm. Find the radius of the cylinder.
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Let the radius of the cylinder be r cm.
Given, sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises cm
Volume of the sphere = Volume of the water in the cylinder.
⇒ ×53 = π × r2 ×
⇒ r2 = 4×52
⇒ r = 10 cm
If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?
Volume of a sphere = (4/3)πr3
Let the radius of first sphere be’r’.
Radius of 2nd sphere = 2r
Ratio of the volume of the first sphere to that of the second sphere = 1 : 8
A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
Volume of a hemisphere = (2/3)πr3
Volume of a cone = (1/3)πr2h
Given, cone and a hemisphere have equal bases which implies they have the same radius.
Height of the hemisphere is its radius.
Let the base radius be ‘r’ and the height of cone be ‘h’.
Given, cone and hemisphere have equal volume.
(2/3)πr3 = (1/3)πr2h
⇒ h : r = 2 : 1
A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.
Volume of a hemisphere = (2/3)πr3
Volume of a cylinder = πr2h
Given, a vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively.
⇒ h = 7/12 cm
A cylinder whose height is two thirds of its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Given, cylinder whose height is two thirds of its diameter, has the same volume as a sphere of radius 4 cm.
⇒ h = (2/3) × 2r = 4r/3
Thus,
⇒ r = 4 cm
A vessel in the form of a hemispherical bowl is full of water. The contents are emptired into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.
Volume of a hemisphere = (2/3)πr3
Volume of a cylinder = πr2h
Given, internal radii of the bowl and cylinder are respectively 6 cm and 4 cm.
⇒ h = 9 cm
A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 9 cm. Find the radius of the ball. (Use π =22/7).
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Given, a cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 9 cm.
Volume of water displaced = Volume of the iron ball
⇒ r3 = 1728
⇒ r = 12 cm
A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball (Use π= 22/7)
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Given, a cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm.
Volume of water displaced = Volume of the iron ball
⇒ r3 = 729
⇒ r = 9 cm
The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Given, diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. The length of the wire is 108 m.
Long wire can be assumed to be a cylinder.
⇒ r = 0.6 cm
A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimetres?
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Given, cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil.
Level of the oil has to rise by 2 cm.
Let the number of spheres required be ‘n’.
⇒ n = 16
A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Amount of water displaced = Volume of the spheres
Given, measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each are dropped in it and they sink down in water completely.
Let the rise in level of water be ‘h’ cm.
⇒ h = 16/75 cm
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Given, diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm.
Long wire can be assumed to be a cylinder.
⇒ l = 3600 cm = 36 m
The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cylinder of height cm. Find the diameter of the cylinder.
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Given, radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid cylinder of height cm.
Volume of material = Volume of the solid cylinder
⇒ r2 = 49
⇒ r = 7 cm
Diameter = 14 cm
A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.
Volume of a hemisphere = (2/3)πr3
Volume of a right circular cone = (1/3)πr2h
Given, hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm
⇒ r2 = 14
⇒ r = 3.74 cm
A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.
Volume of a sphere = (4/3)πr3
Volume of a cone = (1/3)πr2h
Given, radius of the internal and external surfaces of a hollow spherical shell are 2 cm and 4 cm respectively. It is melted into a cone of base radius 4 cm.
⇒ Volume of material in sphere = Volume of the cone
⇒ h = 14 cm
L2 = h2 + r2
⇒ l = √(142 + 42)
⇒ l =√212 = 14.56 cm
A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.
Volume of a sphere = (4/3)πr3
Volume of a cone = (1/3)πr2h
Given, metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm.
Let the number of cones be ‘n’.
⇒ n × (1/3)π × 3.52 × 3 = (4/3) × π × 10.53
⇒ n = 126
A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
Volume of a hemisphere = (2/3)πr3
Volume of a right circular cone = (1/3)πr2h
Given, cone and a hemisphere have equal bases and equal volume
Height of a hemisphere is the radius and equal bases implies equal base radius.
(2/3)πr3 = (1/3)πr2h
⇒ r : h = 1 : 2
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2 : 3.
Volume of a hemisphere = (2/3)πr3
Volume of a right circular cone = (1/3)πr2h
Volume of a cylinder = πr2h
Given, a cone, a hemisphere and a cylinder stand on equal bases and have the same height.
Height of a hemisphere is the radius and equal bases implies equal base radius.
Thus, height of cone = height of cylinder = base radius = r
Ratio of volumes = (1/3)πr2h : (2/3)πr3 : πr2h
⇒ Ratio of volumes = r3 : 2r3 : 3r3 = 1 : 2 : 3
A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Volume of water displaced = Volume of the iron ball
Given, cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm
⇒ r3 = 729
⇒ r = 9 cm
The largest sphere is carved out of a cube of side 10.5 cm. Find the ratio of their volumes.
Largest sphere that can be carved out of a cube will have its diameter as the side of the cube.
Radius of the largest sphere carved out of a cube of side 10.5 cm = 10.5/2 = 5.25 cm
Volume of a cube = side3
Volume of a sphere = (4/3)πr3
Ratio of their volumes
A sphere, a cylinder and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.
Volume of a sphere = (4/3)πr3
Volume of a right circular cone = (1/3)πr2h
Volume of a cylinder = πr2h
Given, sphere, a cylinder and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere
Height of cone and cylinder = 2r
Ratio of their volumes = (4/3)πr3 : πr2h : (1/3)πr2h
⇒ Ratio of their volumes = 4r3 : 6r3 : 2r3 = 2 : 3 : 1
A cube of side 4 cm contains a sphere touching its side. Find the volume of the gap in between.
Volume of a cube = side3
Volume of a sphere = (4/3)πr3
Given, cube of side 4 cm contains a sphere touching its side
Radius of the sphere = 4/2 = 2 cm
Volume of the gap in between = 43 - (4/3)π × 23
⇒ Volume of the gap in between = 30.48 cm3
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Volume of a hemisphere = (2/3)πr3
Volume of material = (2/3)π × (r03 – ri3)
Given, hemispherical tank is made up of an iron sheet 1 cm thick and the inner radius is 1 m
⇒ r0 = 1 + 0.01 = 1.01 m
Volume of iron used
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to full this capsule?
Volume of a sphere = (4/3)πr3
Given, capsule of medicine is in the shape of a sphere of diameter 3.5 mm.
Radius = 3.5/2 = 1.75 mm
Volume of medicine filled inside = (4/3) × π × 1.753 = 22.458 mm3
The diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Volume of a sphere = (4/3)πr3
Given, diameter of the moon is approximately one fourth of the diameter of the earth
Radius of moon = 1/4 × radius of the earth.
Ratio of their volume
Volume of the moon is 1/64th times the volume of the earth.
In a cone the number of faces is
A. 1
B. 2
C. 3
D. 4
Faces are the flat surface on a 3D figure.
Thus, cone has one flat surface, i.e., the base.
Number of faces of a cone = 1
Find the surface area of a sphere of radius 14 cm.
Surface area of a sphere = 4πr2
Surface area of a sphere of radius 14 cm = 4 × (22/7) × 142 = 2464 cm2
The total surface area of a hemisphere of radius r is
A. πr2
B. 2πr2
C. 3πr2
D. 4πr2
Total surface area of a hemisphere = curved surface area + base area
⇒ total surface area of a hemisphere of radius r is = 2πr2 + πr2 = 3πr2
Find the total surface area of a hemisphere of radius 10 cm.
Total surface area of a hemisphere = 3πr2
Total surface area of a hemisphere of radius 10 cm = 3 × π × 102 = 942 cm2
Find the radius of a sphere whose surface area is 154 cm2.
Surface area of a sphere = 4πr2
Given, surface area is 154 cm2
⇒ 4 × (22/7) × r2 = 154
⇒ r2 = 49/4
⇒ r = 3.5 cm
The ratio of the total surface area of a sphere and a hemisphere of same radius is
A. 2 : 1
B. 3 : 2
C. 4 : 1
D. 4 : 3
Total surface area of a sphere = 4πr2
Total surface area of a hemisphere = 2πr2 + πr2 = 3πr2
Ratio of the total surface area of a sphere and a hemisphere of same radius = 4 : 3
A sphere and a cube are of the same height. The ratio of their volumes is
A. 3 : 4
B. 21: 11
C. 4 : 3
D. 11 : 21
Volume of a cube = side3
Volume of a sphere = (4/3)πr3
Given, sphere and a cube are of the same height.
Side = diameter = 2r
Ratio of their volumes
The hollow sphere, in which the circus motor cyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.
Surface area of a sphere = 4πr2
Given, hollow sphere, in which the circus motor cyclist performs his stunts, has a diameter of 7 m
Area available to the motorcyclist for riding = 4 × (22/7) × 3.52 = 154 m2
Find the volume of a sphere whose surface area is 154 cm2.
Surface area of a sphere = 4πr2
Given, surface area is 154 cm2
⇒ 4 × (22/7) × r2 = 154
⇒ r2 = 49/4
⇒ r = 3.5 cm
Volume of a sphere = (4/3)πr3
⇒ Volume of the given sphere = (4/3) π × 3.53 = 179.66 cm3
The largest sphere is cut off from a cube of side 6 cm. The volume of the sphere will be
A. 27π cm3
B. 36π cm3
C. 108π cm3
D. 12π cm3
Largest sphere that can be cut out of a cube will have its diameter as the side of the cube.
Radius of the largest sphere cut out of a cube of side 6 cm = 6/2 = 3 cm
Volume of a sphere = (4/3)πr3
⇒ Volume of the largest sphere that is cut off from a cube of side 6 cm
How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?
Volume of a cube = side3
Volume of a sphere = (4/3)πr3
Given, spherical bullets are to be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter.
Let the number of bullets be ‘a’.
⇒ 443 = a × (4/3) × (22/7) × 23
⇒ a = 2541
A cylindrical rod whose height is 8 times of its radius is melted and recast into spherical balls of same radius. The number of balls will be
A. 4
B. 3
C. 6
D. 8
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Given, cylindrical rod whose height is 8 times of its radius is melted and recast into spherical balls of same radius.
Let the number of such balls be ‘a’.
⇒ π × r2 × 8r = a × (4/3)π × r3
⇒ a = 6
If a sphere of radius 2r has the same volume as that of a cone with circular base of radius r, then find the height of the cone.
Volume of a sphere = (4/3)πr3
Volume of a right circular cone = (1/3)πr2h
Given, a sphere of radius 2r has the same volume as that of a cone with circular base of radius r
⇒ h = 32r
If the ratio of volumes of two spheres is 1: 8, then the ratio of their surface areas is
A. 1 : 2
B. 1 : 4
C. 1 : 8
D. 1 : 16
Ratio of volume of spheres = (ratio of radius)3
Given, ratio of volumes of two spheres is 1: 8
⇒ (ratio of radius)3 = 1 : 8
⇒ ratio of radius = 1 : 2
Ratio of surface area = (ratio of radius)2
⇒ Ratio of surface area = 1 : 4
If a hollow sphere of internal and external diamaters 4 cm and 8 cm respectively melted into a cone of base diameter 8 cm, then find the height of the cone.
Volume of a sphere = (4/3)πr3
Volume of a cone = (1/3)πr2h
Given, diameter of the internal and external surfaces of a hollow spherical shell are 4 cm and 8 cm respectively. It is melted into a cone of base diameter 8 cm.
⇒ Volume of material in sphere = Volume of the cone
⇒ h = 14 cm
If the surface area of a sphere is 144πm2, then its volume (in m3) is
A. 288π
B. 316π
C. 300π
D. 188π
Surface area of a sphere = 4πr2
Given, surface area of a sphere is 144πm2
⇒ 4πr2 = 144π
⇒ r = 6 m
Volume of the sphere = (4/3) × π × 63
⇒ Volume of the sphere = 288π m3
The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height of the cone.
Surface area of sphere = 4πr2
Curved syrface area of a cone = πrl
Given, surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm
⇒ 4 × π × 52 = 5 × π × 4 × l
⇒ l = 5 cm
L2 = h2 + r2
⇒ 52 = h2 + 42
⇒ h2 = 9
⇒ h = 3 cm
If a solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then the surface area of each ball (in sq. cm) is
A. 100π
B. 75π
C. 60π
D. 50π
Volume of sphere = (4/3)πr3
Given, solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius
⇒ (4/3)π × 103 = 8 × (4/3)π × r3
⇒ r= 10/2 = 5cm
Surface area of a sphere = 4πr2
Thus, the surface area of each sphere = 4 × π × 52 = 100π
If a sphere is inscribed in a cube, find the ratio of the volume of cube to the volume of the sphere.
Volume of a cube = side3
Volume of a sphere = (4/3)πr3
Given, sphere is inscribed in a cube
Diameter of sphere = side of the cube
Side of cube = 2r
Ratio of the volume of cube to the volume of the sphere
The ratio between the volume of a sphere and volume of a circumscribing right circular cylinder is
A. 2 : 1
B. 1 : 1
C. 2 : 3
D. 1 : 2
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
If a cylinder circumscrimbes a sphere of radius r , then its base radius is ‘r’ and height is diameter = 2r
Ratio between the volume of a sphere and volume of a circumscribing right circular cylinder
If a sphere is inscribed in a cube, then the ratio of the volume of the sphere to the volume of the cube is
A. π : 2
B. π : 3
C. π : 4
D. π : 6
Volume of a cube = side3
Volume of a sphere = (4/3)πr3
Given, sphere is inscribed in a cube
Diameter of sphere = side of the cube
Side of cube = 2r
Ratio of the volume of the sphere to the volume of cube
If a solid sphere of radius r is melted and cast into the shape of a solid cone of height r, then the radius of the base of the cone is
A. 2r
B. 3r
C. r
D. 4r
Volume of a sphere = (4/3)πr3
Volume of a solid cone = (1/3)πr2h
Given, solid sphere of radius r is melted and cast into the shape of a solid cone of height r
Let the base radius be A.
⇒ (4/3)πr3 = (1/3)π × A2 × r
⇒ A = 2r
A sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder. If the radius of the sphere is r, then the volume of the cylinder is
A. 4πr3
B. πr3
C. 2πr3
D. 8πr3
Volume of a sphere = (4/3)πr3
Volume of a cylinder = πr2h
Given, sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder and the radius of the sphere is r
Thus, height of the cylinder = diameter = 2r and base radius = r
Volume of the cylinder = π × r2 × 2r = 2πr3
A cone and a hemisphere have equal bases and equal volumes the ratio of their heights is
A. 1 : 2
B. 2 : 1
C. 4 : 1
D. : 1
Volume of a hemisphere = (2/3)πr3
Volume of a right circular cone = (1/3)πr2h
Given, cone and a hemisphere have equal bases and equal volume
Height of a hemisphere is the radius and equal bases implies equal base radius.
(2/3)πr3 = (1/3)πr2h
⇒ r : h = 1 : 2
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is
A. 1 : 2 : 3
B. 2 : 1 : 3
C. 2 : 3 : 1
D. 3 : 2 : 1
Volume of a hemisphere = (2/3)πr3
Volume of a right circular cone = (1/3)πr2h
Volume of a cylinder = πr2h
Given, a cone, a hemisphere and a cylinder stand on equal bases and have the same height.
Height of a hemisphere is the radius and equal bases implies equal base radius.
Thus, height of cone = height of cylinder = base radius = r
Ratio of volumes = (1/3)πr2h : (2/3)πr3 : πr2h
⇒ Ratio of volumes = r3 : 2r3 : 3r3 = 1 : 2 : 3