Surface Area And Volume Of A Sphere

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

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

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

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

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

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

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

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

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

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

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

∴ r_m = 1/4 × r_e

⇒ r_m : r_e = 1 : 4

Ratio of their surface area = 1 : 16

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Volume of a hemisphere = (2/3)πr³

Volume of material = (2/3)π × (r₀³ – r_i³)

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

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

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 1/64^th times the volume of the earth.

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

Surface area of a sphere = 4πr²

Surface area of a sphere of radius 14 cm = 4 × (22/7) × 14² = 2464 cm²

Total surface area of a hemisphere = curved surface area + base area

⇒ total surface area of a hemisphere of radius r is = 2πr² + πr² = 3πr²

Total surface area of a hemisphere = 3πr²

Total surface area of a hemisphere of radius 10 cm = 3 × π × 10² = 942 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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