Chapter 21: Surface Area and Volume of a Sphere Exercise – 21.2

Question: 1Find the volume of a sphere whose radius is:

(i) 2 cm

(ii) 3.5 cm

(iii) 10.5 cm.

Solution:(i) Radius (r) = 2 cm

Therefore volume = 4/3πr

^{3}= 4/3 × 22/7 × (2)

^{3}= 33.52 cm

^{3}(ii) Radius (r) = 3.5 cm

Therefore volume = 4/3πr

^{3}= 4/3 × 22/7 × (3.5)

^{3}= 179.666 cm^{3}(iii) Radius (r) = 10.5 cm

Therefore volume = 4/3πr

^{3}= 4/3 × 22/7 × (10.5)

^{3}= 4851 cm^{3}

Question: 2Find the volume of a sphere whose diameter is:

(i) 14 cm

(ii) 3.5 dm

(iii) 2.1 m

Solution:(i) Diameter = 14 cm, Radius(r) = 14/2 = 7 cm

Therefore volume = 4/3πr

^{3}= 4/3 × 22/7 × (7)

^{3}= 1437.33 cm^{3}(ii) Diameter = 3.5 dm, Radius (r) = 3.52 = 1.75 dm

Therefore volume = 4/3πr

^{3}= 4/3 × 22/7 × (1.75)

^{3}= 22.46 dm

^{3}(iii) Diameter = 2.1m, Radius(r) = 2.1/2 = 1.05 m

Therefore volume = 4/3πr

^{3}= 4/3 × 22/7 × (1.05)

^{3}= 4.851m^{3}

Question: 3A hemispherical tank has the inner radius of 2.8 m. Find its capacity in liters.

Solution:Radius of the tank = 2.8 m

Therefore Capacity = 2/3πr

^{3}= 2/3 × 22/7 × (2.8)

^{3}= 45.994 m^{3}1m

^{3}= 1000lTherefore capacity in litres = 45994 litres

Question: 4A 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.

Solution:Inner radius = 5 cm

Outer radius = 5 + 0.25 = 5.25

Volume of steel used = Outer volume-Inner volume

= 2/3 × π × ((5.25)

^{3}− (5)^{3})= 2/3 × 22/7 × ((5.25)

^{3}− (5)^{3})= 41.282 cm

^{3}

Question: 5How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?

Solution:Cube edge = 22 cm

Therefore volume of the cube = (22)

^{3}= 10648 cm^{3}And,

Volume of each bullet = 4/3πr

^{3}= 4/3 × 22/7 × (1)

^{3}= 4/3 × 22/7

= 88/21cm

^{3}

Question: 6A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made?

Solution:Volume of laddoo having radius = 5 cm

i.e Volume (V

_{1}) = 4/3πr^{3}(V

_{1}) = 4/3 × 22/7 × (5)^{3}(V

_{1}) = 11000/21 cm^{3}Also Volume of laddoo having radius 2.5 cm

i.e Volume (V

_{2}) = 4/3πr^{3}(V

_{2}) = 4/3 × 22/7 × (2.5)^{3}(V

_{2}) = 1375/21 cm^{3}Therefore number of laddoos = V

_{1}/V_{2 }= 11000/1375 = 8

Question: 7A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 3/2cm and 2 cm, find the diameter of the third ball.

Solution:Volume of lead ball = 4/3πr

^{3}= 4/3 × 22/7 × (3/2)

^{3}Diameter of first ball d

_{1}= 3/2 cmRadius of first ball r

_{1}Diameter of second ball d

_{2}= 2 cmRadius of second ball r

_{2}= 2/2cm = 1 cmDiameter of third ball d

_{3}= dRadius of third ball r

_{3}= d/2 cmd = 2.5 cm

Question: 8A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises 5/3 cm. Find the radius of the cylinder.

Solution:Radius of cylinder = r

Radius of sphere = 5cm

Volume of sphere = 4/3πr

^{3}= 4/3 × π × (5)

^{3}Height of water rised = 5/3cm

Volume of water rised in cylinder = πr

^{2}hTherefore, Volume of water rises in cylinder = Volume of sphere

Let r be the radius of the cylinder

πr

^{2}h = 4/3πr^{3}r

^{2}× 5/3 = 4/3 × π × (5)^{3}r

^{2}× 5/3 = 4/3 × 22/7 × 125r

^{2}= 20 × 5r = √100

r = 10 cm

Question: 9If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?

Solution:Let v

_{1}and v_{2}be the volumes of the first and second sphere respectivelyRadius of the first sphere = r

Radius of the second sphere = 2r

= 1/8

Question: 10A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Solution:Given that

Volume of the cone = Volume of the hemisphere

1/3πr

^{2}h = 2/3πr^{3}r

^{2}h = 2r^{3}h = 2r

h/r = 1/1 × 2 = 2

Therefore

Ratio of their heights = 2:1

Question: 11A 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.

Solution:Given that

Volume of water in the hemispherical bowl = Volume of water in the cylinder

Let h be the height to which water rises in the cylinder

Inner radii of the bowl = r

_{1}= 3.5 cmInner radii of the bowl = r

_{2}= 7 cmh = 7/12 cm

Question: 12A 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.

Solution:Given that

Height of the cylinder = 2/3 diameter

We know that

Diameter = 2(radius)

h = 2/3 × 2r = 4/3r

Volume of the cylinder=Volume of the sphere

πr

^{2}h = 4/3πr^{3}π × r

^{2}× (4/3r) = 4/3π(4)^{3}(r)

^{3}= (4)^{3}r = 4 cm

Question: 13A vessel in the form of a hemispherical bowl is full of water. The contents are emptied 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.

Solution:It is given that

Volume of water in hemispherical bowl = Volume of cylinder

h = 9 cm

Question: 14A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?

Solution:Let r be the radius of the iron ball

Radius of the cylinder = 16 cm

Then,

Volume of iron ball = Volume of water raised in the hub

r

^{3}= 1728r = 12 cm

Therefore radius of the ball = 12 cm.

Question: 15A 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 = 227).

Solution:Given that:

Radius of the cylinder = 12cm = r

_{1}Raised in raised = 6.75 cm = r

_{2}Volume of water raised = Volume of the sphere

= r

_{2}= 9 cmRadius of the sphere is 9 cm

Question: 16The 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.

Solution:Given that diameter of a copper sphere = 18 cm

Radius of the sphere = 9 cm

Length of the wire = 108 m = 10800 cm

Volume of cylinder = Volume of sphere

r

_{1}= 0.3 cmTherefore Diameter = 2 × 0.3 = 0.6 cm

Question: 17A 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 centimeters?

Solution:Given that,

Radius of the cylinder jar = 6 cm = r

_{1}Level to be rised = 2 cm

Radius of each iron sphere = 1.5 cm = r

_{2}Number of sphere = 16

Question: 18A 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?

Solution:Given that,

Diameter of jar = 10 cm

Radius of jar = 5 cm

Let the level of water be raised by h

Diameter of the spherical bowl = 2 cm

Radius of the ball = 1 cm

Volume of jar = 4 (Volume of spherical ball)

Height of water in jar = 16/75 cm

Question: 19The 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.

Solution:Given that

Diameter of sphere = 6 cm

Radius of sphere = d/2 = 6/2 = 3 cm = r

_{1}Diameter of the wire = 0.2 cm

Radius of the wire = 0.1 cm = r

_{2}Volume of sphere = Volume of wire

h = 3600 cm

h = 36 m

Therefore length of wire = 36 m

Question: 20The 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 22/3 cm. Find the diameter of the cylinder.

Solution:Given that,

Internal radius of the sphere = 3 cm = r

_{1}External radius of the sphere = 5 cm = r

_{2}Height of the cylinder = 8/3cm = h

Volume of the spherical shell = Volume of cylinder

r

_{3}= 7cmTherefore diameter of the cylinder = 2(radius) = 14 cm

Question: 21A hemisphere of the lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.

Solution:Given

Radius of the hemisphere = Volume of cone

r

_{2}= 3.47 cmTherefore radius of the base = 3.74 cm

Question: 22A 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.

Solution:Given that

Hollow sphere external radii = r

_{2}= 4 cmInternal radii = r

_{1}= 2 cmCone base radius (R) = 4 cm

Height = h

Volume of cone = Volume of sphere

h = 14 cm

Question: 23A 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.

Solution:Given that

Metallic sphere of radius = 10.5 cm

Cone radius = 3.5 cm

Height of radius = 3 cm

Let the number of cones obtained be x

x = 126

Therefore number of cones = 126

Question: 24A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Solution:Given that

A cone and a hemisphere have equal bases and volumes

v

_{cone}= v_{hemisphere}1/3πr

^{2}h = 2/3πr^{3}r

^{2}h = 2r^{3}h = 2r

hr = 2/1

h:r = 2:1

Therefore the ratio is 2:1

Question: 25A 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.

Solution:Given that

A cone, a hemisphere and a cylinder stand on one equal bases and have the same weight

We know that

v

_{cone}:v_{hemisphere}:v_{cylinder}1/3 πr

^{2}h: 2/3 πr^{3}: πr^{2}hMultiplying by 3

πr

^{2}h: 2πr^{3}: 3πr^{2}hπr

^{3}: 2πr^{3}: 3πr^{3 }(∴ r = h and r^{2}h = r^{3})1:2:3 Therefore the ratio is 1: 2: 3.

Question: 26A 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?

Solution:Radius of cylindrical tub = 12 cm

Depth = 20 cm

Let r be the radius of the ball

Then

Volume of the ball = Volume of water raised

r = 9 cm

Therefore radius of the ball = 9 cm

Question: 27The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.

Solution:Side of cube = 10.5 cm

Volume of sphere = v

Diameter of the largest sphere = 10.5 cm

2r = 10.5

r = 5.25 cm

Volume of sphere = 4/3πr

^{3}= 4/3 × 22/7 × 5.25 × 5.25 × 5.25v = 606.375 cm

^{3}

Question: 28A 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.

Solution:Let r be the common radius

Height of the cone = height of the cylinder = 2r

Let

v

_{1}= Volume of sphere = 4/3 πr^{3}v

_{1}= Volume of cylinder = πr^{2}h = πr^{2}× 2rv

_{1}= Volume of cone = 1/3πr^{2}h = 1/3πr^{3}Now

v

_{1}:v_{2}:v_{3}= 4/3πr^{3}:2πr^{3}: 2/3πr^{3}= 4:6:2 = 2:3:1

Question: 29A cube of side 4 cm contains a sphere touching its side. Find the volume of the gap in between.

Solution:It is given that

Cube side = 4cm

Volume of cube = (4 cm)

^{3}= 64 cm^{3}Diameter of the sphere = Length of the side of the cube = 4cm

Therefore radius of the sphere = 2cm

Volume of the sphere = 4/3πr

^{3 }= 4/3 × 22/7 × (2)^{3}= 33.52 cm^{3}Volume of gap = Volume of cube - Volume of sphere

= 64 cm

^{3}- 33.52 cm^{3}= 30.48 cm^{3}

Question: 30A 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.

Solution:Given that,

Inner radius of the hemispherical tank = 1 m = r

_{1}Thickness of the hemispherical tank = 1 cm = 0.01 m

Outer radius of hemispherical tank = (1 + 0.01) = 1.01 m = r

_{2}Volume of iron used to make the tank

= 0.06348 m

^{3}

Question: 31A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (mm

^{3}) is needed to fill this capsule?

Solution:Given that

Diameter of capsule = 3.5 mm

Radius = 3.5/2 = 1.75 mm

Volume of spherical sphere = 4/3πr

^{3}= 4/3 × 22/7 × (1.75)

^{3}= 22.458 mm

^{3}Therefore 22.46 mm

^{3}of medicine is required

Question: 32The diameter of the moon is approximately one-fourth of the diameter of the earth. What is the earth the volume of the moon?

Solution:Diameter of moon = 1/4th diameter of earth

Let the diameter of earth be d, so radius = d/2

Then diameter of moon = d/4

= 1/64

Thus the volume of the moon is 1/64 of volume of earth