Chapter 21: Surface Area and Volume of a Sphere Exercise – 21.2
Question: 1
Find 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 = 4/3πr3
= 4/3 × 22/7 × (2)3
= 33.52 cm3
(ii) Radius (r) = 3.5 cm
Therefore volume = 4/3πr3
= 4/3 × 22/7 × (3.5)3 = 179.666 cm3
(iii) Radius (r) = 10.5 cm
Therefore volume = 4/3πr3
= 4/3 × 22/7 × (10.5)3 = 4851 cm3
Question: 2
Find 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 = 4/3πr3
= 4/3 × 22/7 × (7)3 = 1437.33 cm3
(ii) Diameter = 3.5 dm, Radius (r) = 3.52 = 1.75 dm
Therefore volume = 4/3πr3
= 4/3 × 22/7 × (1.75)3
= 22.46 dm3
(iii) Diameter = 2.1m, Radius(r) = 2.1/2 = 1.05 m
Therefore volume = 4/3πr3
= 4/3 × 22/7 × (1.05)3 = 4.851m3
Question: 3
A 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 = 2/3πr3
= 2/3 × 22/7 × (2.8)3 = 45.994 m3
1m3 = 1000l
Therefore capacity in litres = 45994 litres
Question: 4
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.
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 cm3
Question: 5
How 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 cm3
And,
Volume of each bullet = 4/3πr3
= 4/3 × 22/7 × (1)3
= 4/3 × 22/7
= 88/21cm3

Question: 6
A 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 (V1) = 4/3πr3
(V1) = 4/3 × 22/7 × (5)3
(V1) = 11000/21 cm3
Also Volume of laddoo having radius 2.5 cm
i.e Volume (V2) = 4/3πr3
(V2) = 4/3 × 22/7 × (2.5)3
(V2) = 1375/21 cm3
Therefore number of laddoos = V1/V2 = 11000/1375 = 8
Question: 7
A 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 = 4/3πr3
= 4/3 × 22/7 × (3/2)3
Diameter of first ball d1 = 3/2 cm
Radius of first ball r1
Diameter of second ball d2 = 2 cm
Radius of second ball r2 = 2/2cm = 1 cm
Diameter of third ball d3 = d
Radius of third ball r3 = d/2 cm

d = 2.5 cm
Question: 8
A 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 = 4/3πr3
= 4/3 × π × (5)3
Height of water rised = 5/3cm
Volume of water rised in cylinder = πr2h
Therefore, Volume of water rises in cylinder = Volume of sphere
Let r be the radius of the cylinder
πr2h = 4/3πr3
r2 × 5/3 = 4/3 × π × (5)3
r2 × 5/3 = 4/3 × 22/7 × 125
r2 = 20 × 5
r = √100
r = 10 cm
Question: 9
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?
Solution:
Let v1 and v2 be the volumes of the first and second sphere respectively
Radius of the first sphere = r
Radius of the second sphere = 2r

= 1/8
Question: 10
A 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
1/3πr2h = 2/3πr3
r2h = 2r3
h = 2r
h/r = 1/1 × 2 = 2
Therefore
Ratio of their heights = 2:1
Question: 11
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.
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 = r1 = 3.5 cm
Inner radii of the bowl = r2 = 7 cm

h = 7/12 cm
Question: 12
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.
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
πr2h = 4/3πr3
π × r2 × (4/3r) = 4/3π(4)3
(r)3 = (4)3
r = 4 cm
Question: 13
A 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: 14
A 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

r3 = 1728
r = 12 cm
Therefore radius of the ball = 12 cm.
Question: 15
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 = 227).
Solution:
Given that:
Radius of the cylinder = 12cm = r1
Raised in raised = 6.75 cm = r2
Volume of water raised = Volume of the sphere

= r2 = 9 cm
Radius of the sphere is 9 cm
Question: 16
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.
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

r1 = 0.3 cm
Therefore Diameter = 2 × 0.3 = 0.6 cm
Question: 17
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 centimeters?
Solution:
Given that,
Radius of the cylinder jar = 6 cm = r1
Level to be rised = 2 cm
Radius of each iron sphere = 1.5 cm = r2

Number of sphere = 16
Question: 18
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?
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: 19
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.
Solution:
Given that
Diameter of sphere = 6 cm
Radius of sphere = d/2 = 6/2 = 3 cm = r1
Diameter of the wire = 0.2 cm
Radius of the wire = 0.1 cm = r2
Volume of sphere = Volume of wire

h = 3600 cm
h = 36 m
Therefore length of wire = 36 m
Question: 20
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 22/3 cm. Find the diameter of the cylinder.
Solution:
Given that,
Internal radius of the sphere = 3 cm = r1
External radius of the sphere = 5 cm = r2
Height of the cylinder = 8/3cm = h
Volume of the spherical shell = Volume of cylinder

r3 = 7cm
Therefore diameter of the cylinder = 2(radius) = 14 cm
Question: 21
A 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

r2 = 3.47 cm
Therefore radius of the base = 3.74 cm
Question: 22
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.
Solution:
Given that
Hollow sphere external radii = r2 = 4 cm
Internal radii = r1 = 2 cm
Cone base radius (R) = 4 cm
Height = h
Volume of cone = Volume of sphere

h = 14 cm

Question: 23
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.
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: 24
A 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
vcone = vhemisphere
1/3πr2h = 2/3πr3
r2h = 2r3
h = 2r
hr = 2/1
h:r = 2:1
Therefore the ratio is 2:1
Question: 25
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.
Solution:
Given that
A cone, a hemisphere and a cylinder stand on one equal bases and have the same weight
We know that
vcone:vhemisphere:vcylinder
1/3 πr2h: 2/3 πr3: πr2h
Multiplying by 3
πr2h: 2πr3: 3πr2h
πr3: 2πr3: 3πr3 (∴ r = h and r2h = r3)
1:2:3 Therefore the ratio is 1: 2: 3.
Question: 26
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?
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: 27
The 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 = 4/3πr3 = 4/3 × 22/7 × 5.25 × 5.25 × 5.25

v = 606.375 cm3
Question: 28
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.
Solution:
Let r be the common radius
Height of the cone = height of the cylinder = 2r
Let
v1 = Volume of sphere = 4/3 πr3
v1 = Volume of cylinder = πr2h = πr2 × 2r
v1 = Volume of cone = 1/3πr2h = 1/3πr3
Now
v1:v2:v3 = 4/3πr3:2πr3: 2/3πr3
= 4:6:2 = 2:3:1
Question: 29
A 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 cm3
Diameter of the sphere = Length of the side of the cube = 4cm
Therefore radius of the sphere = 2cm
Volume of the sphere = 4/3πr3 = 4/3 × 22/7 × (2)3 = 33.52 cm3
Volume of gap = Volume of cube - Volume of sphere
= 64 cm3 - 33.52 cm3 = 30.48 cm3
Question: 30
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.
Solution:
Given that,
Inner radius of the hemispherical tank = 1 m = r1
Thickness of the hemispherical tank = 1 cm = 0.01 m
Outer radius of hemispherical tank = (1 + 0.01) = 1.01 m = r2
Volume of iron used to make the tank

= 0.06348 m3
Question: 31
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (mm3) 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 = 4/3πr3
= 4/3 × 22/7 × (1.75)3
= 22.458 mm3
Therefore 22.46 mm3 of medicine is required
Question: 32
The 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
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