RD Sharma Solutions Class 9 Chapter 21 Surface Area and Volume of A Sphere

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Exercise 21.1 
 
Question 1: Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm 
Solution:
(i) It is give that,
Radius = 10.5 cm
Surface area of a sphere = 4πr2
Surface area = 4 × 22/7 × (10.5)2
Surface area = 4 × 22/7 × 10.5 × 10.5
Surface area = 4 × 22 × 1.5 × 10.5
Surface area = 1386
Surface area is 1386 cm2
 
(ii) It is given that,
Radius = 5.6 cm
Surface area = 4 × 22/7 × (5.6)2
Surface area = 4 × 22/7 × 5.6 × 5.6
Surface area = 4 × 22 × 0.8 × 5.6
Surface area = 394.24
Surface area is 394.24 cm2
 
(iii) It is given that,
Radius = 14 cm
Surface area = 4 × 22/7  × (14)2
Surface area = 4 × 22/7  × 14 × 14
Surface area = 4 × 22 × 2 × 14
Surface area = 2464
Surface area is 2464 cm2
 
Question 2: Find the surface area of a sphere of diameter:
(i) 14 cm (ii) 21 cm (iii) 3.5 cm 
Solution: 
Where, r = radius of a sphere
(i) Diameter= 14 cm
Radius = diameter/2 
Radius = 14/2 cm 
Radius = 7 cm
Surface area of a sphere = 4πr2
Surface area = 4 × 22/7 × (7)2
Surface area = 4 × 22/7 × 7 × 7
Surface area = 4 × 22 × 7 
Surface area = 616
Surface area is 616 cm2
 
(ii) Diameter = 21cm
Radius = diameter/2 
Radius = 21/2 cm 
Radius = 10.5 cm
Surface area of a sphere = 4πr2
Surface area = 4 × 22/7 × (10.5)2
Surface area = 4 × 22/7 × 10.5 × 10.5
Surface area = 4 × 22 × 1.5 × 10.5
Surface area = 1386
Surface area is 1386 cm2
 
(iii) Diameter= 3.5cm
Radius = diameter/2 
Radius = 3.5/2 cm 
Radius = 1.75 cm
Surface area of a sphere = 4πr2
Surface area = 4 × 22/7  × (1.75)2
Surface area = 4 × 22/7  × 1.75 × 1.75
Surface area = 4 × 22 × 0.25 × 1.75
Surface area = 38.5
Surface area is 38.5 cm2
 
Question 3: Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. (π = 3.14) 
Solution:
It is given that, the Radius of hemisphere is 10cm.
Surface area of the hemisphere = 2πr2
Surface area of the hemisphere = 2 × 3.14 × (10)
Surface area of the hemisphere = 2 × 3.14 × 10 × 10 
Surface area of the hemisphere = 628cm2
 
Surface area of solid hemisphere = 3πr2
Surface area of solid hemisphere = 3 × 3.14 × (10)2
Surface area of solid hemisphere = 3 × 3.14 × 10 × 10
Surface area of solid hemisphere = 942cm2
 
Question 4: The surface area of a sphere is 5544 cm2, find its diameter. 
Solution:
It is given that,
Surface area of a sphere is 5544cm2
Surface area of a sphere = 4πr2
4πr2 = 5544
4 × 22/7 × (r)2 = 5544
r2 = (5544 × 7)/(22 × 4)
r2 = (5544 × 7)/88
r2 = (5544 × 7)/88
r2 = 63×7
r2 = 441
r = 21cm
Diameter = 2(radius) 
Diameter = 2(21) 
Diameter = 42cm
Hence, the diameter is surface area of a sphere is 42cm.
 
Question 5: A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin plating it on the inside at the rate of Rs.4 per 100 cm2. 
Solution:
It is given that,
Inner diameter of hemispherical bowl = 10.5 cm
Radius = Diameter/2
Radius = 10.5/2cm 
Radius = 5.25 cm
 
Surface area of hemispherical bowl = 2πr2
Surface area of hemispherical bowl = 2 × 3.14 × (5.25)2
Surface area of hemispherical bowl = 173.25
Surface area of hemispherical bowl is 173.25cm2
 
It is also given that,
Cost of tin plating 100cm2 area is Rs.4
Cost of tin plating 173.25cm2 area = Rs. 4 × 173.25/100 = Rs. 6.93
Hence, cost of tin plating the inner side of hemispherical bowl is Rs.6.93.
 
Question 6: 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. 
Solution:
It is given that, Radius of hemispherical = 63dm  
Change in meters,
Radius of hemispherical dome = 63/10
Radius of hemispherical dome = 6.3m
 
Inner surface area = 2πr2
Inner surface area = 2 × 3.14 × (6.3)2
Inner surface area = 2 × 3.14 × 6.3 × 6.3 
Inner surface area = 249.48
Hence, the inner surface area of dome is 249.48 m2
 
It is also given that,
Cost of painting 1m2 = Rs.2
Cost of painting 249.48 m2 
Cost of painting = Rs. 249.48 × 2 
Cost of painting = Rs. 498.96
Hence, the cost of painting is Rs. 498.96.  
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) It is given that, 
Radius = 2 cm
Volume of a sphere = 4/3πr3 
Volume of a sphere = 4/3 × 22/7 × (2)3
Volume of a sphere = 33.52
Volume of a sphere = 33.52 cm3
 
(ii) It is given that, 
Radius = 3.5cm
Volume of a sphere = 4/3πr
Volume of a sphere = 4/3 × 22/7 × (3.5)3
Volume of a sphere = 4/3 × 22/7 × 3.5 × 3.5
Volume of a sphere = 179.666
Volume of a sphere = 179.666 cm3
 
(iii) It is given that,
Radius = 10.5 cm
Volume of a sphere = 4/3πr
Volume of a sphere = 4/3 × 22/7 × (10.5)3
Volume of a sphere = 4/3 × 22/7 × 10.5 × 10.5
Volume of a sphere = 4/3 × 22 × 1.5 × 10.5
Volume of a sphere = 4851
Volume of a sphere = 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) It is given that,
Diameter = 14 cm
Radius = diameter/2 = 14/2 = 7cm
Volume of a sphere = 4/3πr
Volume of a sphere = 4/3 × 22/7 × (7)3
Volume of a sphere = 4/3 × 22/7 × 7 × 7 × 7
Volume of a sphere = 1437.33
Volume of a sphere = 1437.33 cm3
 
(ii) It is given that,
Diameter = 3.5dm
Radius = diameter/2 
Radius = 3.5/2 
Radius = 1.75dm 
Volume of a sphere = 4/3πr3 
Volume of a sphere = 4/3 × 22/7  × (1.75)3
Volume of a sphere = 4/3 × 22/7  × 1.75 × 1.75 × 1.75 
Volume of a sphere = 22.46
Volume of a sphere = 22.46 dm3
 
(iii) It is given that,
Diameter = 2.1m
Radius = diameter/2 
Radius = 2.1/2 
Radius = 1.05 m
Volume of a sphere = 4/3πr3 
Volume of a sphere = 4/3  × 22/7 × (1.05)3
Volume of a sphere = 4/3  × 22/7 × 1.05 × 1.05 × 1.05
Volume of a sphere = 4 × 22 × 0.05 × 1.05 × 1.05
Volume of a sphere = 4.851
Volume of a sphere = 4.851 m3
 
Question 3: A hemispherical tank has the inner radius of 2.8 m. Find its capacity in litre. 
Solution:
It is given that, the radius of hemispherical tank is 2.8 m
Capacity of hemispherical tank = 2/3 πr3
Capacity of hemispherical tank = 2/3 × 22/7 × (2.8)
Capacity of hemispherical tank = 2/3 × 22/7 × 2.8 × 2.8 × 2.8
Capacity of hemispherical tank = 2/3 × 22 × 0.4 × 2.8 × 2.8
Capacity of hemispherical tank = 45.997 m3
 
Change in litres
1m3 = 1000 litres
Capacity in litres = 45.997 × 1000
Capacity in litres = 45997 litres
Hence, the capacity in litres is 45997litres.
 
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:
It is given that, the inner radius of a hemispherical bowl is 5cm
The outer radius of a hemispherical bowl is 5 cm + 0.25 cm = 5.25 cm
Volume of steel used = Outer volume – Inner volume
Volume of hemisphere = 2/3 πR3 - 2/3 πr
Volume of hemisphere = 2/3 π(R3 - r3)
Volume of steel used = 2/3 × π × ((5.25)3 − (5)3)
Volume of steel used = 2/3 × 22/7 × (144.70 − 125)
Volume of steel used = 2/3 × 22/7 × (19.70)
Volume of steel used = 41.282cm3
Hence, the volume of steel used in making the bowl is 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:
Edge of a cube = 22cm
Diameter of bullet = 2cm
Radius of bullet (r) = 2/2 cm = 1cm
Volume of the cube = (side)
Volume of the cube = (22)3 cm
Volume of the cube = 10648 cm3
 
Volume of sphere = 4/3πr3
Volume of sphere = 4/3 × 22/7 × (1)3 cm3
Volume of sphere = 4/3 × 22/7 cm3
Volume of sphere = 88/21 cm3
 
Number of bullets =( (Volume of cube) )/((Volume of bullet)) 
Number of bullets = (10648/88)/21
Number of bullets = (10648  × 21)/88
Number of bullets = 2541
Hence, 2541 bullets can be made out of a cube of lead.
 
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 (of Radius 5 cm) = 4/3πr3
Volume of laddoo (of Radius 5 cm) = 4/3  × 22/7 × (5)3
Volume of laddoo (of Radius 5 cm) = 4/3  × 22/7 × 5 × 5 × 5
Volume of laddoo (of Radius 5 cm) = 4/3  × 22/7 × 125
Volume of laddoo (of Radius 5cm) = 11000/21 cm3
 
Volume of laddoo (of Radius 2.5) = 4/3πr3
Volume of laddoo (of Radius 2.5) = 4/3 × 22/7 × (2.5)
Volume of laddoo (of Radius 2.5) = 4/3 × 22/7 × 2.5 × 2.5 × 2.5 
Volume of laddoo (of Radius 2.5) = 4/3 × 22/7 × 15.625 
Volume of laddoo (of Radius 2.5) = 1375/21 cm3
 
Number of laddoos of radius 2.5 cm that can be made with laddoos of radius 5cm 
Number of laddoos = (Volume of Laddoo of R=5cm)/(Volume of Laddoo of R=2.5cm) 
Number of laddoos = (11000/21)/(1375/21) 
Number of laddoos = 11000/1375
Number of laddoos = 8
Hence, the Number of laddoos is 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:
It is given that,
Diameter of spherical ball = 3cm
Radius of spherical ball = 3/2cm
Volume of lead ball = 4/3πr3
Volume of lead ball = 4/3 × π × (3/2)3
 
It is also given that,
Diameter of first ball = 3/2cm
Radius of first ball = 3/4 cm
Diameter of second ball = 2 cm
Radius of second ball = 2/2 cm = 1 cm
Let us assumed that,
Diameter of third ball = d
Radius of third ball = d/2 cm
We know that,
Volume of Lead ball = Volume of First Ball + Volume of Second Ball + Volume of Third Ball
 
RD Sharma Solutions Class 9 Chapter 21 Surface Area and Volume of A Sphere
 
Question 8: A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises 5/3cm. Find the radius of the cylinder. 
Solution:
It is given that, Radius of sphere is 5 cm
Let ‘r’ be the radius of cylinder
Volume of sphere = 4/3πr3
Volume of sphere = 4/3  × π × (5)3
Volume of sphere = 4/3  × π × 5 × 5 × 5
Volume of sphere = 4/3  × π × 125
It is also given that,
Height (h) of water rises = 5/3 cm
Volume of water rises in cylinder = πr2h
Volume of water rises in cylinder = Volume of sphere
πr2h = 4/3πr3
π × r2 × 5/3 = 4/3 × π × 125
r2 = 4/3 × 125 × 3/5
r2 = 4 × 25
r2 = 100
r = 10
Hence, the radius of the cylinder is 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 us assumed that, the radius of the first sphere = r  
Radius of the second sphere = 2r
Volume of sphere = 4/3 πr3
RD Sharma Solutions Class 9 Chapter 21 Surface Area and Volume of A Sphere
 
 
Question 10: A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights. 
Solution:
It is given that,
Volume of the cone = Volume of the hemisphere
1/3 πr2 h = 2/3 πr3 
r2h = 2r3
h = (2r3)/r2 
h = 2r
Height and radius of a hemisphere is equal. 
h/r = 2/1
h : r = 2 : 1
Hence, the ratio of their heights is 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:
It is given that,
Volume of water in the hemispherical bowl = Volume of water in the cylinder
Inner radius of the bowl (r) = 3.5cm
Inner radius of cylinder (R) = 7cm
Volume of water in the hemispherical bowl = Volume of water in the cylinder
2/3 π(r)3 = π(R)2 h
2/3 π(3.5)3 = π(7)2 h
2/3×3.5×3.5×3.5 = 7×7×7×h
(2 × 3.5 × 3.5 × 3.5)/(3 × 7 × 7 × 7) = h
(2 × 0.5 × 0.5 × 0.5)/3 = h
0.25/3=h 
0.0833…. cm = h
Hence, the height to which water rises in the cylinder is 0.0833cm.
 
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:
It is given that,
Radius of a sphere = 4cm
Let the radius of cylinder be r
Height of the cylinder = 2/3 diameter
h = 2/3 × (2r) 
h = 4r/3
Volume of the cylinder = Volume of the sphere
πr2h = 4/3πR3
π × r2 × (4r/3) = 4/3 π (4)3
r2 × (r) = (4)3
r3 = (4)3
r3 = 4×4×4
r = 4
Hence, the radius of the base of the cylinder is 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,
Radius of a bowl (r1) = 6cm
Radius of a cylinder (r2) = 4cm
Let 
Height of a cylinder = h
Volume of water in hemispherical bowl = Volume of cylinder
2/3πR3 = πr2 h
2/3 × π × (6)3 = π × (4)2 × h
2 × 2 × 6 × 6 = 4 × 4 × 4 × h
(2 × 2 × 6 × 6)/(4 × 4 × 4) = h
3 × 3 = h
h = 9
Hence, the height of water in the cylinder 9cm.
 
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:
It is given that,
Radius of the cylinder (R) = 16cm
Let us assumed that, 
Radius of the iron ball = r 
Height = 9 cm
Volume of iron ball = Volume of water raised in the hub
4/3πr3 = πR2h
4/3× π × r3 = π × (16)2 × 9
4/3 × r3 = 16 × 16 × 9
r3 = (16 × 16 × 9 × 3)/4
r3 = 16 × 4 × 9 × 3
r3 = 1728
r = 12
Hence, the radius of the ball is 12cm.
 
Exercise VSAQs ..........................
 
Question 1: Find the surface area of a sphere of radius 14 cm. 
Solution:
It is given that,
Radius of a sphere = 14 cm
Surface area of a sphere = 4πr2
Surface area of a sphere = 4 × 22/7 × 14 × 14
Surface area of a sphere = 2464 cm2
 
Question 2: Find the total surface area of a hemisphere of radius 10 cm. 
Solution:
It is given that,
Radius of a hemisphere = 10 cm
TSA of a hemisphere = 3πr2
TSA of a hemisphere = 3 × (22/7) × 10 × 10 
TSA of a hemisphere = 942.85cm2
 
Question 3: Find the radius of a sphere whose surface area is 154 cm2. 
Solution:
It is given that,
Surface area of a sphere = 154cm2
Surface area of a sphere = 4πr2
4πr2 = 154
4 × 22/7 × r2 = 154
r2 = (154 × 7)/(22 × 4)
r2 = (7 × 7)/4
r2 = 49/4
r2 = √(49/4)
r = 7/2 
r = 3.5
Hence, the radius of a sphere is 3.5 cm.
 
Question 4: 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. 
Solution:
It is given that,
Diameter of hollow sphere = 7 m
Radius of hollow sphere = 7/2m = 3.5 cm
Surface area of a sphere = 4πr2
Surface area of a sphere = 4 × 22/7 × 3.5 × 3.5
Surface area of a sphere = 4 × 22 × 0.5 × 3.5
Surface area of a sphere = 154 m2
Hence, the area available to the motorcyclist for riding is 154 m2.
 
Question 5: Find the volume of a sphere whose surface area is 154 cm2. 
Solution:
It is given that,
Surface area of a sphere = 154 cm2
Surface area of a sphere = 4πr2
4πr2 = 154
4 × 22/7 × r2 = 154
r2 = (154 × 7)/(22 × 4)
r2 = (7 × 7)/4
r2 = 49/4
r2 = √(49/4)
r = 7/2 
r = 3.5
Radius = 3.5 cm
Volume of sphere = 4/3πr3
Volume of sphere = 4/3 × 22/7 × 3.5 × 3.5 × 3.5 
Volume of sphere = 4/3 × 22 × 0.5 × 3.5 × 3.5
Volume of sphere = 179.66
Hence, the volume of sphere is 179.66 cm3.
 
NCERT Exemplar Solutions Class 9 Maths Areas of Parallelogram and Triangle
NCERT Exemplar Solutions Class 9 Maths Circle
NCERT Exemplar Solutions Class 9 Maths Constructions
NCERT Exemplar Solutions Class 9 Maths Coordinate Geometry
NCERT Exemplar Solutions Class 9 Maths Euclids Geometry
NCERT Exemplar Solutions Class 9 Maths Herons Formula
NCERT Exemplar Solutions Class 9 Maths Linear Equations in two variables
NCERT Exemplar Solutions Class 9 Maths Lines and Angles
NCERT Exemplar Solutions Class 9 Maths Number System
NCERT Exemplar Solutions Class 9 Maths Polynomials
NCERT Exemplar Solutions Class 9 Maths Probability
NCERT Exemplar Solutions Class 9 Maths Quadrilaterals
NCERT Exemplar Solutions Class 9 Maths Statistics
NCERT Exemplar Solutions Class 9 Maths Surface areas and Volumes
NCERT Exemplar Solutions Class 9 Maths Triangles
RS Aggarwal Class 9 Mathematics Solutions Chapter 1 Real Numbers
RS Aggarwal Class 9 Mathematics Solutions Chapter 2 Polynomials
RS Aggarwal Class 9 Mathematics Solutions Chapter 3 Introduction to Euclids Geometry
RS Aggarwal Class 9 Mathematics Solutions Chapter 4 Lines and Triangles
RS Aggarwal Class 9 Mathematics Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle
RS Aggarwal Class 9 Mathematics Solutions Chapter 6 Coordinate Geometry
RS Aggarwal Class 9 Mathematics Solutions Chapter 7 Areas by herons formula
RS Aggarwal Class 9 Mathematics Solutions Chapter 8 Linear Equations in Two Variables
RS Aggarwal Class 9 Mathematics Solutions Chapter 9 Quadrilaterals and Parallelograms
RS Aggarwal Class 9 Mathematics Solutions Chapter 10 Area
RS Aggarwal Class 9 Mathematics Solutions Chapter 11 Circle
RS Aggarwal Class 9 Mathematics Solutions Chapter 12 Geometrical Constructions
RS Aggarwal Class 9 Mathematics Solutions Chapter 13 Volume and Surface Area
RS Aggarwal Class 9 Mathematics Solutions Chapter 14 Statistics
RS Aggarwal Class 9 Mathematics Solutions Chapter 15 Probability
RD Sharma Solutions Class 9 Maths
RD Sharma Solutions Class 9 Chapter 1 Number System
RD Sharma Solutions Class 9 Chapter 2 Exponents of Real Number
RD Sharma Solutions Class 9 Chapter 3 Rationalisation
RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities
RD Sharma Solutions Class 9 Chapter 5 Factorization of Algebraic Expressions
RD Sharma Solutions Class 9 Chapter 6 Factorization of Polynomials
RD Sharma Solutions Class 9 Chapter 7 Introduction To Euclids Geometry
RD Sharma Solutions Class 9 Chapter 8 Lines and Angles
RD Sharma Solutions Class 9 Chapter 9 Triangle and its Angles
RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles
RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry
RD Sharma Solutions Class 9 Chapter 12 Herons Formula
RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables
RD Sharma Solutions Class 9 Chapter 14 Quadrilaterals
RD Sharma Solutions Class 9 Chapter 15 Area of Parallelograms and Triangles
RD Sharma Solutions Class 9 Chapter 16 Circles
RD Sharma Solutions Class 9 Chapter 17 Construction
RD Sharma Solutions Class 9 Chapter 18 Surface Area and Volume of Cuboid and Cube
RD Sharma Solutions Class 9 Chapter 19 Surface Area and Volume of A Right Circular Cylinder
RD Sharma Solutions Class 9 Chapter 20 Surface Area and Volume of A Right Circular Cone
RD Sharma Solutions Class 9 Chapter 21 Surface Area and Volume of A Sphere
RD Sharma Solutions Class 9 Chapter 22 Tabular Representation of Statistical Data
RD Sharma Solutions Class 9 Chapter 23 Graphical Representation of Statistical Data
RD Sharma Solutions Class 9 Chapter 24 Measure of Central Tendency
RD Sharma Solutions Class 9 Chapter 25 Probability

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