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

Read RD Sharma Solutions Class 9 Chapter 21 Surface Area and Volume of A Sphere below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations

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,
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,
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,
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
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
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
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(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

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,
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,
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,
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
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
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.

Solution:
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 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

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

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

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