Volumes and Surface Area of Solids - Exercise 15D

# Volumes and Surface Area of Solids - Exercise 15D | Extra Documents & Tests for Class 9 PDF Download

``` Page 1

Question:83
Find the volume and surface area of a sphere whose radius is:
i 3.5 cm
ii 4.2 cm
iii 5 m
Solution:
i Radius of the sphere = 3.5 cm
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×3. 5 ×3. 5 ×3. 5 = 179. 67 cm
3
? Surface area = 4 pr
2
= 4 ×
22
7
×3. 5 ×3. 5 = 154 cm
2
ii Radius of the sphere=4.2 cm
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×4. 2 ×4. 2 ×4. 2 = 310. 46 cm
3
? Surface area = 4 pr
2

= 4 ×
22
7
×4. 2 ×4. 2 = 221. 76 cm
2
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×5
3
= 523. 81 cm
3
? Surface area = 4 pr
2

= 4 ×
22
7
×5
2
= 314. 29 cm
2
Question:84
The volume of a sphere is 38808 cm
3
. Find its radius and hence its surface area.
Solution:
Volume of the sphere = 38808 cm
3
Suppose that r cm is the radius of the given sphere.
?
4
3
pr
3
= 38808 ? r
3
=
38808×3×7
4×22
= 9261 ? r =
3
v 9261 = 21 cm
? Surface area of the sphere =4 pr
2

= 4 ×
22
7
×21 ×21 = 5544 cm
2
Question:85
Find the surface area of a sphere whose volume is 606.375 m
3
.
Solution:
Volume of the sphere = 606.375 m
3
Then
4
3
pr
3
= 606. 375 ? r
3
=
606.375×3×7
4×22
= 144. 703 ? r = 5. 25 m
? Surface area = 4 pr
2

= 4 ×
22
7
×5. 25 ×5. 25 = 346. 5 m
2
Question:86
Note Take p =
22
7
, unless stated otherwise.
Find the volume of a sphere whose surface area is 154 cm
2
.
Solution:
Let the radius of the sphere be r cm.
Surface area of the sphere = 154 cm
2
? 4 pr
2
= 154 ? 4 ×
22
7
×r
2
= 154 ? r =
154×7
4×22
= v 12. 25 ? r = 3. 5 cm
? Volume of the sphere =
4
3
pr
3
=
4
3
×
22
7
×(3. 5)
3
˜ 179. 67 m
3
Thus, the volume of the sphere is approximately 179.67 m
3
.
Question:87
The surface area of a sphere is 576 p cm
2
. Find its volume.
v
Page 2

Question:83
Find the volume and surface area of a sphere whose radius is:
i 3.5 cm
ii 4.2 cm
iii 5 m
Solution:
i Radius of the sphere = 3.5 cm
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×3. 5 ×3. 5 ×3. 5 = 179. 67 cm
3
? Surface area = 4 pr
2
= 4 ×
22
7
×3. 5 ×3. 5 = 154 cm
2
ii Radius of the sphere=4.2 cm
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×4. 2 ×4. 2 ×4. 2 = 310. 46 cm
3
? Surface area = 4 pr
2

= 4 ×
22
7
×4. 2 ×4. 2 = 221. 76 cm
2
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×5
3
= 523. 81 cm
3
? Surface area = 4 pr
2

= 4 ×
22
7
×5
2
= 314. 29 cm
2
Question:84
The volume of a sphere is 38808 cm
3
. Find its radius and hence its surface area.
Solution:
Volume of the sphere = 38808 cm
3
Suppose that r cm is the radius of the given sphere.
?
4
3
pr
3
= 38808 ? r
3
=
38808×3×7
4×22
= 9261 ? r =
3
v 9261 = 21 cm
? Surface area of the sphere =4 pr
2

= 4 ×
22
7
×21 ×21 = 5544 cm
2
Question:85
Find the surface area of a sphere whose volume is 606.375 m
3
.
Solution:
Volume of the sphere = 606.375 m
3
Then
4
3
pr
3
= 606. 375 ? r
3
=
606.375×3×7
4×22
= 144. 703 ? r = 5. 25 m
? Surface area = 4 pr
2

= 4 ×
22
7
×5. 25 ×5. 25 = 346. 5 m
2
Question:86
Note Take p =
22
7
, unless stated otherwise.
Find the volume of a sphere whose surface area is 154 cm
2
.
Solution:
Let the radius of the sphere be r cm.
Surface area of the sphere = 154 cm
2
? 4 pr
2
= 154 ? 4 ×
22
7
×r
2
= 154 ? r =
154×7
4×22
= v 12. 25 ? r = 3. 5 cm
? Volume of the sphere =
4
3
pr
3
=
4
3
×
22
7
×(3. 5)
3
˜ 179. 67 m
3
Thus, the volume of the sphere is approximately 179.67 m
3
.
Question:87
The surface area of a sphere is 576 p cm
2
. Find its volume.
v
Solution:
Surface area of the sphere = 576 p cm
2
Suppose that r cm is the radius of the sphere.
Then 4 pr
2
= 576 p ? r
2
=
576
4
= 144 ? r = 12 cm
? Volume of the sphere=
4
3
× p× 12× 12× 12 cm
3
= 2304 p cm
3
Question:88
How many lead shots, each 3 mm in diameter, can be made from a cuboid with dimensions (12 cm × 11 cm × 9 cm)?
Solution:
Here, l = 12 cm, b = 11 cm and h = 9 cm
Volume of the cuboid = l ×b ×h                                         = 12 ×11 ×9                                        = 1188 cm
3
0.3
2
cm
Volume of one lead shot =
4
3
×
22
7
×
0.3
2
3
=
11×9
7000
= 0. 014 cm
3
? Number of lead shots =
volume of the cuboid
=
1188
0.014
= 84857. 14 ˜ 84857
Question:89
Solution:
Radius of the sphere = 8 cm
Volume of the sphere =
4
3
pr
3
=
4
3
×
22
7
×8
3
= 2145. 52 cm
3
Volume of one lead ball =
4
3
×
22
7
×1
3
= 4. 19 cm
3
? Number of lead balls =
volume of the sphere
=
2145.52
4.19
= 512. 05 ˜ 512
Question:90
A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained.
Solution:
Radius of the solid sphere = 3 cm
Volume of the solid sphere =
4
3
pr
3

=
4
3
×
22
7
×3
3
cm
3
Diameter of the spherical ball = 0.6 cm
Radius of the spherical ball = 0.3 cm
Volume of the spherical ball =
4
3
×
22
7
×(0. 3)
3
cm
3
Now, number of small spherical balls =
volume of the sphere
volume of the spherical ball
=
4
3
p ×27
4
3
p × 0.3
3
= 1000
? The number of small balls thus obtained is 1000
Question:91
A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?
Solution:
Radius of the metallic sphere, r = 10.5 cm
Volume of the sphere =
4
3
pr
3
=
4
3
p 10. 5)
3
cm
3
Radius of each smaller cone, r
2
= 3.05 cm
Height of each smaller cone = 3 cm
Volume of each smaller cone =
1
3
pr
2
2
h =
1
3
p 3. 05)
2
×3 cm
3
Number of cones obtained =
volume of the sphere
volume of each smaller cone
=
4
3
p r
3
1
3
p r 2
2
h
=
4×10.5×10.5×10.5
3.5×3.5×3
= 126. 006 ˜ 126
? 126 cones are obtained.
Question:92
How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and height 90 cm?
Solution:
Diameter of each sphere, d=12 cm
Radius of each sphere, r=6 cm
Volume of each sphere =
4
3
pr
3
=
4
3
p 6)
3
cm
3
Diameter of base of  the cylinder, D=8 cm
( )
( )
(
(
(
Page 3

Question:83
Find the volume and surface area of a sphere whose radius is:
i 3.5 cm
ii 4.2 cm
iii 5 m
Solution:
i Radius of the sphere = 3.5 cm
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×3. 5 ×3. 5 ×3. 5 = 179. 67 cm
3
? Surface area = 4 pr
2
= 4 ×
22
7
×3. 5 ×3. 5 = 154 cm
2
ii Radius of the sphere=4.2 cm
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×4. 2 ×4. 2 ×4. 2 = 310. 46 cm
3
? Surface area = 4 pr
2

= 4 ×
22
7
×4. 2 ×4. 2 = 221. 76 cm
2
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×5
3
= 523. 81 cm
3
? Surface area = 4 pr
2

= 4 ×
22
7
×5
2
= 314. 29 cm
2
Question:84
The volume of a sphere is 38808 cm
3
. Find its radius and hence its surface area.
Solution:
Volume of the sphere = 38808 cm
3
Suppose that r cm is the radius of the given sphere.
?
4
3
pr
3
= 38808 ? r
3
=
38808×3×7
4×22
= 9261 ? r =
3
v 9261 = 21 cm
? Surface area of the sphere =4 pr
2

= 4 ×
22
7
×21 ×21 = 5544 cm
2
Question:85
Find the surface area of a sphere whose volume is 606.375 m
3
.
Solution:
Volume of the sphere = 606.375 m
3
Then
4
3
pr
3
= 606. 375 ? r
3
=
606.375×3×7
4×22
= 144. 703 ? r = 5. 25 m
? Surface area = 4 pr
2

= 4 ×
22
7
×5. 25 ×5. 25 = 346. 5 m
2
Question:86
Note Take p =
22
7
, unless stated otherwise.
Find the volume of a sphere whose surface area is 154 cm
2
.
Solution:
Let the radius of the sphere be r cm.
Surface area of the sphere = 154 cm
2
? 4 pr
2
= 154 ? 4 ×
22
7
×r
2
= 154 ? r =
154×7
4×22
= v 12. 25 ? r = 3. 5 cm
? Volume of the sphere =
4
3
pr
3
=
4
3
×
22
7
×(3. 5)
3
˜ 179. 67 m
3
Thus, the volume of the sphere is approximately 179.67 m
3
.
Question:87
The surface area of a sphere is 576 p cm
2
. Find its volume.
v
Solution:
Surface area of the sphere = 576 p cm
2
Suppose that r cm is the radius of the sphere.
Then 4 pr
2
= 576 p ? r
2
=
576
4
= 144 ? r = 12 cm
? Volume of the sphere=
4
3
× p× 12× 12× 12 cm
3
= 2304 p cm
3
Question:88
How many lead shots, each 3 mm in diameter, can be made from a cuboid with dimensions (12 cm × 11 cm × 9 cm)?
Solution:
Here, l = 12 cm, b = 11 cm and h = 9 cm
Volume of the cuboid = l ×b ×h                                         = 12 ×11 ×9                                        = 1188 cm
3
0.3
2
cm
Volume of one lead shot =
4
3
×
22
7
×
0.3
2
3
=
11×9
7000
= 0. 014 cm
3
? Number of lead shots =
volume of the cuboid
=
1188
0.014
= 84857. 14 ˜ 84857
Question:89
Solution:
Radius of the sphere = 8 cm
Volume of the sphere =
4
3
pr
3
=
4
3
×
22
7
×8
3
= 2145. 52 cm
3
Volume of one lead ball =
4
3
×
22
7
×1
3
= 4. 19 cm
3
? Number of lead balls =
volume of the sphere
=
2145.52
4.19
= 512. 05 ˜ 512
Question:90
A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained.
Solution:
Radius of the solid sphere = 3 cm
Volume of the solid sphere =
4
3
pr
3

=
4
3
×
22
7
×3
3
cm
3
Diameter of the spherical ball = 0.6 cm
Radius of the spherical ball = 0.3 cm
Volume of the spherical ball =
4
3
×
22
7
×(0. 3)
3
cm
3
Now, number of small spherical balls =
volume of the sphere
volume of the spherical ball
=
4
3
p ×27
4
3
p × 0.3
3
= 1000
? The number of small balls thus obtained is 1000
Question:91
A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?
Solution:
Radius of the metallic sphere, r = 10.5 cm
Volume of the sphere =
4
3
pr
3
=
4
3
p 10. 5)
3
cm
3
Radius of each smaller cone, r
2
= 3.05 cm
Height of each smaller cone = 3 cm
Volume of each smaller cone =
1
3
pr
2
2
h =
1
3
p 3. 05)
2
×3 cm
3
Number of cones obtained =
volume of the sphere
volume of each smaller cone
=
4
3
p r
3
1
3
p r 2
2
h
=
4×10.5×10.5×10.5
3.5×3.5×3
= 126. 006 ˜ 126
? 126 cones are obtained.
Question:92
How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and height 90 cm?
Solution:
Diameter of each sphere, d=12 cm
Radius of each sphere, r=6 cm
Volume of each sphere =
4
3
pr
3
=
4
3
p 6)
3
cm
3
Diameter of base of  the cylinder, D=8 cm
( )
( )
(
(
(
Radius of base of cylinder, R=4 cm
Height of the cylinder, h = 90 cm
Volume of the cylinder = pR
2
h = p 4)
2
×90 cm
3
Number of spheres=
volume of the cylinder
volume of the sphere
=
p R
2
h
4
3
pr
3
=
4
2
×90×3
4×6
3
=
12×90
216
= 5
? Five spheres can be made.
Question:93
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire.
Solution:
Let the length of the wire be h cm.
Radius of the wire, r = 1 mm = 0.1 cm
Radius of the sphere, R = 3 cm
Now, volume of the sphere = volume of the cylindrical wire
?
4
3
pR
3
= pr
2
h ? 4 ×3
2
= (0. 1)
2
×h ? h =
4×9
0.1×0.1
= 3600 cm = 36 m
? Length of the wire = 36 m
Question:94
The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter.
Solution:
Radius of the copper sphere, R = 9 cm
Length of the wire, h = 108 m = 10800 cm
Volume of the sphere = volume of the wire
Suppose that r cm is the radius of the wire.
Then
4
3
pR
3
= pr
2
h ?
4
3
×9
3
= r
2
×10800 ? r
2
=
4×729
3×10800
=
4×81
3×1200
=
9
100
? r =
3
10
= 0. 3 cm
? Diameter of the wire = 0.6 cm
Question:95
A sphere of diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm. Find the diameter of the base of the cone.
Solution:
Radius of the sphere, r = 7.8 cm
Height of the cone, h = 31.2 cm
Suppose that R cm is the radius of the cone.
Now, volume of the sphere = volume of the cone
?
4
3
pr
3
=
1
3
pR
2
h ? 4 ×(7. 8)
3
= R
2
×31. 2 ? R
2
=
4×7.8×7.8×7.8
31.2
= 60. 84 ? R = 7. 8 cm
? The diameter of the cone is 15.6 cm.
Question:96
A spherical cannonball 28 cm in diameter is melted and cast into a right circular cone would, whose base is 35 cm in diameter. Find the height of the cone.
Solution:
Radius of the spherical cannonball, R = 14 cm
Radius of the base of the cone, r = 17.5 cm
Let h cm be the height of the cone.
Now, volume of the sphere = volume of the cone
?
4
3
pR
3
=
1
3
pr
2
h ? 4 ×14
3
= (17. 5)
2
×h ? h =
4×14×14×14
17.5×17.5
= 35. 84 cm
? The height of the cone is 35.84 cm.
Question:97
A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball.
Solution:
Radius of the original spherical ball = 3 cm
Suppose that the radius of third ball is r cm.
Then volume of the original spherical ball = volume of the three spherical balls
?
4
3
p ×3
3
=
4
3
p ×1. 5
3
+
4
3
p ×2
3
+
4
3
p ×r
3
? 27 = 3. 375 +8 +r
3
? r
3
= 27 -11. 375 = 15. 625 ? r = 2. 5 cm
? The radius of the third ball is 2.5 cm.
Question:98
The radii of two spheres are in the ratio 1 : 2. Find the ratio of their surface areas.
Solution:
Suppose that the radii are r and 2r.
Now, ratio of the surface areas  =
4 p r
2
4 p (2r)
2
=
r
2
4r
2
=
1
4
= 1:4
? The ratio of their surface areas is 1 : 4.
Question:99
The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes.
Solution:
Suppose that the radii of the spheres are r and R.
We have:
4 p r
2
4 p R
2
=
1
4
?
r
R
=
1
4
=
1
2
(
v
Page 4

Question:83
Find the volume and surface area of a sphere whose radius is:
i 3.5 cm
ii 4.2 cm
iii 5 m
Solution:
i Radius of the sphere = 3.5 cm
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×3. 5 ×3. 5 ×3. 5 = 179. 67 cm
3
? Surface area = 4 pr
2
= 4 ×
22
7
×3. 5 ×3. 5 = 154 cm
2
ii Radius of the sphere=4.2 cm
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×4. 2 ×4. 2 ×4. 2 = 310. 46 cm
3
? Surface area = 4 pr
2

= 4 ×
22
7
×4. 2 ×4. 2 = 221. 76 cm
2
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×5
3
= 523. 81 cm
3
? Surface area = 4 pr
2

= 4 ×
22
7
×5
2
= 314. 29 cm
2
Question:84
The volume of a sphere is 38808 cm
3
. Find its radius and hence its surface area.
Solution:
Volume of the sphere = 38808 cm
3
Suppose that r cm is the radius of the given sphere.
?
4
3
pr
3
= 38808 ? r
3
=
38808×3×7
4×22
= 9261 ? r =
3
v 9261 = 21 cm
? Surface area of the sphere =4 pr
2

= 4 ×
22
7
×21 ×21 = 5544 cm
2
Question:85
Find the surface area of a sphere whose volume is 606.375 m
3
.
Solution:
Volume of the sphere = 606.375 m
3
Then
4
3
pr
3
= 606. 375 ? r
3
=
606.375×3×7
4×22
= 144. 703 ? r = 5. 25 m
? Surface area = 4 pr
2

= 4 ×
22
7
×5. 25 ×5. 25 = 346. 5 m
2
Question:86
Note Take p =
22
7
, unless stated otherwise.
Find the volume of a sphere whose surface area is 154 cm
2
.
Solution:
Let the radius of the sphere be r cm.
Surface area of the sphere = 154 cm
2
? 4 pr
2
= 154 ? 4 ×
22
7
×r
2
= 154 ? r =
154×7
4×22
= v 12. 25 ? r = 3. 5 cm
? Volume of the sphere =
4
3
pr
3
=
4
3
×
22
7
×(3. 5)
3
˜ 179. 67 m
3
Thus, the volume of the sphere is approximately 179.67 m
3
.
Question:87
The surface area of a sphere is 576 p cm
2
. Find its volume.
v
Solution:
Surface area of the sphere = 576 p cm
2
Suppose that r cm is the radius of the sphere.
Then 4 pr
2
= 576 p ? r
2
=
576
4
= 144 ? r = 12 cm
? Volume of the sphere=
4
3
× p× 12× 12× 12 cm
3
= 2304 p cm
3
Question:88
How many lead shots, each 3 mm in diameter, can be made from a cuboid with dimensions (12 cm × 11 cm × 9 cm)?
Solution:
Here, l = 12 cm, b = 11 cm and h = 9 cm
Volume of the cuboid = l ×b ×h                                         = 12 ×11 ×9                                        = 1188 cm
3
0.3
2
cm
Volume of one lead shot =
4
3
×
22
7
×
0.3
2
3
=
11×9
7000
= 0. 014 cm
3
? Number of lead shots =
volume of the cuboid
=
1188
0.014
= 84857. 14 ˜ 84857
Question:89
Solution:
Radius of the sphere = 8 cm
Volume of the sphere =
4
3
pr
3
=
4
3
×
22
7
×8
3
= 2145. 52 cm
3
Volume of one lead ball =
4
3
×
22
7
×1
3
= 4. 19 cm
3
? Number of lead balls =
volume of the sphere
=
2145.52
4.19
= 512. 05 ˜ 512
Question:90
A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained.
Solution:
Radius of the solid sphere = 3 cm
Volume of the solid sphere =
4
3
pr
3

=
4
3
×
22
7
×3
3
cm
3
Diameter of the spherical ball = 0.6 cm
Radius of the spherical ball = 0.3 cm
Volume of the spherical ball =
4
3
×
22
7
×(0. 3)
3
cm
3
Now, number of small spherical balls =
volume of the sphere
volume of the spherical ball
=
4
3
p ×27
4
3
p × 0.3
3
= 1000
? The number of small balls thus obtained is 1000
Question:91
A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?
Solution:
Radius of the metallic sphere, r = 10.5 cm
Volume of the sphere =
4
3
pr
3
=
4
3
p 10. 5)
3
cm
3
Radius of each smaller cone, r
2
= 3.05 cm
Height of each smaller cone = 3 cm
Volume of each smaller cone =
1
3
pr
2
2
h =
1
3
p 3. 05)
2
×3 cm
3
Number of cones obtained =
volume of the sphere
volume of each smaller cone
=
4
3
p r
3
1
3
p r 2
2
h
=
4×10.5×10.5×10.5
3.5×3.5×3
= 126. 006 ˜ 126
? 126 cones are obtained.
Question:92
How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and height 90 cm?
Solution:
Diameter of each sphere, d=12 cm
Radius of each sphere, r=6 cm
Volume of each sphere =
4
3
pr
3
=
4
3
p 6)
3
cm
3
Diameter of base of  the cylinder, D=8 cm
( )
( )
(
(
(
Radius of base of cylinder, R=4 cm
Height of the cylinder, h = 90 cm
Volume of the cylinder = pR
2
h = p 4)
2
×90 cm
3
Number of spheres=
volume of the cylinder
volume of the sphere
=
p R
2
h
4
3
pr
3
=
4
2
×90×3
4×6
3
=
12×90
216
= 5
? Five spheres can be made.
Question:93
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire.
Solution:
Let the length of the wire be h cm.
Radius of the wire, r = 1 mm = 0.1 cm
Radius of the sphere, R = 3 cm
Now, volume of the sphere = volume of the cylindrical wire
?
4
3
pR
3
= pr
2
h ? 4 ×3
2
= (0. 1)
2
×h ? h =
4×9
0.1×0.1
= 3600 cm = 36 m
? Length of the wire = 36 m
Question:94
The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter.
Solution:
Radius of the copper sphere, R = 9 cm
Length of the wire, h = 108 m = 10800 cm
Volume of the sphere = volume of the wire
Suppose that r cm is the radius of the wire.
Then
4
3
pR
3
= pr
2
h ?
4
3
×9
3
= r
2
×10800 ? r
2
=
4×729
3×10800
=
4×81
3×1200
=
9
100
? r =
3
10
= 0. 3 cm
? Diameter of the wire = 0.6 cm
Question:95
A sphere of diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm. Find the diameter of the base of the cone.
Solution:
Radius of the sphere, r = 7.8 cm
Height of the cone, h = 31.2 cm
Suppose that R cm is the radius of the cone.
Now, volume of the sphere = volume of the cone
?
4
3
pr
3
=
1
3
pR
2
h ? 4 ×(7. 8)
3
= R
2
×31. 2 ? R
2
=
4×7.8×7.8×7.8
31.2
= 60. 84 ? R = 7. 8 cm
? The diameter of the cone is 15.6 cm.
Question:96
A spherical cannonball 28 cm in diameter is melted and cast into a right circular cone would, whose base is 35 cm in diameter. Find the height of the cone.
Solution:
Radius of the spherical cannonball, R = 14 cm
Radius of the base of the cone, r = 17.5 cm
Let h cm be the height of the cone.
Now, volume of the sphere = volume of the cone
?
4
3
pR
3
=
1
3
pr
2
h ? 4 ×14
3
= (17. 5)
2
×h ? h =
4×14×14×14
17.5×17.5
= 35. 84 cm
? The height of the cone is 35.84 cm.
Question:97
A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball.
Solution:
Radius of the original spherical ball = 3 cm
Suppose that the radius of third ball is r cm.
Then volume of the original spherical ball = volume of the three spherical balls
?
4
3
p ×3
3
=
4
3
p ×1. 5
3
+
4
3
p ×2
3
+
4
3
p ×r
3
? 27 = 3. 375 +8 +r
3
? r
3
= 27 -11. 375 = 15. 625 ? r = 2. 5 cm
? The radius of the third ball is 2.5 cm.
Question:98
The radii of two spheres are in the ratio 1 : 2. Find the ratio of their surface areas.
Solution:
Suppose that the radii are r and 2r.
Now, ratio of the surface areas  =
4 p r
2
4 p (2r)
2
=
r
2
4r
2
=
1
4
= 1:4
? The ratio of their surface areas is 1 : 4.
Question:99
The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes.
Solution:
Suppose that the radii of the spheres are r and R.
We have:
4 p r
2
4 p R
2
=
1
4
?
r
R
=
1
4
=
1
2
(
v
Now, ratio of the volumes =
4
3
p r
3
4
3
p R
3
=
r
R
3
=
1
2
3
=
1
8
? The ratio of the volumes of the spheres is 1 : 8.
Question:100
A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron 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:
Suppose that the radius of the ball is r cm.
Radius of the cylindrical tub =12 cm
Depth of the tub= 20 cm
Now, volume of the ball = volume of water raised in the cylinder
?
4
3
pr
3
= p ×12
2
×6. 75 ? r
3
=
144×6.75×3
4
= 36 ×6. 5 ×3 = 729 ? r = 9 cm
? The radius of the ball is 9 cm.
Question:101
A cylindrical bucket with base radius 15 cm is filled with water up to a height of 20 cm. A heavy iron spherical ball of radius 9 cm is dropped into the bucket to submerge completely in
the water. Find the increase in the level of water.
Solution:
Let h cm be the increase in the level of water.
Radius of the cylindrical bucket = 15 cm
Height up to which water is being filled = 20 cm
Radius of the spherical ball = 9 cm
Now, volume of the sphere = increased in volume of the cylinder
?
4
3
p ×9
3
= p ×15
2
×h ? h =
4×729
3×15×15
=
4×27
25
= 4. 32 cm
? The increase in the level of water is 4.32 cm.
Question:102
The outer diameter of a spherical shell is 12 cm and its inner diameter is 8 cm. Find the volume of metal contained in the shell. Also, find its outer surface area.
Solution:
Outer radius of the spherical shell = 6 cm
Inner radius of the spherical shell = 4 cm
Volume of metal contained in the shell =
4
3
×
22
7
6
3
-4
3
=
88
21
×(216 -64)                                                             =
88
21
×152
? Outer surface area = 4 ×
22
7
×6 ×6                                 = 452. 57 cm
2
Question:103
A hollow spherical shell is made of a metal of density 4.5 g per cm
3
. If its internal and external radii are 8 cm and 9 cm respectively, find the weight of the shell.
Solution:
Internal radius of the hollow spherical shell, r= 8 cm
External radius of the hollow spherical shell, R= 9 cm
Volume of the shell =
4
3
p R
3
-r
3

=
4
3
p 9
3
-8
3
=
4
3
×
22
7
×(729 -512) =
4×22×217
21
=
88×31
3
=
2728
3
cm
3
Weight of the shell = volume of the shell × density per cubic cm
=
2728
3
×4. 5 ˜ 4092 g = 4. 092 kg
? Weight of the shell = 4.092 kg
Question:104
A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.
Solution:
Radius of the hemisphere = 9 cm
Height of the right circular cone = 72 cm
Suppose that the radius of the base of the cone is r cm.
Volume of the hemisphere = volume of the cone
?
2
3
p ×9
3
=
1
3
p ×r
2
×72 ? r
2
=
2×9×9×9
72
=
81
4
? r =
9
2
= 4. 5 cm
? The radius of the base of the cone is 4.5 cm.
Question:105
A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are
required to empty the bowl?
Solution:
Internal radius of the hemispherical bowl = 9 cm
Radius of a cylindrical shaped bottle = 1.5 cm
Height of a bottle = 4 cm
Number of bottles required to empty the bowl  =
volume of the hemispherical bowl
volume of a cylindrical shaped bottle
=
2
3
p ×9
3
p ×1.5
2
×4
=
2×9×9×9
3×1.5×1.5×4
= 54
( ) ( )
( )
( )
( )
Page 5

Question:83
Find the volume and surface area of a sphere whose radius is:
i 3.5 cm
ii 4.2 cm
iii 5 m
Solution:
i Radius of the sphere = 3.5 cm
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×3. 5 ×3. 5 ×3. 5 = 179. 67 cm
3
? Surface area = 4 pr
2
= 4 ×
22
7
×3. 5 ×3. 5 = 154 cm
2
ii Radius of the sphere=4.2 cm
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×4. 2 ×4. 2 ×4. 2 = 310. 46 cm
3
? Surface area = 4 pr
2

= 4 ×
22
7
×4. 2 ×4. 2 = 221. 76 cm
2
Now, volume =
4
3
pr
3

=
4
3
×
22
7
×5
3
= 523. 81 cm
3
? Surface area = 4 pr
2

= 4 ×
22
7
×5
2
= 314. 29 cm
2
Question:84
The volume of a sphere is 38808 cm
3
. Find its radius and hence its surface area.
Solution:
Volume of the sphere = 38808 cm
3
Suppose that r cm is the radius of the given sphere.
?
4
3
pr
3
= 38808 ? r
3
=
38808×3×7
4×22
= 9261 ? r =
3
v 9261 = 21 cm
? Surface area of the sphere =4 pr
2

= 4 ×
22
7
×21 ×21 = 5544 cm
2
Question:85
Find the surface area of a sphere whose volume is 606.375 m
3
.
Solution:
Volume of the sphere = 606.375 m
3
Then
4
3
pr
3
= 606. 375 ? r
3
=
606.375×3×7
4×22
= 144. 703 ? r = 5. 25 m
? Surface area = 4 pr
2

= 4 ×
22
7
×5. 25 ×5. 25 = 346. 5 m
2
Question:86
Note Take p =
22
7
, unless stated otherwise.
Find the volume of a sphere whose surface area is 154 cm
2
.
Solution:
Let the radius of the sphere be r cm.
Surface area of the sphere = 154 cm
2
? 4 pr
2
= 154 ? 4 ×
22
7
×r
2
= 154 ? r =
154×7
4×22
= v 12. 25 ? r = 3. 5 cm
? Volume of the sphere =
4
3
pr
3
=
4
3
×
22
7
×(3. 5)
3
˜ 179. 67 m
3
Thus, the volume of the sphere is approximately 179.67 m
3
.
Question:87
The surface area of a sphere is 576 p cm
2
. Find its volume.
v
Solution:
Surface area of the sphere = 576 p cm
2
Suppose that r cm is the radius of the sphere.
Then 4 pr
2
= 576 p ? r
2
=
576
4
= 144 ? r = 12 cm
? Volume of the sphere=
4
3
× p× 12× 12× 12 cm
3
= 2304 p cm
3
Question:88
How many lead shots, each 3 mm in diameter, can be made from a cuboid with dimensions (12 cm × 11 cm × 9 cm)?
Solution:
Here, l = 12 cm, b = 11 cm and h = 9 cm
Volume of the cuboid = l ×b ×h                                         = 12 ×11 ×9                                        = 1188 cm
3
0.3
2
cm
Volume of one lead shot =
4
3
×
22
7
×
0.3
2
3
=
11×9
7000
= 0. 014 cm
3
? Number of lead shots =
volume of the cuboid
=
1188
0.014
= 84857. 14 ˜ 84857
Question:89
Solution:
Radius of the sphere = 8 cm
Volume of the sphere =
4
3
pr
3
=
4
3
×
22
7
×8
3
= 2145. 52 cm
3
Volume of one lead ball =
4
3
×
22
7
×1
3
= 4. 19 cm
3
? Number of lead balls =
volume of the sphere
=
2145.52
4.19
= 512. 05 ˜ 512
Question:90
A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained.
Solution:
Radius of the solid sphere = 3 cm
Volume of the solid sphere =
4
3
pr
3

=
4
3
×
22
7
×3
3
cm
3
Diameter of the spherical ball = 0.6 cm
Radius of the spherical ball = 0.3 cm
Volume of the spherical ball =
4
3
×
22
7
×(0. 3)
3
cm
3
Now, number of small spherical balls =
volume of the sphere
volume of the spherical ball
=
4
3
p ×27
4
3
p × 0.3
3
= 1000
? The number of small balls thus obtained is 1000
Question:91
A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?
Solution:
Radius of the metallic sphere, r = 10.5 cm
Volume of the sphere =
4
3
pr
3
=
4
3
p 10. 5)
3
cm
3
Radius of each smaller cone, r
2
= 3.05 cm
Height of each smaller cone = 3 cm
Volume of each smaller cone =
1
3
pr
2
2
h =
1
3
p 3. 05)
2
×3 cm
3
Number of cones obtained =
volume of the sphere
volume of each smaller cone
=
4
3
p r
3
1
3
p r 2
2
h
=
4×10.5×10.5×10.5
3.5×3.5×3
= 126. 006 ˜ 126
? 126 cones are obtained.
Question:92
How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and height 90 cm?
Solution:
Diameter of each sphere, d=12 cm
Radius of each sphere, r=6 cm
Volume of each sphere =
4
3
pr
3
=
4
3
p 6)
3
cm
3
Diameter of base of  the cylinder, D=8 cm
( )
( )
(
(
(
Radius of base of cylinder, R=4 cm
Height of the cylinder, h = 90 cm
Volume of the cylinder = pR
2
h = p 4)
2
×90 cm
3
Number of spheres=
volume of the cylinder
volume of the sphere
=
p R
2
h
4
3
pr
3
=
4
2
×90×3
4×6
3
=
12×90
216
= 5
? Five spheres can be made.
Question:93
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire.
Solution:
Let the length of the wire be h cm.
Radius of the wire, r = 1 mm = 0.1 cm
Radius of the sphere, R = 3 cm
Now, volume of the sphere = volume of the cylindrical wire
?
4
3
pR
3
= pr
2
h ? 4 ×3
2
= (0. 1)
2
×h ? h =
4×9
0.1×0.1
= 3600 cm = 36 m
? Length of the wire = 36 m
Question:94
The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter.
Solution:
Radius of the copper sphere, R = 9 cm
Length of the wire, h = 108 m = 10800 cm
Volume of the sphere = volume of the wire
Suppose that r cm is the radius of the wire.
Then
4
3
pR
3
= pr
2
h ?
4
3
×9
3
= r
2
×10800 ? r
2
=
4×729
3×10800
=
4×81
3×1200
=
9
100
? r =
3
10
= 0. 3 cm
? Diameter of the wire = 0.6 cm
Question:95
A sphere of diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm. Find the diameter of the base of the cone.
Solution:
Radius of the sphere, r = 7.8 cm
Height of the cone, h = 31.2 cm
Suppose that R cm is the radius of the cone.
Now, volume of the sphere = volume of the cone
?
4
3
pr
3
=
1
3
pR
2
h ? 4 ×(7. 8)
3
= R
2
×31. 2 ? R
2
=
4×7.8×7.8×7.8
31.2
= 60. 84 ? R = 7. 8 cm
? The diameter of the cone is 15.6 cm.
Question:96
A spherical cannonball 28 cm in diameter is melted and cast into a right circular cone would, whose base is 35 cm in diameter. Find the height of the cone.
Solution:
Radius of the spherical cannonball, R = 14 cm
Radius of the base of the cone, r = 17.5 cm
Let h cm be the height of the cone.
Now, volume of the sphere = volume of the cone
?
4
3
pR
3
=
1
3
pr
2
h ? 4 ×14
3
= (17. 5)
2
×h ? h =
4×14×14×14
17.5×17.5
= 35. 84 cm
? The height of the cone is 35.84 cm.
Question:97
A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball.
Solution:
Radius of the original spherical ball = 3 cm
Suppose that the radius of third ball is r cm.
Then volume of the original spherical ball = volume of the three spherical balls
?
4
3
p ×3
3
=
4
3
p ×1. 5
3
+
4
3
p ×2
3
+
4
3
p ×r
3
? 27 = 3. 375 +8 +r
3
? r
3
= 27 -11. 375 = 15. 625 ? r = 2. 5 cm
? The radius of the third ball is 2.5 cm.
Question:98
The radii of two spheres are in the ratio 1 : 2. Find the ratio of their surface areas.
Solution:
Suppose that the radii are r and 2r.
Now, ratio of the surface areas  =
4 p r
2
4 p (2r)
2
=
r
2
4r
2
=
1
4
= 1:4
? The ratio of their surface areas is 1 : 4.
Question:99
The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes.
Solution:
Suppose that the radii of the spheres are r and R.
We have:
4 p r
2
4 p R
2
=
1
4
?
r
R
=
1
4
=
1
2
(
v
Now, ratio of the volumes =
4
3
p r
3
4
3
p R
3
=
r
R
3
=
1
2
3
=
1
8
? The ratio of the volumes of the spheres is 1 : 8.
Question:100
A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron 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:
Suppose that the radius of the ball is r cm.
Radius of the cylindrical tub =12 cm
Depth of the tub= 20 cm
Now, volume of the ball = volume of water raised in the cylinder
?
4
3
pr
3
= p ×12
2
×6. 75 ? r
3
=
144×6.75×3
4
= 36 ×6. 5 ×3 = 729 ? r = 9 cm
? The radius of the ball is 9 cm.
Question:101
A cylindrical bucket with base radius 15 cm is filled with water up to a height of 20 cm. A heavy iron spherical ball of radius 9 cm is dropped into the bucket to submerge completely in
the water. Find the increase in the level of water.
Solution:
Let h cm be the increase in the level of water.
Radius of the cylindrical bucket = 15 cm
Height up to which water is being filled = 20 cm
Radius of the spherical ball = 9 cm
Now, volume of the sphere = increased in volume of the cylinder
?
4
3
p ×9
3
= p ×15
2
×h ? h =
4×729
3×15×15
=
4×27
25
= 4. 32 cm
? The increase in the level of water is 4.32 cm.
Question:102
The outer diameter of a spherical shell is 12 cm and its inner diameter is 8 cm. Find the volume of metal contained in the shell. Also, find its outer surface area.
Solution:
Outer radius of the spherical shell = 6 cm
Inner radius of the spherical shell = 4 cm
Volume of metal contained in the shell =
4
3
×
22
7
6
3
-4
3
=
88
21
×(216 -64)                                                             =
88
21
×152
? Outer surface area = 4 ×
22
7
×6 ×6                                 = 452. 57 cm
2
Question:103
A hollow spherical shell is made of a metal of density 4.5 g per cm
3
. If its internal and external radii are 8 cm and 9 cm respectively, find the weight of the shell.
Solution:
Internal radius of the hollow spherical shell, r= 8 cm
External radius of the hollow spherical shell, R= 9 cm
Volume of the shell =
4
3
p R
3
-r
3

=
4
3
p 9
3
-8
3
=
4
3
×
22
7
×(729 -512) =
4×22×217
21
=
88×31
3
=
2728
3
cm
3
Weight of the shell = volume of the shell × density per cubic cm
=
2728
3
×4. 5 ˜ 4092 g = 4. 092 kg
? Weight of the shell = 4.092 kg
Question:104
A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.
Solution:
Radius of the hemisphere = 9 cm
Height of the right circular cone = 72 cm
Suppose that the radius of the base of the cone is r cm.
Volume of the hemisphere = volume of the cone
?
2
3
p ×9
3
=
1
3
p ×r
2
×72 ? r
2
=
2×9×9×9
72
=
81
4
? r =
9
2
= 4. 5 cm
? The radius of the base of the cone is 4.5 cm.
Question:105
A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are
required to empty the bowl?
Solution:
Internal radius of the hemispherical bowl = 9 cm
Radius of a cylindrical shaped bottle = 1.5 cm
Height of a bottle = 4 cm
Number of bottles required to empty the bowl  =
volume of the hemispherical bowl
volume of a cylindrical shaped bottle
=
2
3
p ×9
3
p ×1.5
2
×4
=
2×9×9×9
3×1.5×1.5×4
= 54
( ) ( )
( )
( )
( )
? 54 bottles are required to empty the bowl.
Question:106
A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Find the volume of steel used in making the bowl.
Solution:
Internal radius of the hemispherical bowl = 4 cm
Thickness of a the bowl = 0.5 cm
Now, external radius of the bowl = 4 +0.5 cm = 4.5 cm
Now, volume of steel used in making the bowl =  volume of the shell
=
2
3
p 4. 5
3
-4
3
=
2
3
×
22
7
×(91. 125 -64) =
2
3
×
22
7
×27. 125 = 56. 83 cm
3
? 56.83 cm 3 of steel is used in making the bowl .
Question:107
Note Take p =
22
7
, unless stated otherwise.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
Inner radius of the bowl, r = 5 cm
Let the outer radius of the bowl be R cm.
Thickness of the bowl = 0.25 cm         Given
? R - r = 0.25 cm
? R = 0.25 + r = 0.25 + 5 = 5.25 cm
? Outer curved surface area of the bowl = 2 pr
2
= 2 ×
22
7
×(5. 25)
2
= 173.25 cm
2
Thus, the outer curved surface area of the bowl is 173.25 cm
2
.
Question:108
Note Take p =
22
7
, unless stated otherwise.
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  32 per 100 cm
2
.
Solution:
Inner radius of the bowl, r =
10.5
2
= 5.25 cm
? Inner curved surface area of the bowl = 2 pr
2
= 2 ×
22
7
×(5. 25)
2
= 173.25 cm
2
Rate of tin-plating =  32 per 100 cm
2
? Cost of tin-plating the bowl on the inside
= Inner curved surface area of the bowl × Rate of tin-plating
= 173. 25 ×
32
100
=  55.44
Thus, the cost of tin-plating the bowl on the inside is  55.44.
Question:109
Note Take p =
22
7
, unless stated otherwise.
The diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Solution:
Let the radius of the moon and earth be r units and R units, respectively.
? 2r =
1
4
× 2R        Given
? r =
R
4
.....1
?
Volume of the moon
Volume of the earth
=
4
3
pr 3
4
3
pR 3
=
R
4
3
R
3
=
1
64

Using(1)
Thus, the volume of the moon is
1
64
of the volume of the earth.
Question:110
Note Take p =
22
7
, unless stated otherwise.
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere?
Solution:
Let the radius of the solid hemisphere be r units.
( )
( )
```

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