The document contains solutions to 12 math questions about calculating properties of shapes like cuboids, cylinders, and boxes. The key details are: 1) Question 1 involves finding the surface area and volume of a cuboid with dimensions 5 cm x 9 cm x 12 cm. 2) Question 8 asks how many bricks are needed to build a wall based on its dimensions and the size of each brick. 3) Several questions involve finding surface areas, volumes, or radii of cylinders given various measurements like height, diameter, curved surface area, or total surface area.

1. Rajasthan Board RBSE Class 10 Maths Chapter 16 solution by Arvind Kumar (arvind.saini@outlook.com)
+91-9460755486
Question 1.
The height, breadth and length of a cuboid are 5 cm, 9 cm and 12 cm respectively. Find the total
surface area and the volume of cuboid.
Solution :
Given
Length of cuboid (l) = 12 cm
Breadth (b) =9 cm
and Height (h) = 5 cm
Total surface area of cuboid = 2(lb + bh + hl)
= 2[(12 × 9) + (9 × 5) +(5 × 12)]
= 2[108 + 45 + 60]
2 × 213 = 426 cm2
Volume = l × b × h
= 12 × 9 × 5 = 540 cm3
Hence, total surface area of cuboid 426 cm2
And volume of the cuboid 540 cm3
Question 2.
The cores of three cubes are 8 cm, 6 cm and 1 cm respectively. Having melted these cubes a new cube
is recastesd. Find the total surface area of the new cube recanted.
Solution :
Volume of the cube with core of 8 cm = (core)3
= 83
= 512 cm3
Volume of the cube with core 6 cm = (core)3
= (6)3
= 216 cm3
.
Volume of the cube with core 1 cm = (core)3
= (1)3
= 1 cm3
.
The total volume of three cubes 512 + 216 + 1 = 729 cm3
.
Having melted these cubes, a new cube is recasted
∴ The volume of the cube recasted = 729 cm3
⇒ (core)3
= 729
⇒ core =
= (93
)1/3
= 9 cm
Total surface area of cuboid recasted =6 (core)2
6 × 9 × 9 = 486 cm2
.
Hence the surface area of the new cube = 486 cm2
.
Question 3.
The measures of a box are 50 cm × 36 cm × 25 cm. To make the cover of the box. How much cloth is
needed?
Solution :
Given
Length of box (l) = 50 cm
Breadth of box (b) = 36 cm
Height of box (h) = 25 cm
The necessary cloth to make its cover = surface area of the box
= 2(lb + bh + hl)
= 2[(50 × 36) + (36 + 25) + (25 + 50)]

2. = 2[1800 + 900 + 1250]
= 2 × 3950 = 7900 cm2
.
Question 4.
Each of the face of cube is 100 sq. cm. If the cube Is cut of parallel to its base and divided into two
equal parts. Find the total surface area of each part separately.
Solution :
Given
The area of each face of cube = 100 cm2
∴ core of the cube = = 10 cm
When a cube is cut of parallel to its base and divided into two equal parts, then two cuboids are formed.
In which the length of each cube (l) = 10 cm
Breadth (b) = 5 cm
Height (h) = 10 cm
∴ And the surface area of each cuboid formed
= 2(lb + bh + hl)
= 2[(10 × 5) + (5 × 10) + (10 × 10)]
= 2[50 + 50 + 100]
= 2 × 200=400 cm2
Hence, the surface area of each cuboid formed = 400 cm2
.
Question 5.
A box with out upper lid is made of wood of width 3 cm. Its external length, breadth and height are 146
cm, 116 cm and 83 cm respectively.
Find cost of painting inside if the rate of painting = ₹ 2 per 1000 sq.cm.
Solution : Given
The external length of box = 146 cm
breadth = 116 cm
height = 83 cm
The width of the wood = 3 cm
∴ The internal length of box = (146 – 3 – 3) = 140 cm.
Internal breadth of box = (116 – 3 – 3) = 110 cm
Internal height of box = 83 – 3 = 80 cm.
Besides of the lid, the internal total surface area of the box
= 2(l + b) × h + l × b
= 2(140 + 110) × 80 + (140 × 110)
= 160 × 250 + 15400
= 55400 sq. cm.
The cost of painting of 1000 sq. cm. = ₹ 2.
∴ The cost of painting of 55400 = = ₹ 110.80
Hence the cost of painting the box = ₹ 110.80.

3. Question 6.
The sum of the length, breadth and height of a cuboid is 19 cm and the length of its diagonal is 11 cm.
Find the total surface area of the cuboid.
Solution :
Given, the sum of the length, breadth and height of cuboid = 19 cm.
And the length of its diagonal = 11 cm.
Let the length, breadth and height of the cuboid be l, b and h respectively.
∴ l + b + h = 19 cm …(i)
and = 11 cm. …(ii)
Squarnng both the sides, we get
l2
+ b2
+ h2
= (11)2
= 121 …(iii)
We know that,
(l + b + h)2
=[l2
+ b2
+ h2
] + 2(lb + bh + hl)
⇒ (19)2
= 121 + 2(lb + bh + hl)
⇒ 361 = 121 + 2(lb + bh + hl)
⇒ 2(lb + bh + hl) = 361 – 121
⇒ 2(lb + bh + hl) = 240 sq. cm.
Hence, the total surface area of cuboid = 240 cm2
.
Question 7.
A room contains 180 m3
air in it. Find the height of the room if its floor is a square with side 6 meter.
Solution :
Let the height of the room be h meter.
Since the floor of the room is square shaped.
∴ the length of room (l) = 6 m
also the breadth of room (b) = 6 m
The volume of the room = 180 m3
⇒ l × b × h = 180
6 × 6 × h = 180
h = = 5 m.
Hence the height of the room = 5 m.
Question 8.
How many brisks are need to build a wall of length 44 meter, height 1.5 meter and broad 85 cm. The
measurement of each brick is 20 cm × 10 cm × 17 cm.
Solution :
Length of the wall = 44 meter
=44 × 100 cm = 4400 cm
height = 1.5 meter = 1.5 × 100 = 150 cm
width of the wall = 85 cm
volume of the wall = length × breadth × height
=4400 × 85 × 150 cm3
.
length of one brick = 22 cm
breadth = 10 cm
height = 17 cm
The volume of a brick 22 × 10 × 17 cm3
.
Hence, 15000 bricks will be needed.

4. Question 9.
Find the length of the longest rod can be put in a room with dimensions 10 meter, 8 meter and 6 meter.
Solution :
Given
The length of the room (l) = 10 meter
breath (b) = 8 meter
height (h) = 6 meter
The longest rod that can be put in the room will be equal of its diagonal.
∴ Diagonal =
=
=
= 10√2 m
∴ The length of the longest rod = 10√2 m.
Question 10.
The volume of a cube ¡s 512 m3
. Find its side.
Solution :
Given:
The volume of cube = 512 cm3
.
⇒ (side)3
= 512
side =
side = (8 × 8 × 8)1/3
= (83
)1/3
= 8 m.
Hence side of the cube = 8 m
Question 11.
The length, breath and height of a wall are 5 m, 30 cm and 3 m. How many bricks will be needed to
build of the wall. The dimensions of one brick are 20 cm × 10 cm × 7.5 cm.
Solution :
Given
The length of wall = 5 m
= 5 × 100 = 500 cm
breadth = 30 cm
height = 3 m
= 3 × 1000 = 300 cm3
The volume of the wall = length × breadth × height
= 500 × 30 × 300 cm3
.
For a brick = 20 cm
breadth = 10 cm
and height = 7.5 cm
Volume of one brick = l × b × h
= 20 × 10 × 7.5 cm3
∴ Number of bricks needed
= 3000
Hence, 3000 bricks are needed.

5. Question 12.
The ratio of the length, breadth and height of a cuboid is 5 : 3 : 2. If the total surface area of the cuboid
in 558 cm2
. Find the length of its sides.
Solution :
Let for the cuboid,
length = 5x
breadth = 3x
and height = 2x
Total surface area of cuboid = 558 cm2
⇒ 2(lb + bh + hl) = 558
⇒ (5x × 3x) + (3x × 2x) + (2x × 5x) =
⇒ 15x2
+ 6x2
+ 10x2
= 279
31x2
= 279
x2
=
x2
= 9
x = √9
= 3 cm
∴ The length of the cuboid 5 × 3 = 15 cm
breadth = 3 × 3 = 9 cm
height = 2 × 3 = 6 cm
Hence, the length = 15 cm, breadth = 9 cm and height = 6 cm.
Ex 16.2
Question 1.
The diameter and height of a cylinder are 14 cm and 15 cm respectively. Find the total surface area and
the cylinder.
Solution :
Given,
Diameter of cylinder = 14 cm
∴ Radius of cylinder (r) = = 7 cm
The height of cylinder (h) = 15 cm
Total surface area of cylinder = 2πr(h + r)
2 × × 7(15 + 7)
= 2 × 22 × 22
= 968 cm2
Volume of cylinder = πr2
h
= × (7)2
× 15
=
= 2310 cm3
.
Hence, Total surface area of cylinder is 968 cm2
and Volume is 2310 cm3
.
Question 2.
The height of a right circular cylinder is 7 cm and radius is 3 cm. Find its curved surface area, total
and volume.
Solution :
Given :
Height of cylinder (h) = 7 cm
Radius of base (r) = 3 cm
Curved surface area of cylinder = 2πrh
= 2 × × 3 × 7

6. = 22 × 6
= 132 cm2
.
Total surface area of cylinder = 2πr(h + r)
= 2 × × 3(7 + 3)
2 × × 3 × 10
=
= 188.57 cm2
Volume of cylinder = πr2
h
= × 3 × 3 × 7
= 22 × 3 × 3
= 198 cm3
Hence, curved surface area of cylinder = 132 cm2
.
Total surface area = 188.57 cm2
and Volume = 198 cm3
.
Question 3.
The area of one end of a cylinder is 154 cm2
and its height is 21 cm. Find the volume and curved
cylinder.
Solution:
Given
Height of cylinder (h) = 21 cm
Area of its one end = 154 cm2
πr2
= 154
⇒ × r2
= 154
r2
=
= = 49
r = √49 =
r = 7 cm
Volume of cylinder = πr2
h
= × 7 × 7 × 21
= 22 × 7 × 21
= 3234 cm3
Curved surface area = 2πrh
= 2 × × 7 × 21
= 2 × 22 × 21
= 924 cm2
.
Hence, the volume of cylinder= 3234 cm3
curve surface area = 924 cm2
Question 4.
The ratio of radii of two right circular cylinder is 2 : 3 and the ratio of their heights is 5 : 4. Find the ratio
curved surface areas and volume.
Solution :
Let the radius of first cylinder be r1 and height be h1 and let the radius of second cylinder be r2 and
Then according to question
=
and =
Curved surface area of first cylinder S1 = 2πr1h1
and curved surface area of second cylinder S2 = 2πr2h2

7. Hence, ratio of curved surface area = 5 : 6
and ratio of volumes = 5 : 9
Question 5.
The total surface area of a solid cylinder is 462 cm2
. Its curved surface area is one third of its total
Find the volume of the cylinder.
Solution :
Let radius of cylinder be r and height be h. Then
Given, Total surface area = 262 cm2
.
⇒ 2πr(h+r) = 264
According to question.
Curved surface area
= (total surface area)
= × 462
= 2πrh = 154 …(ii)
on dividing equation (i) by equation (ii) we get.

8. Hence the volume of cylinder = 539 cm3
Question 6.
The curved surface area and height of cylinder are 660 cm2
and 15 cm respectively. Find its volume.
Solution :
Let radius of cylinder be r.
Given, height of cylinder = 15cm
and curved surface area = 660 cm2
⇒ 2πrh = 660

9. Hence, the volume of cylinder = 2310 cm3
.
Question 7.
Volume of a cylinder is 30π cm3
and the base is 6π cm2
. Find the height of the cylinder.
Solution :
Given,
Volume of cylinder =30π cm3
and base area = 6π cm2
:. Volume of cylinder = base area × height
⇒ 30π = 6π × height
⇒ height =
= 5 cm.
Hence, the height of the cylinder 5 cm.
Question 8.
Volume and curved surface area of a cylinder are 1650 cm3
and 660 cm2
respectively. Find the radius
of the cylinder.
Solution :
Let radius of cylinder be r. and height be h.
Given volume of cylinder 1650 cm3
⇒ πr2
h = 1650 ….(i)
Curved surface area of cylinder = 660 cm2
⇒ 2πrh = 660 …(ii)
On dividing (i) by (ii), we get
substituting this value of r in equation (ii), we get

10. Hence, the radius is 5 cm and height is 21 cm.
Question 9.
The height and the radius of a cylinder are 7.5 cm and 3.5 cm respectively. Find the ratio of its total
and curved surface area.
Solution :
Given,
Height of cylinder (h) = 7.5 cm
Radius (r) = 3.5 cm
Total surface area = 2πr(h + r)
Hence, required ratio = 22 : 15
Question 10.
A well 20 m deep and 7 m ¡n diameter is dug. The earth taken out is spread to form a 22 m × 14 m
Find the height of the embankment.
Solution :
Given,
Diameter of well = 7 m

11. ∴ Radius of well (r) = m
Depth of well (h) = 20m
The volume of the soil taken out from well = πr2
h
= × × 20
= 770 cm3
.
The length of plateform (L) = 22 m
The breadth of plateform (B) = 14 m
Let the height of plateform be H.
Volume of plateform = L x B x H m3
.
= 22 × 14 × H m3
According to question
Volume of plateform = volume of earth digut 22 × 14 × H = 770
H =
= 2.5 m.
Hence the height of plate form is 2.5 m.
Question 11.
In a cylindrical vessel 30800 cm3
water can be filled. If the internal radius of vessel is 14 cm. Find it
surface area.
Solution :
Given
The internal radius of vessel r = 14 cm
Volume of the vessel 30800 cm3
Let the height of the vessel be h.
∴ πr2
h = 30800 cm3
.
Hence the internal curved surface area of cylindrical vessel = 4400 cm2
Question 12.
If the width of a hollow cylinder is 2 cm. Its internal diameter is 14 cm and height 26 cm. The two ends of
are opened. Find the total surface area of the hollow cylinder.
Solution :
Given internal diameter of hollow cylinder =14 cm.
∴ Internal radius (r2) = 7 cm
Width of the cylinder = 2 cm
∴ External radius (r1) = 7 + 2 = 9 cm
Height (h) 26 cm.
Total surface area of cylinder

12. Hence, the total surface area of cylinder = 2816 cm2
Question 13.
A hollow cylinder ¡s open at both ends. Its height 20 cm and internal and external diameters arc 26 cm
respectively. Find the volume of the metal by which cylinder is made of.
Solution :
Given,
Height of the cylinder(h) = 20 cm
Internal diameter = 26 cm
Internal radius (r1) = = 13 cm
External diameter = 30 cm
External radius (r2) = = 15 cm
Hence, the volume of the metal = 3520 cm3
Rajasthan Board RBSE Class 10 Maths Chapter 16
Surface Area and Volume Ex 16.3
Question 1.
The height and radius of base of a cone are 28 cm and 21 cm respectively. Find its curved surface area,
surface area and the volume.
Solution :
Given, height of cone (h) = 28 cm.

13. Radius of base (r) = 21 cm
Question 2.
The volume of a right circular cone is 1232 cm3
and its height is 24 cm. Find the slant height of the
Solution :
Given, height of cone (h) = 24 cm
Let radius of the cone be r.
Volume of the cone = 1232 cm3
∴ radius of the cone = 7 cm

14. ∵ slant height of cone (l)
Hence, the slant height of the cone = 25 cm.
Question 3.
Diameter and slant height of a cone are 14 m and 25 m respectively. Find it total surface area.
Solution :
Given, the base diameter of cone = 14 m
∴ Radius (r) = = 7 m
Slant height (d) = 25 m
Total surface area of the cone = πr(l+r)
= × 7 × (25 + 7)
= × 7 × 32
= 22 × 32
= 744 m2
Hence, total surface area of cone 704 m2
Question 4.
The radius of the base of a cone is 14 cm and its slant height is 50 cm. Find the curved surface area
surface area of the cone.
Solution :
Given
The base radius of cone (r) = 14 cm.
Slant height (l) = 50 cm
∴Curved surface area of cone = πrl
= × 14 × 50
= 2200 cm2
And total surface area of cone = πr (l + r)
= × 14 × (50+14)
= 22 × 2 × 64
= 2816 cm2
Hence, curved surface area 2200 cm2
and total surface area = 2816 cm2
Question 5.
The height of a right cone is 8 cm and its radius of the base is 6 cm. Find the volume of the cone.
Solution :
Given,
Height of cone (h) = 8 cm
base radius (r) = 6 cm
Volume of the cone (V) = πr2
h
= × × 6 × 6 × 8

15. =
= 301.7 cm3
Hence, volume of the cone = 301.7 cm3
Question 6.
The lateral surface area and the slant height of a cone are 1884.4 m2
and 12 m respectively. Find its
Solution :
Given,
Slant height of cone (l) = 12 meter
Lateral surface area = 1884.4 m2
Let the radius of base be r.
∴ πrl = 1884.4
× 12 × r = 1884.4
r =
= 50 m(approx)
Question 7.
Base area of a right cone is 154 cm2
. If its slant height is 25 cm. Find the height of the cone.
Solution :
Given,
Slant height of the cone (l) = 25 cm
Base area = 154 cm2
A cone has a circular base,
Let the height of the cone be h,
Hence, the height of the cone = 24 cm.
Question 8.
Two cones have same diameters of the base. The ratio of their slant heights is 5 : 4. If the lateral
smaller cone is 400 cm2
. find the lateral surface area of the larger cone.
Solution :
Given,

16. Two cones have same base diameters. So their radii will also be equal.
Let r1 = r2 = r.
Let the slant heights S be l1 and l1 then according to the question.
=
And lateral surface area of smaller cone (πrl2) = 400 cm2
lateral surface area of larger cone = × 400
= × 100 = 500 cm2
Question 9.
The ratio of slant height and the radius of a cone is 7 : 4. If its lateral surface area is 792 cm2
, then find
Solution :
Given, slant height (l) : radius (r) = 7 : 4
Hence, radius of the cone = 12 cm.
Question 10.
The circumference of the base of conical tent with a height 9 m is 44 m. Find the volume of the air inside
Solution :
Given,
Height of the cane (h) = 9 m
Circumference of the base = 44 m

17. ∴ 2πr = 44
Question 11.
A conical vessel with base radius 10 cm. contains some water in it. The water level in the vessel is 18
is pored into another cylindrical vessel with radius 5 cm. Find the water level in the cylindrical vessel.
Solution :
Base radius of conical vessel (R) = 10 cm
And the height (H) = 18 cm
Volume of conical vessel = πR2
H
= × π × (10)2
× 18
= π ×100 × 6
= 600 π cm3
Let the water lavel in the cylindrical vessel be h.
Radius (r) = 5 cm.
Now according to the problem
Volume of water in the cylindrical vessel = Volume of water in conical vessel
πr2
h = 600 π
or r2
h = 600
⇒ (5)2
h = 600
⇒ 25 h = 600
h =
= 24 cm
Hence the water level in the cylindrical vessel = 24 cm.
Question 12.
A largest right circular cone ¡s formed by cutting a wooden cubical piece of 14 cm side. Find the volume
cone.
Solution :
Given,
Side of the cube (a) = 14 cm.
The diameter of the largest cube cut off = 14 cm

18. The radius of ths cone r = = 7 cm
The height of the cone cut off (h) = 14 cm.
∴ Volume of cone = πr2
h
× × 7 × 7 × 14
=
Hence, the volume of cubical piece cut off = 718.67 cm3
.
Question 13.
The base radius and height of a cube are 7 cm and 24 cm respectively. Find it slant height, lateral
total surface area and volume.
Solution :
Given,
The base radius of cone (e) = 7 cm
Height (h) = 24 cm

19. Hence, slant height of cone 25 cm, curved surface area = 550 cm2
, total surface area = 704 cm2
and
cm3
Question 14.
Radius and the angle of sector are 12 cm and 120° respectively. By making its straight cores a cone is
Find its volume.
Solution :
Given,
Radius of the sector OA = OB (r) = 12 cm
Angle of the sector = 120°

20. Rajasthan Board RBSE Class 10 Maths Chapter 16
Surface Area and Volume Ex 16.4
Question 1.
Find the surface area and the volume of a sphere with radius 1.4 cm.
Solution :
Given
Radius of sphere (r) = 1.4 cm
Surface area = 4πr2
= 4 × × 1.4 × 1.4
= 24.64 cm2
Volume of the sphere = πr2
= × × 1.4 × 1.4 × 1.4
= 11.5 cm3
Hence, surface area of sphere = 24.64 cm2
and its volume = 11.5 cm3

21. Question 2.
The surface area of a sphere is 616 cm2
. Find the volume of the sphere.
Solution :
Let radius of sphere be r.
Given, Surface area of sphere = 616 cm2
∴ 4πr2
= 616
Hence, volume of sphere = 1437.33 cm3
Question 3.
Radius of a hemisphere is 4.5 cm. Find its total surface area and volume.
Solution :
Given,
Radius of hemisphere (r) = 4.5 cm
Surface area of hemisphere = 3πr2
Let

22. Hence, total surface area = 190.93 cm2
and volume = 190.93 cm3
Question 4.
The volume of a sphere is 38808 cm3
. Find its surface area.
Solution :
Let radius of the sphere be r,
Volume of sphere = 38808 cm3
Hence, the surface area of sphere = 5544 cm2
Question 5.
A cylinder is made of glass, whose radius and height are 4 cm and 10 cm respectively. By melting it,

23. sphere with radii 2 cm each can be recasted?
Solution :
Given, for a cylinder
Radius = 4 cm
And height = 10 cm
Volume = πr2
h
= π × 4 × 4 × 10 cm3
.
Again the radius of a sphere recasted (r) = 2 cm.
Volume of sphere = πr2
= × π × (2)3
Let the number of sphere recasted be n, then
Volume of the cylinder = n × volume of sphere
Hence, the number of sphere recasted = 15
Question 6.
The thickness of hollow spherical shell is 2 cm. If its external radius is 8 cm find the volume of metal
Solution :
Given,
External radius of hollow spherical shell r1 = 8 cm.
Internal radius r2 = 8 – 2 = 6
Hence, volume of the metal used = 1240.38 cm3
Question 7.
How many cones with radius 3 cm and height 6 cm can be formed by melting a metallic sphere with
Solution :
Given,
Radius of sphere (r) = 9 cm
Volume of the sphere = πr3

24. = × × 9 × 9 × 9 cm3
For a cone,
Radius (R) = 3 cm
Height (h) = 6 cm
Volume of the cone = πr2
h
= × × 3 × 3 × 6 cm3
Let the number of cones recasted be n.
∴ Volume of sphere = n × volume of a cones
Hence, the number of cones recasted = 54.
Question 8.
Eight spheres of same volumes are recasted by melting a metallic sphere with radius of 10 cm. Find the
the recasted sphere.
Solution :
Given,
Radius of the metallic sphere (r) = 10 cm
∴ Volume of sphere = πr3
= × π × r × (10)3
Let the radius of each recasted spheres be R.
∴ The volume of large sphere =8 × volume of spheres with radius R
Hence, radius of the sphere casted = 5 cm.
Surface area of sphere = 4πR2
= 4 × π × (5)2
=4 × π × 25
= 100π cm2
Hence, surface area of casted sphere = 100π cm2
Question 9.
If surface area of a sphere ¡n 5544 cm2
, then find its volume.
Solution :
Let radius of sphere be r, then
Given, surface area of sphere = 5544 cm2
4πr2
= 5544

25. Hence, volume of sphere = 38808 cm3
Question 10.
The measures of a solid cuboid are 66 cm, 42 cm and 21 cm respectively. How many spheres with
each can be recasted by melting it.
Solution :
Length (l) = 66 cm
Breadth (b) = 42 cm
and height (h) = 21 cm
Volume of cuboid = l × b × h
= 66 × 42 × 21 cm3
Diameter of sphere recasted = 4.2 cm
radius (r) = = 2.1 cm
Volume of sphere = π × (2.1)3
Let by melting the cuboid n spheres are formed.
∴ Volume of cuboid = n × volume of sphere
Hence, the number of sphere reformed = 1500
Question 11.
A sphere with diameter 6 cm is put into the water filled in a cylindrical vessel with diameter 12 cm. How
water level in the vessel will rise?
Solution :
Given,
Diameter of sphere = 6 cm
radius (r) = = 3 cm
Volume of sphere = πr3
= × × (3)3
Diameter of cylindrical vessel = 12 cm
radius (R) = = 6 cm

26. When the sphere is put into the vessel, the water level rises by h.
∴ Volume of sphere = volume of water rise into cylinder.
Hence, water level will rise in vessel = 1 cm.
Question 12.
A hemispherical bowl with internal radius 9 cm is filled with liquid This liquid is to fill into the smaller
bottle with diameter 3 cm and height 4 cm. How many bottle will be needed to fill whole liquid?
Solution :
Given,
Radius of hemispherical bowl (r) = 9 cm
Volume of the bowl = × π × r3
= × π × (9)3
cm3
Diameter of cylindrical bottle = 3 cm
Radius = cm
And height = 4 cm
Volume of the bottle = πr2
h
= π × 4
Let n bottles will be needed to be filled the whole Liquid
∴ According to the question.
Volume of hemispherical bowl = n × volume of cylindrical bottles.
Hence, 54 bottles will be needed.
Question 13.
The diameter of a sphere is 0.7 cm. From a water tank, 3000 spheres completely filled with water is
Find the volume of the
water thrown out.
Solution :
Given,
Diameter of sphere = 0.7 cm
Radius (r) = cm
∴ Volume of sphere V = πr3
= π cm3
Volume of water thrown out = 3000 × volume of sphere
= 3000 × π

27. = 539 cm3
Question 14.
The external and internal diameters of hemispheric bowl are 43 cm and 42 cm respectively. If the
paise 7 per cm2
, find the total cost of coloring the bowl.
Solution :
Given,
External diameter of hemispherical bowl = 43 cm
∴ External radius (r1) = cm
Internal diameter = 42 cm
∴ Internal radius (r2) = = 21 cm
External surface area of bowl = 2πr1
2
And internal surface area = 2πr2
2
Rajasthan Board RBSE Class 10 Maths Chapter 16
Surface Area and Volume Miscellaneous Exercise
Multiple Choice Questions
Question 1.
Total surface area of cube is 486 cm2
. Its side will be :
(A) 6 cm
(B) 8 cm
(C) 9 cm
(D) 7 cm

28. Solution :
Total surface area of cube = 486 cm2
6a2
= 2 m
a2
= = 81
a2
= = 9 cm
∴ side of the cube is 9 cm
Hence, option (C) is correct.
Question 2.
Length, breadth and height of a cuboid are 9 m, 2 m and 1 m respectively. Surface area of cuboid is:
(A) 12 m2
(B) 1 m2
(C) 21 m2
(D) 22 m2
Solution :
Given,
l = 9 m
b = 2 m
h = 1 m
Surface area of cuboid = 2(l + b)h
= 2(9 + 2) × 1
=2 × 11 × 1
22 m2
Hence, option (D) is correct.
Question 3.
If diameter of a sphere is 6 cm, them volume is:
(A) 16π cm3
(B) 20π cm3
(C) 36π cm3
(D) 30π cm3
Solution :
Given,
Diameter of sphere = 6 cm
Radius (r) = = 3 cm
Volume V = πr3
= × π × 3 × 3 × 3
36π cm3
Hence, option (C) is correct.
Question 4.
The radius of a cylinder is 14 cm and its height is 10 cm. Curved surface area of cylinder is:
(A) 881 cm2
(B) 880 cm2
(C) 888 cm2
(D) 890 cm2
Solution :
Given,
Radius of cylinder (r) = 14 cm
Height (h) = 10 cm
Curved Surface area = 2πrh
= 2 × × 14 × 10

29. = 880 cm2
Hence, option (B) is correct.
Question 5.
The volume and height of a cone are 308 cm3
and 6 cm respectively. Its radius will be:
(A) 7 cm
(B) 8 cm
(C) 6 cm
(D) None of these
Solution :
Given.
Height of cone = 6 cm
Volume = 308 cm3
Hence, option (A) is correct.
Question 6.
The diameter of a metallic hemisphere is 42 cm. Find the cost of polishing its total surface at the rate of
per cm2
.
Solution :
Diameter of hemisphere = 42 cm
Radius (r) = = 21 cm
Total surface area of hemisphere = 3πr2
= 3 × × 21 × 21
= 3 × 22 × 3 × 21
= 4158 cm2
∵ Cost of polishing 1 cm2
paise 20 = ₹ 0.20
∴ Cost of polishing 4158 cm2
= 4158 × 0.20 = ₹ 831.60
Question 7.
A cone, a hemisphere and a cylinder have same base and equal to their heights. Find the ratio among
volumes.
Solution :
Given, a cone, hemisphere and a cylinder have same base, their base and heights are equal.

30. Hence, the ratio of their volumes = 1 : 2 : 3
Question 8.
Left side of a solid is cylindrical and right side is conical. If diameter and length of cylindrical portion are
and 40 cm respectively and diameter and length of the conical portion are 14 cm and 12 cm
volume of the solid.
Solution :
Given,
Diameter of cylindrical portion = 14 cm
Radius (r) = = 7 cm
Height (h) = 40 cm
Volume of cylindrical portion V1 = πr2
h
= × 7 × 7 × 40 = 6160 cm3
Again, given that diameter of conical portion = 14 cm.
∴ Its radius(R) = = 7 cm
∴ Height (H) = 12 cm
Volume of conical portion V2 = πR2
H
= × × 7 × 7 × 12 = 616 cm3
∴ Volume of the solid = Volume cylinder + Volume of cone
= 6160 + 616 = 6776 cm3
Question 9.
By melting a metallic solid sphere with radius 9 cm, some cones are recasted. If the radius and height of
recasted are 3 cm and 6 cm
respectively, then find the number of cones recasted.
Solution :
Given,
Radius of metallic sphere (r) = 9 cm
∴ Volume of metalic sphere = πr3
= π × (r)3
And radius of cone recasted (R) = 3 cm
And its height (H) = 6 cm
∴ Volume of cone recasted = πR2
H

31. = × π × 3 × 3 × 6
Let the number of cones recasted be n. Then According to the question
Volume of metallic sphere = n × volume of a cone
Hence 54 cones be recasted
Question 10.
The population of a village is 4000, where each person needs 150 liter of water per day. There is a
dimensions 20 m × 15 m × 6 m in the village for how long will the water of the tank be sufficient?
Solution :
Population of the village = 4000
Each person needs water = 150 lit per day
Quantity of water that 4000 people need = 4000 × 150
= m3
= 600 m3
Volume of the tank = 20 × 15 × 6 m3
= 1800 m3
Number of days for the tank has water in it for the village = = 3
Hence, the water in the tank is sufficient for 3 days.
Question 11.
Three spheres with radius 6 cm, 8 cm and 10 cm respectively are melted and a large sphere is
radius of this sphere.
Solution :
Given
Radius of Ist
sphere (r1) = 6 cm
Radius of IInd
sphere (r2) = 8 cm
Radius of IIIrd
sphere (r3) 10 cm
Let the radius of the sphere recasted be R.
Since, by melting three given sphere into larger sphere is recasted
∴ Volume of sphere recasted = sum of volumes of three spheres
Hence radius of sphere recasted = 12 cm

32. Question 12.
The radius of a conical vessel is 10 cm and its height is 18 cm. It is completely filled with water. The
pored into another cylindrical
vessel with radius 5 cm. Find the height of water in this vessel.
Solution :
Given,
Radius of conical vessel (R) = 10 cm
and its height = 18 cm
Volume of conical vessel = r2
h
= × π × (10)2
× 18
= × π × 100 × 18
π × 100 × 6
= 600 π cm3
.
Let the height of water in cylindrical vessel be H and its radius = 5 cm.
Now, according to question.
Volume of cylindrical vessel = Volume of water in the conical vessel.
πR2
H = 600 π
π(5)2
H = 600 π
H = = 24 cm
Hence, height of water in cylindrical vessel = 24 cm
Question 13.
A candle with diameter 2.8 cm is formed by melting wax cuboid with dimensions 11 cm × 3.5 cm × 2.5
the length of candle.
Solution :
Given,
Length of cuboid (l) = 11 cm.
Breadth (b) = 3.5 cm
Height (h) = 2.5 cm
Volume of wax-cuboid = l × b × h
=11 × 3.5 × 2.5 = 96.25 cm3
Diameter of candle = 2.8 cm
Radius = = 1.4 cm
Let the height of the candle of h
Volume of the candle πr2
h.
= × (1.4)2
× h
According to the question Volume of wax cuboid = Volume of candle
Hence, the length of candle = 15.625 cm.
Question 14.
Diameter of a metallic sphere is 6 cm. By melting it an another circular wire ¡s recasted. If the length of
m, find its radius.
Solution :
Given,
Diameter of sphere = 6 cm
Radius of sphere (r) = = 3 cm

33. Volume of metallic sphere = πr3
= × π × (3)3
cm3
Let the radius of the wire be R.
According to question
Length of wire = 36 m = 3600 cm.
∴ Volume of wire = πR2
h.
π × R2
× 3600 cm3
Again according to question
Volume of sphere = volume of wire
Rajasthan Board RBSE Class 10 Maths Chapter 16 solution by Arvind Kumar (arvind.saini@outlook.com)
+91-9460755486