NCERT Exemplar Class 10 Maths Solutions Chapter 12 Surface Areas and Volumes

Edited By Safeer PP | Updated on Mar 28, 2025 01:08 PM IST | #CBSE Class 10th

NCERT Exemplar Class 10 Maths Solutions chapter 12 discusses the method to determine the surface area and volume of various three-dimensional objects. This chapter helps us to understand how to calculate the area related to covering a solid object and the space it occupies. The concepts we study in this chapter are crucial for solving practical problems related to geometry and real-life applications. These three-dimensional objects can be cones, spheres, or cylinders. In this article, we will deal with many questions such as, 'How do we calculate the surface area and volume of different solids like cubes, cylinders, cones, and spheres?', What are the formulas for surface area and volume?', and 'How do we apply these formulas in different scenarios?' These consist of detailed solutions to study and understand the NCERT class 10 Maths. These NCERT exemplar class 10 Maths chapter 12 solutions are prepared by our skilled subject matter experts.

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The NCERT Exemplar Class 10 Maths chapter 12 solutions are beneficial for understanding of concepts of surface area and volume due to their comprehensive nature. The CBSE Syllabus Class 10 Maths is the reference for NCERT Exemplar Class 10 Maths solutions chapter 12.

Class 10 Maths Chapter 12 exemplar solutions Exercise: 12.1
Page number: 138-140
Total questions: 20

Question:1

A cylindrical pencil sharpened at one edge is the combination of
(A) a cone and a cylinder (B) frustum of a cone and a cylinder
(C) a hemisphere and a cylinder (D), two cylinders.

Answer: (A) a cone and a cylinder
(A) A cone –A cone is a three-dimensional geometric shape that tapers smoothly from a flat base to a point called the apex or vertex.
A cylinder – A cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface at a field distance.

Therefore, A cylindrical pencil sharpened at one edge is the combination of a cone and a cylinder.

Question:2

A Surahi is the combination of
(A) a sphere and a cylinder (B) a hemisphere and a cylinder
(C) two hemispheres (D) a cylinder and a cone.

Answer [A]
Solution:
(A) A Sphere – The set of all points in three-dimensional space lying at the same distance from a given point.
A cylinder – A cylinder is a three-dimensional solid that has two parallel bases joined by a curved surface at a fixed distance. The neck of Surahi is in the shape of a cylinder, and the bottom forms a shape of sphere.

Question:3

A plumbline (sahul) is the combination of

(A) a cone and a cylinder (B) a hemisphere and a cone
(C) frustum of a cone and a cylinder (D) sphere and cylinder

Answer(B) A hemisphere and a cone

Question:4

The shape of a glass (tumbler) (see figure) is usually in the form of

(A) a cone (B) frustum of a cone
(C) a cylinder (D) a sphere

Answer (B) frustum of a cone
Solution
A cone – A cone is a three-dimensional geometrical shape that tapers smoothly from a flat base to a point called the apex or vertex.

(B) Frustum of a cone – It is a portion of a solid that lies between one or two parallel planes cutting it.

(C) A cylinder – A cylinder is a three-dimensional solid that has two parallel bases joined by a curved surface at a fixed distance.

(D) A sphere – The set of all points in three-dimensional space lying the same distance from a given point.

Hence the shape of glass is usually in the form of the frustum of a cone.

Question:5

The shape of a gilli, in the gilli-danda game (See figure), is a combination of

(A) two cylinders (B) a cone and a cylinder
(C) two cones and a cylinder (D) two cylinders and a cone

Answer(C) two cones and a cylinder

Question:6

A shuttle cock used for playing badminton has the shape of the combination of
(A) a cylinder and a sphere
(B) a cylinder and a hemisphere
(C) a sphere and a cone
(D) frustum of a cone and a hemisphere

Answer: (D) frustum of a cone and a hemisphere
Shuttle cock which is used for playing badminton has the shape of a combination of a frustum of a cone and hemisphere.

Question:7

A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called
(A) a frustum of a cone (B) cone
(C) cylinder (D) sphere

Answer: (A) a frustum of a cone

Frustum of a cone – It is a portion of a solid that lies between one or two parallel planes cutting it

Question:8

A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that $\frac{1}{8}$ space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is
(A) 142296 (B) 142396 (C) 142496 (D) 142596

Answer(A) 142296
Solution
Diameter of marble = 0.5 cm
Radius

\frac{0.5}{2} = \frac{5}{20} = \frac{1}{4} c m

Volume of marble

\frac{4 π r^{3}}{3} = \frac{4}{3} \times \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{11}{168} c m^{3}

Edge of cube = 22 cm
Volume (V)

22 \times 22 \times 22

Space occupied by marble = total volume

\frac{1}{8}

part of volume

= v - \frac{1}{8} v = \frac{7 v}{8}

Number of marble

= \frac{s p a c e o c c u p i e d}{v o l u m e o f m a r b l e}

= \frac{7 v \times 168}{8 \times 11}

= \frac{7 \times 22 \times 22 \times 22 \times 22 \times 168}{8 \times 11}

= 142296

Question:10

A solid piece of iron in the form of a cuboid of dimensions $49 c m \times 33 c m \times 24 c m$ , is moulded to form a solid sphere. The radius of the sphere is
(A) 21cm (B) 23cm (C) 25cm (D) 19cm

Answer (A) 21cm
Solution
Length of cuboid = 49 cm
Breadth of cuboid = 33 cm
Height of cuboid = 24 cm
Volume

= l \times b \times h

49 \times 33 \times 24 = 38808

We know that volume of sphere

\frac{4}{3} π r^{3}

Also volume of sphere 38808 (given)

r^{3} = \frac{38808 \times 7 \times 3}{4 \times 22} = \frac{882 \times 7 \times 3}{2} = 9261

\Rightarrow r^{3} = 9261

\Rightarrow r = \sqrt[3]{3 \times 3 \times 3 \times 7 \times 7 \times 7}

\Rightarrow r = 3 \times 7 = 21

Hence radius of the sphere = 21 cm

Question:11

A mason constructs a wall of dimensions $270 c m \times 300 c m \times 350 c m$ with the bricks each of size $22.5 c m \times 11.25 c m \times 8.75 c m$ and it is assumed that $\frac{1}{8}$ space is covered by the mortar. Then the number of bricks used to construct the wall is
(A) 11100 (B) 11200 (C) 11000 (D) 11300

Answer(B) 11200
Solution
Length of wall = 270 cm
Breadth = 300 cm
Height = 350 cm
Volume

= l \times b \times h

= 270 \times 300 \times 350 = 28350000 c m^{3}

Length of brick = 22.5 cm
Breadth = 11.25 cm
Height = 8.75 cm
Volume

= l \times b \times h

= 22.5 \times 11.25 \times 8.75 = 2214.84375 c m^{3}

\frac{1}{8}

Space is covered by mortar (given)
Remaining space

= \frac{v o l u m e o f w a l l}{8}

= \frac{2835000}{8} = 3543750 c m^{3}

Surface constructed

= 28350000 - 3543750 = 24806250 c m^{3}

Number of bricks used

= \frac{surface contracted}{volume of brick}

\Rightarrow \frac{24806250}{2214.84375} = 11200

Question:12

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is
(A) 4 cm (B) 3 cm (C) 2 cm (D) 6 cm

Answer(C) 2 cm
Solution
Diameter of metallic cylinder = 2 cm
Radius

= \frac{2}{2} = 1 c m

Height = 16cm
Volume

= π r^{2} h

= π \times 1 \times 1 \times 16 = 16 π

We know that twelve solid spheres are made by melting of solid metallic cylinder
Volume of sphere

= \frac{4}{3} π r^{3}

Hence the volume of 12 spheres

= 16 π

\Rightarrow 12 \times \frac{4}{3} π r^{3} = 16 π

\Rightarrow 16 π r^{3} = 16 π

\Rightarrow r^{3} = 1

\Rightarrow r = 1

Radius = 1 cm
Diameter

2 \times 1 = 2 c m

Question:13

The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is
(A) 4950 cm² (B) 4951 cm² (C) 4952 cm² (D) 4953 cm²

Answer (A) 4950 cm²
Solution
Slant height of a bucket = 45 cm
Top radius

= r_{1} = 28 c m

Bottom radius

= r_{2} = 7 c m

Curved surface area of bucket is

= π l (r_{1} + r_{2})

= \frac{22}{7} \times 45 \times (28 + 7)

= \frac{22}{7} \times 45 \times 35

= 4950 c m^{2}

Question:14

A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is
(A) 0.36 cm³ (B) 0.35 cm³ (C) 0.34 cm³ (D) 0.33 cm³

Answer:

Diameter of hemisphere = 0.5 cm
$R a d i u s$ $= \frac{0.5}{2} = \frac{5}{20} = \frac{1}{4} = 0.25 c m$
$Volume of hemisphere$ $= \frac{2}{3} π r^{3}$
$= \frac{2}{3} \times \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{11}{336}$
$Volume of two hemispheres$ $= \frac{2 \times 11}{336} = \frac{11}{168}$
Similarly radius of cylinder = 0.25
Height = 2- 0.25 -0.25
= 2- 0.5
=1.5 cm
$V o l u m e$ $= π r^{2} h$
$= \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times \frac{15}{10} \Rightarrow \frac{33}{112}$
The total volume of capsule = volume of two hemispheres + volume of the cylinder
$= \frac{11}{168} + \frac{33}{112}$
$= 0.065 + 0.294$
= 0.359cm³
=0.36cm³ (approximate)

Question:15

If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is
(A) $4 π r^{2}$ (B) $6 π r^{2}$ (C) $3 π r^{2}$ (D) $8 π r^{2}$

The radius of the hemisphere = r

Curved surface area

= 2 π r^{2}

The curved surface area of two solid hemispheres

= 2 \times 2 π r^{2}

= 4 π r^{2}

Question:16

A right circular cylinder of radius r cm and height h cm (h > 2r) just encloses a sphere of diameter
(A) r cm (B) 2r cm (C) h cm (D) 2h cm

Answer (B) 2r cm
Solution

Here we found that the right circular cylinder of radius r cm and height h cm (h > 2r) can enclose a sphere of radius up to r cm.

∴

The required sphere is of diameter 2r.

Question:17

During conversion of a solid from one shape to another, the volume of the new shape will
(A) increase (B) decrease
(C) remain unaltered (D) be doubled

Answer (C) remains unaltered
Solution
Volume – Volume is defined as the amount of space the object takes up that is the amount of fluid that the container could hold.
During the conversion of a solid from one shape to another, the volume of the new shape remains unchanged.
That is when you convert one solid shape to another then the volume of the original as well as the new solid remains the same.

Question:18

The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is
(A) 32.7 litres (B) 33.7 litres (C) 34.7 litres (D) 31.7 litres

Answer(A) 32.7 litres
Solution
Volume of frustum of cone

= \frac{π h}{3} (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2})

Diameter of the first end = 44 cm
Radius

(r_{1}) = \frac{44}{2} = 22 c m

Diameter of the second end = 24 cm
Radius

(r_{2}) = \frac{24}{2} = 12 c m

Height (h) = 35cm
Volume

= \frac{π h}{3} (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2})

= \frac{22 \times 35}{7 \times 3} (22^{2} + 12^{2} + 22 \times 12)

= \frac{110}{3} [484 + 144 + 264]

= \frac{110}{3} [892]

= 32706.67 c m^{3}

We know that 1 liter = 1000 cm³

∴ 32706.67 c m^{3} = 32.7 l

Question:19

In a right circular cone, the cross-section made by a plane parallel to the base is a
(A) circle (B) frustum of a cone
(C) sphere (D) hemisphere

Answer:

According to the question if a right circular cone is cut by a plane parallel to its base the figure formed is

Here BECD is not a circle, not a sphere not a hemisphere but it is a frustum of a cone.
Hence in a right circular cone, the cross-section made by a plane parallel to the base is a frustum of a cone.

Question:20

Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is
(A) 3 : 4 (B) 4 : 3 (C) 9 : 16 (D) 16 : 9

Answer (D) 16: 9
Let two sphere having radius

r_{1}

and

r_{2}

According to question

\frac{volume of first sphere}{volume of second sphere} = \frac{64}{27}

\Rightarrow \frac{\frac{4}{3} π r_{1}^{3}}{\frac{4}{3} π r_{2}^{3}} = \frac{64}{27}

\Rightarrow {(\frac{r_{1}}{r_{2}})}^{3} = \frac{64}{27}

\Rightarrow \frac{r_{1}}{r_{2}} = \sqrt[3]{\frac{64}{27}} = \frac{4}{3}

Ratio of their surface area is

= \frac{4 π r_{1}^{2}}{4 π r_{2}^{2}}

= \frac{r_{1}^{2}}{r_{2}^{2}} = {(\frac{r_{1}}{r_{2}})}^{2}

= {(\frac{4}{3})}^{2} = \frac{16}{9}

Hence required ratio is 16: 9

Class 10 Maths chapter 12 exemplar solutions Exercise: 12.2
Page number: 142-143
Total questions: 8