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

Edited By Safeer PP | Updated on Sep 06, 2022 11:58 AM 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. These three-dimensional objects can be cones, spheres, or cylinders. The NCERT exemplar Class 10 Maths solutions chapter 12 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 with 10+ years of teaching and content development experience. The CBSE 10 Maths Syllabus acts as the outline of these NCERT exemplar Class 10 Maths solutions chapter 12.

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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
Solution
(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 curve surface at a field distance.

(B) Frustum of a cone – It is the portion of a solid that lies between one or two parallel planes cutting it.
A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

(C) Hemisphere – In geometry it is an exact half of a sphere and it is a three dimensional geometric.
A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

(D) Cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

(Combination of two cylinders)
Hence form the above diagrams we conclude that option (A) is correct.
Therefore, A cylindrical pencil sharped 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 dimensions space lying the same distance from a given point.
A cylinder – A cylinder is a three dimensional solid that holds two parallel bases joined by a curve surface at a fixed distance.

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 based 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
Solution
Cylinder – A cylinder is a three dimensional solid that has two parallel based joined by a curved surface at a fixed distance.

Sphere – The set of all points in thee dimensional space lying the same distance from a given point.

Similarly hemisphere

Cone – A cone is a three dimensional geometrical shape that tappers smoothly from a flat base to a point called the apex or vertex.

Similarly frustum of cone

Hence we conclude that shuttle cock which is used for playing badminton has the shape of combination of 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 otherside of the plane is called
(A) a frustum of a cone (B) cone
(C) cylinder (D) sphere

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

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

Cylinder – A cylinder is a three dimensional solid that has two parallel based joined by a curved surface at a fixed distance.

Sphere – The set of all points in three dimensional space lying the same distance from a given point.
Thus A cone which is cut through a plane parallel to its base and then the cone that is formed one side of that plane is removed then the new part which is formed is called frustum of a cone.

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 $\frac{0.5}{2}=\frac{5}{20}=\frac{1}{4}cm$
Volume of marble $\frac{4\pi r^{3}}{3}=\frac{4}{3}\times \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} =\frac{11}{168}cm^{3}$

Edge of cube = 22 cm
Volume (V) $22 \times 22 \times 22$
Space occupied by marble = total volume $\frac{1}{8}$ part of volume
$=v -\frac{1}{8}v=\frac{7v}{8}$
Number of marble $=\frac{space occupied }{volume of marble}$
$=\frac{7v \times 168 }{8 \times 11}$
$=\frac{7 \times 22\times 22\times 22\times 22\times 168 }{8 \times 11}$
$=142296$

Question:10

A solid piece of iron in the form of a cuboid of dimensions $49cm \times 33cm \times 24cm$ , 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 $\frac{4}{3}\pi r^{3}$
Also volume of sphere 38808 (given)
$r^{3}=\frac{38808 \times 7 \times 3}{4 \times 22}=\frac{882 \times 7 \times 3}{2}=9261$
$r^{3}=9261$
$r=\sqrt[3]{3 \times 3 \times 3 \times 7 \times7 \times 7}$
$r= 3 \times 7 = 21$
Hence radius of sphere = 21 cm

Question:11

A mason constructs a wall of dimensions $270cm\times 300cm \times 350cm$ with the bricks each of size $22.5cm \times 11.25cm \times 8.75cm$ 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 $=l \times b\times h$
$=270\times 300 \times 350= 28350000cm^{3}$
Length of brick = 22.5 cm
Breadth = 11.25 cm
Height = 8.75 cm
Volume $=l \times b\times h$
$= 22.5 \times 11.25 \times 8.75 = 2214.84375cm3$
$\frac{1}{8}$ Space is covered by mortar (given)
Remaining space $=\frac{volume of wall}{8}$
$=\frac{2835000}{8}=3543750cm^{3}$
Surface constructed $= 28350000 - 3543750 = 24806250cm^3$
Number of bricks used $= \frac{ \text{surface contracted}} {\text{volume of brick}}$
$\Rightarrow \frac{24806250}{2214.84375}=11200$

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 $=\frac{2}{2}=1cm$
Height = 16cm
Volume $=\pi r^{2}h$
$=\pi\times 1 \times 1 \times 16= 16\pi$
We know that twelve solid sphere are made by melting of solid metallic cylinder
Volume of sphere $=\frac{4}{3}\pi r^{3}$
Hence volume of 12 spheres $=16 \pi$
$=12 \times \frac{4}{3}\pi r^{3}=16 \pi$
$=16\pi r^{3}=16 \pi$
$r^{3}=1$
$r=1$
Radius = 1 cm
Diameter $2 \times 1 = 2cm$

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 $=r_{1}=28cm$
Bottom radius $=r_{2}=7cm$
Curved surface area of bucket is $=\pi l(r_{1}+r_{2})$
$=\frac{22}{7}\times 45 \times (28+7)$
$=\frac{22}{7}\times 45 \times 35$
$=4950cm^{2}$

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
$Radius$ $=\frac{0.5}{2}=\frac{5}{20}=\frac{1}{4}=0.25cm$
$\text{ Volume of hemisphere }$ $=\frac{2}{3}\pi r^{3}$
$= \frac{2}{3}\times\frac{22}{7}\times \frac{1}{4}\times\frac{1}{4}\times \frac{1}{4}=\frac{11}{336}$
$\text{ Volume of two hemispheres}$ $=\frac{2 \times11}{336} =\frac{11}{168}$
Similarly radius of cylinder = 0.25
Height = 2- 0.25 -0.25
= 2- 0.5
=1.5 cm
$Volume$ $= \pi r ^{2}h$
$= \frac{22}{7}\times \frac{1}{4}\times\frac{1}{4}\times\frac{15}{10}\Rightarrow \frac{33}{112}$
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) $4 \pi r^{2}$ (B) $6 \pi r^{2}$ (C) $3 \pi r^{2}$ (D) $8 \pi r^{2}$

The radius of hemisphere = r
$\text{ Curved surface area}$ $=2 \pi r^{2}$
The curved surface area of two solid hemisphere
$=2 \times 2 \pi r^{2}$
$=4 \pi r^{2}$

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 enclosed a sphere of radius upto r cm.
$\therefore$ 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) remain 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 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 $=\frac{\pi h}{3} (r_{1}^{2}+ r_{2}^{2}+r_{1}r_{2})$
Diameter of first end = 44 cm
Radius $(r_{1})=\frac{44}{2}=22cm$
Diameter of second end = 24 cm
Radius $(r_{2})=\frac{24}{2}=12cm$
Height (h) = 35cm
Volume $=\frac{\pi h}{3} (r_{1}^{2}+ r_{2}^{2}+r_{1}r_{2})$
$=\frac{22 \times 35}{7 \times 3}(22^{2}+12^{2}+22 \times 12)$
$= \frac{110}{3}[484 +144+264]$
$= \frac{110}{3}[892]$
$= 32706.67 cm^{3}$
We know that 1 liter = 1000 cm³
$\therefore 32706.67 cm^{3} =32.7 l$

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
$\frac{\text {volume of first sphere}}{\text {volume of second sphere}}=\frac{64}{27}$
$\frac{\frac{4}{3}\pi r_{1}^{3}}{\frac{4}{3}\pi r_{2}^{3}}=\frac{64}{27}$
$\left ( \frac{r_{1}}{r_{2}} \right )^{3}=\frac{64}{27}$
$\frac{r_{1}}{r_{2}} =\sqrt[3]{\frac{64}{27}}=\frac{4}{3}$
Ratio of their surface area is $=\frac{4\pi r_{1}^{2}}{4\pi r_{2}^{2}}$
$=\frac{r_{1}^{2}}{r_{2}^{2}}=\left ( \frac{r_{1}}{r_{2}} \right )^{2}$
$=\left ( \frac{4}{3} \right )^{2}=\frac{16}{9}$
Hence required ratio is 16 : 9