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

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TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 12
Chapter NameSurface Areas and Volumes
CategoryNCERT Exemplar

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

Exercise 12.1 Multiple Choice Questions (MCQs)

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
Solution:
(a) Because the shape of sharpened pencil is
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-1s

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
Solution:
(a) Because the shape of surahi is
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-2s

Question 3:
A plumbline (sahul) is the combination of (see figure)
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-3Q
(a) a cone and a cylinder
(b) a hemisphere and a cone
(c) frustum of a cone and a cylinder
(d) sphere and cylinder
Solution:
(b)
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-3s

Question 4:
The shape of a glass (tumbler) (see figure) is usually in the form of
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-3s
(a) a cone                                                    (b) frustum of a cone
(c) a cylinder                                               (d) a sphere
Solution:
(b) We know that, the shape of frustum of a cone is
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-4s
So, the given figure is usually in the form of frustum of a cone.

Question 5:
The shape of a gilli, in the gilli-danda game (see figure) is a combination of
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-5Q
(a) two cylinders                                         (b) a cone and a cylinder
(c) two cones and a cylinder                    (d) two cylinders and a cone
Solution:
(c)
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-5s

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
Solution:
(d) Because the shape of the shuttle cock is equal to sum 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 other side of the plane is called
(a) a frustum of a cone                     (b) cone                     (c) cylinder                                 (d) sphere
Solution:

ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-7s

Question 8:
If a hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that – space of the cube remains unfilled. Then, the number of marbles that the cube can accomodate is
(a) 142244                          (b) 142344                      (c) 142444                                   (d) 142544
Solution:
(a) Given, edge of the cube = 22 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-8s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-8s-2

Question 9:
A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form a cone of base diameter 8 cm. The height of the cone is
(a) 12 cm                     (b) 14 cm                  (c) 15 cm                   (d) 18 cm
Solution:
(b) Given, internal diameter of spherical shell = 4 cm
and external diameter of shell = 8 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-9s
Hence, the height of the cone is 14 cm.

Question 10:
If a solid piece of iron in the form of a cuboid of dimensions 49 cm x 33 cm x 24 cm, is moulded to form a solid sphere. Then, radius of the sphere is
(a) 21 cm                           (b) 23 cm                                      (c) 25 cm                                    (d)19cm
Solution:
(a) Given, dimensions of the cuboid = 49 cm x 33 cm x 24 cm
∴                   Volume of the cuboid = 49 x 33 x 24 = 38808 cm3
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-10s
Hence, the radius of the sphere is 21 cm.

Question 11:
A mason constructs a wall of dimensions 270 cmx 300 cm x 350 cm with the bricks each of size 22.5 cm x 11.25 cmx 8.75 cm 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
Solution:
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-11s

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
Solution:
(c) Given, diameter of the cylinder = 2 cm
∴ Radius = 1 cm and height of the cylinder = 16 cm                    [∵ diameter = 2 x radius]
∴ Volume of the cylinder = π x (1)2 x 16 = 16 π cm3
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-12s
∴                    Diameter of each sphere, d=2r = 2×1=2 cm
Hence, the required diameter of each sphere is 2 cm.

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
Solution:
(a) Given, the radius of the top of the bucket, R = 28 cm
and the radius of the bottom of the bucket, r = 7 cm
Slant height of the bucket, l= 45 cm
Since, bucket is in the form of frustum of a cone.
∴  Curved surface area of the bucket = π l (R + r) = π x 45 (28 + 7)
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-13s

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
Solution:
(a) Given, diameter of cylinder = Diameter of hemisphere = 0.5 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-14s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-14s-2
Hence, the capacity of capsule is 0.36 cm3

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) 47πr2                      (b) 6πr2                     (c) 3πr2                    (d) 8πr2
Solution:
(a) Because curved surface area of a hemisphere is 2 w2 and here, we join two solid hemispheres along their bases of radius r, from which we get a solid sphere.
Hence, the curved surface area of new solid = 2 πr2 + 2 πr2 = 4πr2

Question 16:
A right circular cylinder of radius r cm and height h cm (where, h>2r) just encloses a sphere of diameter
(a) r cm                         (b) 2r cm             (c) h cm                   (d) 2h cm
Solution:
(b) Because the sphere encloses in the cylinder, therefore the diameter of sphere is equal to diameter of cylinder which is 2r cm.

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
Solution:(c) During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.

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 L                      (b) 33.7 L            (c) 34.7 L                 (d) 31.7 L
Solution:
(a) Given, diameter of one end of the bucket
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-18s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-18s-2
Hence, the capacity of bucket is 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
Solution:
(b) We know that, if a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of the cone.

Question 20:
If volumes of two spheres are in the ratio 64 : 27, then the ratio of their surface areas is
(a) 3: 4                        (b) 4 : 3                      (c) 9 : 16                     (d) 16 : 9
Solution:
(d) Let the radii of the two spheres are r1 and r2, respectively.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-20s
Hence, the required ratio of their surface area is 16 : 9.

Exercise 12.2 Very Short Answer Type Questions

Write whether True or False and justify your answer.

Question 1:
Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6πr2.
Solution:
False
Curved surface area of a hemisphere = 2 πr2
Here, two identical solid hemispheres of equal radius are stuck together. So, base of both hemispheres is
common.
∴ Total surface area of the combination
= 2 πr2 + 2 πr2 = 4π r2

Question 2:
A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is 4πrh + 4πr2.
Solution:
False
Since, the total surface area of cylinder of radius, rand height, h = 2πrh + 2πr2 When one cylinder is placed over the other cylinder of same height and radius,
then height of the new cylinder = 2 h
and radius of the new cylinder = r
∴ Total surface area of the new cylinder = 2πr(2h) + 2πr2 = 4πrh + 2πr2

Question 3:
A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone The total surface area of thecombined solid is  [sqrt{{r}^{2}+{h}^{2}}+3r + 2h].
Solution:
False
We know that, total surface area of a cone of radius, r
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.2-3s

Question 4:
A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is frac{4}{ 3}πa3.
Solution:
False
Because solid ball is exactly fitted inside the cubical box of side a. So, a is the diameter for . the solid ball.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.2-4s

Question 5:
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.2-5Q
Solution:
False
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.2-5s

Question 6:
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.2-6Q
 
Solution:
True
We know that, capacity of cylindrical vessel = πr2h cm3
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.2-6s

Question 7:
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.2-7Q
Solution:
Fasle
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.2-7s

Question 8:
An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder.
Solution:
True
Because the resulting figure is
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.2-8s
Here, ABCD is a frustum of a cone and CDEF is a hollow cylinder.

Exercise 12.3 Short Answer Type Questions

Question 1:
Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed.
Solution:
Given, edges of three solid cubes are 3 cm, 4 cm and 5 cm, respectively.
∴                    Volume of first cube = (3)3 = 27 cm3
                  Volume of second cube = (4)3 = 64 cm3
and                volume of third cube = (5)3 = 125 cm3
∴      Sum of volume of three cubes = (27 + 64 + 125) = 216 cm3
Let the edge of the resulting cube = R cm
Then, volume of the resulting cube, R3 =216 ⇒ R = 6cm

Question 2:
How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm x 11 cm x 12 cm?
Solution:
Given, dimensions of cuboidal = 9 cm x 11 cm x 12 cm
∴ Volume of cuboidal = 9 x 11 x 12 = 1188 cm3
and diameter of shot = 3 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-2s

Question 3:
A backet is in the form of a frustum of a cone and holds 28.490 L of water. The radii of the top and bottom are 28 cm and 21 cm, respectively. Find the height of the bucket.
Solution:
Given, volume of the frustum = 28.49 L = 28.49 x 1000 cm3                                    [∴ 1 L = 1000 cm3]
= 28490 cm3
and radius of the top (r1) = 28 cm
radius of the bottom (r2) = 21 cm
Let height of the bucket = h cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-3s

Question 4:
A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.
Solution:
Let ORN be the cone then given, radius of the base of the cone r1 = 8cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-4s-1
The plane along CD divides the cone into two parts, namely
(i) a smaller cone of radius 4 cm and height 6cm and (ii) frustum of a cone for which
Radius of the top of the frustum, r1 = 4 cm
Radius of the bottom,r2 = 8 cm
and             height of the frustum, h = 6 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-4s-2
∴  Required ratio = Volume of frustum : Volume of cone = 24 π: 32 π = 1:7

Question 5:
Two identical cubes each of volume 64 cm3 are joined together end to end. What is the surface area of the resulting cuboid?
Solution:
Let the length of side of a cube = a cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-5s

Question 6:
From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid.
Solution:
Given that, side of a solid cube (a) = 7 cm
Height of conical cavity i.e., cone, h = 7 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-6s-1
Since, the height of conical cavity and the side of cube is equal that means the conical cavity fit vertically in the cube.
Radius of conical cavity i.e., cone, r = 3 cm
⇒                                                        Diameter = 2 x r = 2 x 3= 6 cm
Since, the diameter is less than the side of a cube that means the base of a conical cavity is not fit inhorizontal face of cube.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-6s-2
Hence, the required volume of solid is 277 cm³

Question 7:
Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed.
Solution:
If two cones with same base and height are joined together along their bases, then the shape so formed is look like as figure shown.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-7s
Hence, the surface area of shape so formed is 855 cm²

Question 8:
Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacities is 2 : 1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-8Q
Solution:
Let volume of cone A be 2 V and volume of cone B be V. Again, let height of the cone A = h1 cm, then height of cone B = (21 – h1) cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-8s

Question 9:
An ice-cream cone full of ice-cream having radius 5 cm and height 10 cm as shown in figure
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-9Q
Calculate the volume of ice-cream, provided that its frac{1}{6} part is left unfilled with ice-cream.
Solution:
Given, ice-cream cone is the combination of a hemisphere and a cone.
Also , radius of hemisphere = 5 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-9s

Question 10:
Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker, so that the water level rises by 5.6 cm.
Solution:
Given, diameter of a marble = 1.4 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-10s-1
Also, given diameter of beaker = 7 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-10s-2

Question 11:
How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm?
Solution:
Given that, lots of spherical lead shots made from a solid rectangular lead piece.
∴ Number of spherical lead shots
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-11s-
Hence, the required number of special lead shots is 1500.

Question 12:
How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge
measures 44 cm.
Solution:
Given that, lots of spherical lead shots made out of a solid cube of lead.
∴ Number of spherical lead shots
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-12s-1
Given that, diameter of a spherical lead shot i.e., sphere = 4cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-12s-2
Hence, the required number of spherical lead shots is 2541.

Question 13:
A wall 24 m long, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm x 16 cm x 10 cm. If the mortar  occupies frac{1}{6} th of the volume of the wall, then find the number of bricks used in constructing the wall.
Solution:
Given that, a wall is constructed with the help of bricks and mortar.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-13s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-13s-2
Hence, the required number of bricks used in constructing the wall is 12960.

Question 14:
Find the number of metallic circular disc with 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
Given that, lots of metallic circular disc to be melted to form a right circular cylinder. Here, a circular disc work as a circular cylinder.
Base diameter of metallic circular disc = 1.5 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.3-14s
Hence, the required number of metallic circular disc is 450.

Exercise 12.4 Long Answer Type Questions

Question 1:
A solid metallic hemisphere of radius 8 cm is melted and recasted into a right circular cone of base radius 6 cm. Determine the height of the cone.
Solution:
Let height of the cone be h.
Given, radius of the base of the cone = 6 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-1s

Question 2:
A rectangular water tank of base 11 m x 6 m contains water upto a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.
Solution:
Given, dimensions of base of rectangular tank = 11 m x 6 m and height of water = 5 m
Volume of the water in rectangular tank = 11 x 6 x 5 = 330 m3
Also, given radius of the cylindrical tank = 3.5 m
Let height of water level in cylindrical tank be h.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-2s
Hence, the height of water level in cylindrical tank is 8.6 m.

Question 3:
How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm. If one cubic centimetre of iron weights 7.5 g, then find the weight of the box.
Solution:
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-3s

Question 4:
The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pin is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one-fifth of a litre?
Solution:
Given, length of the barrel of a fountain pen = 7 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-4s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-4s-2

Question 5:
Water flows at the rate of 10 m min-1 through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm?
Solution:
Given, speed of water flow = 10 m min-1 = 1000 cm/min
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-5s

Question 6:
A heap of rice is in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover heap?
Solution:
Given that, a heap of rice is in the form of a cone.
Height of a heap of rice i.e., cone (h) = 3.5 m
and diameter of a heap of rice i.e., cone = 9 m
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-6s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-6s-2
Hence, 80.61 m2 canvas cloth is required to just cover heap.

Question 7:
A factory manufactures 120000 pencils daily. The pencils are cylindrical in shape each of length 25 cm and circumference of base as 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at ₹ 0.05 per dm2.
Solution:
Given, pencils are cylindrical in shape.
Length of one pencil = 25 cm
and circumference of base = 1.5 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-7s
Cost of colouring 45000 dm2 curved surface = ₹ 2250

Question 8:
Water is flowing at the rate of 15 kmh-1 through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm?
Solution:
Given, length of the pond= 50 m and width of the pond = 44 m
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-8s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-8s-2
Hence, the required time is 2 h.

Question 9:
A solid iron cuboidal block of dimensions 4.4 m x 2.6m x lm is recast into a hollow cylindrical pipe of
internal radius 30 cm and thickness 5 cm. Find the length of the pipe.
Solution:
Given that, a solid iron cuboidal block is recast into a hollow cylindrical pipe,
Length of cuboidal pipe (l) = 4.4 m
Breadth of cuboidal pipe (b) = 2.6 m and height of cuboidal pipe (h) = 1m
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-9s-1
where, h1 be the length of the hollow cylindrical pipe.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-9s-2
Hence, required length of pipe is 112 m.

Question 10:
500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04 m3?
Solution:
Let the rise of water level in the pond be hm, when 500 persons are taking a dip into a cuboidal pond.
Given that,
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-10s
Hence, the required rise of water level in the pond is 0.5 cm.

Question 11:
glass spheres each of radius 2 cm are packed into a cuboidal box of internal dimensions 16 cm x 8 cm x 8 cm and then the box is filled with water. Find the volume of water filled in the box.
Solution:
Given, dimensions of the cuboidal = 16 cm x 8 cm x 8 cm
Volume of the cuboidal = 16 x 8 x 8 = 1024 cm3
Also, given radius of one glass sphere = 2 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-11s

Question 12:
A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of milk at the rate of ? 22 per L which the container can hold.
Solution:
Given that,height of milk container (h) = 16 cm,
Radius of lower end of milk container (r) = 8 cm
and radius of upper end of milk container (R) = 20 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-12s
Hence, the required cost of milk is ₹ 230.12

Question 13:
A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Given, radius of the base of the bucket = 18 cm
Height of the bucket = 32 cm                                         ‘
So, volume of the sand in cylindrical bucket = πr2h= π (18)2 x 32 = 10368 π
Also, given height of the conical heap (h) = 24 cm
Let radius of heap be r cm.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-13s-1
According to the question,
Volume of the sand in cylindrical bucket = Volume of the sand in conical heap
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-13s-2
Hence, radius of conical heap of sand = 36 cm
and slant height of conical heap = 43.267 cm

Question 14:
A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, then find the total surface area and volume of the rocket, (use n = 3.14J)
Solution:
Since, rocket is the combination of a right circular cylinder and a cone.
Given, diameter of the cylinder = 6 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-14s

Question 15:
A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains 41 frac{19}{21} m3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?
Solution:
Let total height of the building = Internal diameter of the dome = 2r m
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-15s

Question 16:
A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl?
Solution:
Given, radius of hemispherical bowl, r = 9 cm
and radius of cylindrical bottles, R = 1.5 cm and height, h = 4 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-16s

Question 17:
A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm. Such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius to the cone.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-17Q
Solution:
(i)   Whenever we placed a solid right circular cone in a right circular cylinder with full of water, then volume of a solid right circular cone is equal to the volume of water failed from the cylinder.
(ii)  Total volume of water in a cylinder is equal to the volume of the cylinder.
(iii) Volume of water left in the cylinder = Volume of the right circular cylinder – volume of a
right circular cone.
Now, given that
Height of a right circular cone = 120 cm
Radius of a right circular cone = 60 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-17s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-17s-2
Hence, the required volume of water left in the cylinder is 1.584 m3.

Question 18:
Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cms-1 in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour?
Solution:
Given, radius of tank, r1 = 40 cm
Let height of water level in tank in half an hour = 1 cm.
Also, given internal radius of cylindrical pipe, r2 = 1 cm
and speed of water = 80 cm/s i.e., in 1 water flow = 80 cm
In 30 (min) water flow = 80x 60 x 30 = 144000 cm According to the question,
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-18s
Hence, the level of water in cylindrical tank rises 90 cm in half an hour.

Question 19:
The rain water from a roof of dimensions 22 m x 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall (in cm).
Solution:
Given, length of roof = 22 m and breadth of roof = 20 m
Let the rainfall be a cm.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-19s
Hence, the rainfall is 2.5 cm

Question 20:
A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimensions of cubiod are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.
Solution:
Given that, length of cuboid pen stand (l) = 10 cm
Breadth of cubiod pen stand (b) = 5 cm
and height of cuboid pen stand (h) = 4 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.4-20s
So, the required volume of the wood in the entire stand is 170.8 cm3.

All Chapter NCERT Exemplar Problems Solutions For Class 10 maths

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All Subject NCERT Exemplar Problems Solutions For Class 10

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