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

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.

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

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

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

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

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

Question:1

Write ‘True’ or ‘False’ and justify your answer in the following:
Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is $6 \pi r^2$.

Answer:

It is given that there is two hemispheres of radius r.
Let A and B are two hemispheres of radius r.

Join A and B along with their bases

Now it is a full sphere of radius r
The total surface area of a sphere $=4 \pi r^{2}$
Here we found that the total surface area of the combination is $4 \pi r^{2}$ , but not $6 \pi r^{2}$

Question:2

Write ‘True’ or ‘False’ and justify your answer in the following :
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 $4\pi rh + 4\pi r^2.$

Answer:

It is given that there is two cylinders of height h and radius r.

Where one is placed on other then the shape formed is

The total surface area of the shape formed $=2\pi r (r +2h)$
$\text{ [Using the total surface area of cylinder }=2\pi r (r +h)]$
$=2\pi r (r +h)$
$=2\pi r^2 + 4 \pi r h$
So the surface area is not equal to $4\pi rh + 4\pi r^2.$

Question:3

Write ‘True’ or ‘False’ and justify your answer in the following :
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 the combined solid is$\pi r\left [ \sqrt{r^{2}+h^2}+3r+2h \right ]$

Answer:

It is given that there is a cone of radius r and height h and a cylinder of height h and radius r.

Where A is placed on B

Total surface area = Surface area of cone + Total surface area of cylinder – Surface area of part I – the surface area of part II
$=\pi r (r +l)+2\pi r(r+h)-\pi r^{2}- \pi r^{2}$
$= \pi r (r +\sqrt{r^2 +h^2})+2 \pi r (r+h)-2\pi r ^2 (QI= \sqrt{r^2 +h^2})$
$= \pi r (r +\sqrt{r^2 +h^2}+2r+2h -2r)$
$= \pi r (r +\sqrt{r^2 +h^2}+r+2h )$
Total surface area is not equal to
$\pi r\left [ \sqrt{r^{2}+h^2}+3r+2h \right ]$

Question:4

Write ‘True’ or ‘False’ and justify your answer in the following :
A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is $\frac{4}{3}\pi a ^{3}$

Answer:

It is given that the ball is exactly filled inside the cubical box of side a.

Hence the diameter of the sphere = a
$\text{ Radius of sphere}$ $=\frac{a}{2}$
$\text{ Volume of sphere}$ $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \pi \left (\frac{a}{2} \right )^{3}$
$=\frac{4}{3} \pi \times \frac{a^3}{8}$
$\text{Volume of sphere}=\frac{\pi a ^{3}}{6}$
$\text{ Hence the volume of the sphere is not equal to }\frac{4}{3}\pi a ^{3}$
Hence the given statement is False.

Question:5

Write ‘True’ or ‘False’ and justify your answer in the following:
The volume of the frustum of a cone is$\frac{1}{3}\pi h\left [ r_{1}^2 +r_{2}^2 -r_{1} r_{2} \right ]$ where h is vertical height of the frustum and $r_{1},r_{2}$ are the radii of the ends.

Answer False
According to question

In this figure ABCE is a frustum of cone ABD, h is the height of frustum and $r_{1},r_{2}$ are the radii of the frustum.
From the figure
$\Delta ADF:\Delta EDF$
$\frac{DF}{DG}=\frac{AF}{EG}$
$\frac{h+h'}{h'}=\frac{r_{1}}{r_{2}}$
$\frac{h}{h'}+1=\frac{r_{1}}{r_{2}}$
$\frac{h}{h'}=\frac{r_{1}}{r_{2}}-1$
$\frac{h}{h'}=\frac{r_{1}-r_{2}}{r_{2}}$
$h' = \frac{r_{2}h}{r_{1}-r_{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; ......(1)$
Volume of frustum ABCE = volume of ABD – volume of ECD
$=\frac{1}{3}\pi r^{2}\left [ h'+h \right ]=\frac{1}{3}\pi r{_{1}}^{2}h'$ $\left ( \therefore \text {Volume of cone }=\frac{1}{3}\pi r^{2}h \right )$
$=\frac{1}{3}\pi r{_{1}}^{2}h'+\frac{1}{3}\pi r{_{1}}^{2}h'-\frac{1}{3}\pi r{_{2}}^{2}h'$
$=\frac{1}{3}\pi \left [ r{_{1}}^{2}h'+ \pi r{_{1}}^{2}h'-\pi r{_{2}}^{2}h' \right ]$
$=\frac{1}{3}\pi \left [ r{_{1}}^{2}h'+ h' \left ( r_{1}^{2} -r_{2}^{2}\right ) \right ]$
$=\frac{1}{3}\pi \left [ r{_{1}}^{2}h'+ h' \left ( r_{1} -r_{2}\right )\left ( r_{1}+r_{2} \right ) \right ]$ $\left [ Q\left ( a^{2}-b^{2} \right )=(a-b)(a+b) \right ]$
$=\frac{1}{3}\pi \left [ r_{1}^{2}h+\frac{r_{2}h}{\left ( r_{1}-r_{2} \right )}\left ( r_{1}-r_{2} \right ) \left ( r_{1}+r_{2} \right ) \right ]$ [ using (1) ]
$=\frac{1}{3}\pi h\left [ r_{1}^{2}+r_{2}^{2}+r_{1}r_{2} \right ]$
Hence the volume of frustum of cone is not equal to $=\frac{1}{3}\pi h\left [ r_{1}^{2}+r_{2}^{2}-r_{1}r_{2} \right ]$
Hence the given statement is false.

Question:6

Write ‘True’ or ‘False’ and justify your answer in the following:
The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Figure is $\frac{\pi r^{2}}{3}[3h-2r]$

AnswerTrue
Solution
It is given that a cylindrical vessel of height h and radius r is raised upward with a hemispherical portion.
From the figure radius of hemisphere = r cm
The volume of the figure = volume of the cylinder – the volume of the hemisphere
$=\pi r^{2}h - \frac{2}{3}\pi r^{3}$
$=\pi r^{2}\left [h-\frac{2}{3}r \right ]$
$=\pi r^{2}\left [\frac{3h -2r}{3} \right ]$
$=\frac{\pi r^{2}}{3}[3h-2r]$
$\text{ Hence the capacity of the vessel is }\frac{\pi r^{2}}{3}[3h-2r]$
So the given statement is True.

Question:7

Answer True
Solution
According to question, here is a metallic bucket 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 figure (A) is the surface area of (a), (b), (c)
(a) = curved surface area of frustum
(b) = area of circular base
(c) = curved area of cylinder
Total surface area = curved surface area of frustum of cone + area of circular base + curved surface are of cylinder.
So the given statement is True.

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.

Answer 6 cm
Solution
It is given that there are three cubes of sides
$a_{1}=3cm , a_{2}=4cm, a_{3}=5cm$
The volume of the cube formed by melting these three cubes = volume of first + volume of second + volume of the third.
Let the edge of the cube formed by melting these three cubes is $a_{4}$ .
$(a_{4})^3=(a_{1})^3+(a_{2})^3+(a_{3})^3$
$(a_{4})^3=27+64+125$
$(a_{4})^3=216$
$(a_{4})^3=(6)^3$
$a_{4}=6cm$
Hence the edge of the cube formed is 6 cm.

Question:2

How many shots each having a diameter of 3 cm can be made from a cuboidal lead solid of dimensions $9cm \times 11cm \times 12cm\ ?$

Answer 84
Solution
Here the dimensions of a cuboid are
$9cm \times 11cm \times 12cm$
Volume of cuboid $=l \times b \times h$
$=9 \times 11 \times 12$
$=1188cm^3$
The diameter of the shot is 3 cm
Radius of shot $\frac{3}{2}=0.5 cm$
The volume of 1 shot
$=\frac{4}{3}\pi r^3$ $\left (\because \text{volume of sphere}=\frac{4}{3}\pi r^3 \right )$
$=\frac{4}{3}\times (3.14) \times (1.5)^3$
$=\frac{4}{3}\times 3.14 \times (1.5 )^3$
Number of shots
$=\frac{\text{volume of cuboid }}{\text {volume of sphere}}=\frac{1188}{14.13}=84$

Question:3

A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm, respectively. Find the height of the bucket.

Answer15 cm
Solution
Given :
Volume of bucket = 28.490 liters
Radii of the top $r_{1}=28cm$
Radii of the bottom $(r_2) = 21cm$
Volume of bucket = 28.490 liters
Or
Volume of bucket = 28.490 liters × 1000 $(Q1L=1000cm^{3})$
=28490 cm3
$\frac{1}{3}\pi h =\left [ r_{1}^2 +r_{2}^2+r_{3}^2\right ]=28490$
(Q volume of frustum of cone $\frac{1}{3}\pi h \left [ r_{1}^2 +r_{2}^2+r_{3}^2\right ]$ )
$\frac{1}{3} \times 3.14 \times h [(28^2)+(21)^2+28 \times 21]=28490$
$=h[784+441+588]=\frac{28940 \times 3}{3.14}$
$h[1813] = 27219.74$
$h= \frac{27219.74}{1813}=15$
h = 15 cm
Hence the height of the bucket is 15 cm.

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.

Answer 1:7
Solution
When a cone is divided into two parts by a plane through the mid-point the image formed is

In figure $\Delta AGE:\Delta EFC$ $Q\angle E$ is common angle $\angle F=\angle G=90^{\circ}$
So the corresponding sides are in equal ratio.
$\frac{EF}{FG}=\frac{FC}{GA}$
$\frac{ 6}{12} =\frac{ FC}{}8$
FC = 4 cm
Volume of cone $EDC= \frac{1}{3}\pi r_{2}^2h$
$= \frac{1}{3}\times 3.14 \times (4)^2 \times 6=100.48cm^3$
Volume of frustum of cone
$ABCD =\frac{1}{3}\pi h[r_{1}^2+r_{2}^2 +r_{1}r_{2}]$
$=\frac{1}{3}\times 3.14 \times 6[(8)^2+(4)^2 +8 \times 4]$
$=6.28 [64+16+32]$
$= 6.28[112] = 703.36$
Volume of cone EDC : volume of ABCD
100.48 : 703.36
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?

Answer:

It is given that volume of cube = 64 cm3
$a^3=64cm^3$ ( Because the volume of cube = a3 )
$a^3=4^3$
$a=4$
So the side of the two cubes are 4 cm
The cuboid formed by joining two cubes.

The surface area of the cuboid $=2(lb +bh+hl)$
$=2(8 \times 4 + 4 \times + 4 \times 8)$
$=2(8 \times 4 +4 \times 4 + 4 \times 8)$
$=2(32+16 +32)$

$=2(80)=160cm^2$

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.

Answer 277 cm3
Solution
The figure formed when a conical cavity is cut out from a cube.

Volume of cube $=a^3$
$=(7)^3=343 cm^3$
The volume of the conical cavity
$=\frac{1}{3}\pi r^2h$
$=\frac{1}{3}\times 3.14 \times (3)^2 \times 7$ $\left ( \because r=3cm ,h=7cm \right )$
$=21\times 3.14= 66cm^3$
The volume of remaining solid = volume of the cube – the volume of the conical cavity
$=343-66= 277cm^3$

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.

Answer 854 cm2
Solution
According to question

Here are two cones joined together along their bases
Height of both cone = 15 cm
Base radius of both cone = 8 cm
Surface area of combination = 2(surface area of one cone)
(Q both cones are same)
$=2(\pi r l)$
$=2\pi r \sqrt{r^2+h^2}$ $\left (Q1= \sqrt{r^2+h^2} \right )$
$= 2 \times 3.14 \times 8 \times \sqrt{64+225}$
$= 2 \times 3.14 \times 8 \times 17$
$=854cm^2$
Hence the surface area of the combination is 854 cm2

Question:8

Answer 396.18 cm3
Solution
Height of the tube = 21 cm
Base radius of the tube =3cm
Volume of tube $=\pi r ^2h$
$=\frac{22}{7}\times 3\times 3\times21=594cm^3$
Let the height of cone A is h cm
Height of cone $B=21-h cm$
Base radius of both A and B =3 cm
Volume of cone
$A=\frac{1}{3}\pi r^2h=\frac{1}{}3\pi (3)^2h$
$=3 \pi h$
Volume of cone
$B=\frac{1}{3}\pi r^2h=\frac{1}{}3\pi (3)^2(21-h)$
$=3 \pi (21-h)$
It is given that the ratio of the volume is 2: 1
$=\frac{3\pi h}{3 \pi (21-h)}=\frac{2}{1}$
$h=2(21-h)$
$h=42-2h$
$3h=42$
$h=14$
Height of cone A = 14 cm
Height of cone B =21-4=7 cm
Volume of cone $A=3 \pi h=3(3.14)(14)$
$=131.88cm^3$
Volume of cone $B = 3\pi (21-h)=2(3.14)(7)$
$=65.94cm^3$
The volume of remaining portion = Volume of the tube – the volume of cone A – the volume of cone B
$594-131.88 -65.94$
= 396.18 cm3
Volume of remaining portion = 396.18 cm3

Question:9

An ice cream cone full of ice cream having radius 5 cm and height 10 cm as shown in the Figure. Calculate the volume of ice cream, provided that its 1/8 part is left unfilled with ice cream.

Answer:

In this figure, there is a hemisphere of radius of 5 cm
And a cone of radius 5 cm and of height $10-5 = 5 cm$
Volume of cone
$=\frac{1}{3}\pi r^2 h=\frac{1}{3}\times 3.14 \times (5)^2\times 5=130.83cm^3$
Volume of hemisphere
$=\frac{2}{3}\pi r^3$
$=\frac{2}{3}\times 3.14 \times (5)^3=261.66cm^2$
The volume of complete figure = volume of cone + volume of the hemisphere
$= 130.83 +261.66 = 392.49 cm^3$
$\text{ The volume of the unfilled part }=\frac{392.49}{6}=65.41cm^3$
Volume of ice cream = volume of complete figure - volume of unfilled part
$392.49- 65.41=327.08 cm^3$
Hence the volume of ice cream is 327.08 cm3

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.

Answer 150
Solution
Given:- Diameter of marble = 1.4 cm
Diameter of beaker = 7cm
Diameter of marble = 1.4 cm
$\text{ Radius of marble}=\frac{1.4}{2}=0.7 cm$
The volume of 1 marble
$=\frac{4}{3} \pi r^3=\frac{4}{3}\pi (0.7)^3$
$=\frac{4}{3} \times 3.14 \times 0.343=1.43cm^3$
Diameter of beaker = 7 cm
$\text{ Radius of beaker}=\frac{7}{2}=3.5 cm$
Water level rises(h) = 5.6cm
Volume of water
$=\pi r^2 h=(3.14)(3.5)^2(5.6)$
$=215.40$
$\text{ Number of marbles required}=\frac{\text {volume of water }}{\text {volume of 1 marble }}$
$=\frac{215.40}{1.43}=150$
Hence 150 marbles should be dropped into the beaker.
So that the water level rises by 5.6 cm.

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.

Answer 1501
Solution
It is given that the length, breadth and height of rectangular solid is 66cm, 42cm and 21cm respectively.
Volume of solid rectangular lead piece = $l \times b \times h$
$66 \times 42 \times 21$
$=5821 cm^3$
The diameter of spherical lead shot =4.2
The radius of spherical lead shot
$=\frac{4.2}{2}=2.1 cm$
$\text{ The volume of lead shot }=\frac{4}{3}\pi r^3$
$=\frac{4}{3}\times 3.14 \times (2.1)^3=38.77 cm^3$
$\text{ Number of lead shot can be obtained}=\frac{\text {volume of lead piece}}{\text{volume of lead shot}}$
$=\frac{58212}{38.77}=1501$
Hence 1501 lead shot can be obtained from the lead piece of dimensions 66cm, 42cm and 21cm

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.

Answer 2542
Solution
Given : Diameter of spherical lead shot = 4 cm
Edge of cube = 44 cm
Volume of cube =a3
=443 (=44 cm)
=85184cm3
Radius of spherical lead shot =2cm
$\text{ Volume of lead shot}=\frac{4}{3}\pi r^3$
$=\frac{4}{3}\times 3.14 \times (2)^3=33.5$
$\text{ Number of lead shots}=\frac{\text {volume of cube}}{\text{volume of lead shot}}$
$=\frac{85184}{33.5}=2542$
Hence 2542 lead shots can be made out of a cube of lead whose edge measures 44 cm.

Question:13

A wall 24 m long, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions $25 cm \times 16 cm \times 10 cm.$ If the mortar occupies1/10 th of the volume of the wall, then find the number of bricks used in constructing the wall.

Answer 12960
Solution
Length of wall $=24 m =24 \times 100 =2400 cm$ ( because 1m = 100cm )
Breadth of wall $=0.4m =0.4 \times 100 =40 cm$
Height of wall $= 6m = 6 \times 100 = 600cm$
The volume of wall = length × breadth × height
$=2400 \times 40 \times 600$
$=5760000cm^3$
$\text{ Mortar occupied}=\frac{5760000}{10} =5760000cm^3$
Remaining volume = 57600000 – 5760000 = 51840000 cm3
Length of brick = 25 cm
Breadth of brick = 16cm
Height of brick = 10cm
Volume of brick = length × breadth × height
= 25 × 16 × 10 = 4000 cm3
$\text{Number of bricks}=\left [ \frac{\text{remaining volume} }{\text {volume of brick}} \right ]=\frac{5184000}{4000}=12960$
Hence the number of bricks 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.

Answer 450
Solution
Base diameter of disc = 1.5 cm
Radius of disc $=\frac{1.5}{2}=0.75cm$
Height of disc = 0.2 cm
Volume of disc $= \pi r^2 h = (3.14 )(0.75)^2(0.2)$
$= 0.353 cm^3$
Height of required cylinder = 10 cm
Diameter of required cylinder = 4.5cm
Radius of required cylinder $=\frac{4.5}{2}=2.25cm$
Volume of required cylinder $=\pi r^2 h$
$=3.14 \times (2.25)^2(10)$
$=158.96cm^3$
Number of disc $=\frac{\text{volume of required cylinder }}{\text{volume of disc}}$
$=\frac{158.96}{0.353}=450$
Hence 450 metallic disc are required to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

Question:1

A solid metallic hemisphere of radius 8 cm is melted and re casted into a right circular cone of base radius 6 cm. Determine the height of the cone.

Answer h = 28.44 cm
Solution
Radius of hemisphere $=8cm$
Volume
$\frac{2}{3}\times\pi r^3$
$\frac{2}{3}\times\frac{22}{7}\times8\times8\times8$
Radius of cone = 6 cm
Let height = h
Volume = $\frac{1}{3}\times\pi r^2h = \frac{1}{3}\times\frac{22}{7}\times6\times6\times h$
If hemisphere is melted and recast into a right circular cone.
Then, the volume of hemisphere = volume of the cone
$\frac{2}{3}\times\frac{22}{7}\times8\times8\times8 = \frac{1}{3}\times\frac{22}{7}\times6\times6\times h$

$2\times8\times8\times8 = \times6\times6\times h$
$\frac{2\times8\times8\times8 }{6\times6}= h$
$\frac{256 }{9}= h$
28.44 = h

Question:2

A rectangular water tank of base 11 m × 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.

Answer h = 8.579m
Solution
Length of cuboid = 11m
Breadth = 6m
Height = 5m
$Volume=1\times b\times h$
$=11\times6\times5 = 330m^3$
Radius of cylindrical tank = 3.5m
Let height = h
$\text{Volume }\pi r^2 h = \pi (3.5)^2 h$
To find the height of water level
Volume of cuboid = volume of cylindrical tank
$= 330= \pi (3.5)^2 h$
$h = \frac{330\times7\times100}{22\times35\times35} = \frac{600}{70} = 8.579m$

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 cm of iron weighs 7.5 g, find the weight of the box.

Answer 37867.5g or 37.867 kg
Solution
External Length = 36 cm
Breadth = 25 cm
Height = 16.5 cm
Volume $1\times b\times h$
= 36 × 25 × 16.5
$= 14850 cm^3$
Thickness of iron = 1.5 cm
Internal length $=36 -1.5-1.5$ {subtract border from both sides}
= 33 cm
Breadth $= 25-1.5-1.5 = 22cm$
Height $16.5-1.5-1.5=13cm$
Internal volume $1\times b\times h$
$33\times22\times13.5 = 9801cm^3$
Volume of iron = external volume – internal volume
$=14850-9801=5049cm^3$
Weight of one cubic cm of iron = 7.5 g
$\text{Weight of 5049 cm3 of iron }=5049\times7.5 = 37867.5g$

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 pen is used upon writing 3300 words on average. How many words can be written in a bottle of ink containing one-fifth of a litre?
$\\\text{Volume of the barrel}=\pi r^2h=\frac{22}{7}\times(0.25)^2\times7\\=1.375cm^3=\frac{1.375}{1000}L$
3300 words can be written with 0.001375 L of ink
So with 1L of ink 3300/0.001375 words can be written=2400000
So with 1/5th of a litter 2400000/5 words can be written=48000

Answer:

Mastermind of math class 10th

Question:5

Water flows at the rate of 10m/minute 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?

Answer:

$\text{The diameter of the conical vessel}= 40cm$
$Radius= \frac{40cm}{2} = 20cm$
$Height=24cm$
$Volume=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi\times20\times20\times24 = 3200\pi cm^3$
Diameter of water that flows out of cylinder = 5mm
$Radius=\frac{5}{10\times2}=\frac{5}{20}$
(Because 1 cm = 10mm)
$Height=10m=1000cm$ ( Because 1m =100cm)
$\text{Volume}=\pi r^2h=\pi\times\frac{5}{20}\times\frac{5}{20}\times1000\Rightarrow \frac{125}{2}\pi cm^3$
$\text{Time required}=\frac{\text {volume of conical flask}}{\text{volume of cylindrical water}}=\frac{3200\pi}{125\pi}\times2$
$\Rightarrow \frac{640\times2}{25}\Rightarrow \frac{1280}{25}=51.2\ minutes$

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 the heap?

Answer 74.18 m3, 67.11 m2
Solution
$\text{Height of cone}=3.5 m$
$Diameter=9 m$
$Radius=\frac{9}{2}= 4.5m$
$Volume=\frac{1}{3}\pi r^2h=\frac{1}{3}\times\frac{22}{7}\times(4.5)^{2}\times3.5=74.18m^3$
$\text{Canvas cloth required to covers the heap}=\pi r l$
Here l is slant height
$1=\sqrt{r^2+h^2}=\sqrt{(4.5)^2+(3.5)^2}\Rightarrow 4.75$
Curved surface area is $=\pi r l$
$=3.14\times4.5\times4.75 = 67.11m^2$
Hence 67.11m2 canvas cloth is required to just cover the 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 Rs 0.05 per dm2.

Answer Rs. 2250
Solution
Given
$\text{Circumference of base}=1.5cm$
$2\pi r=1.5cm$
$r=\frac{1.5}{2\times22}\times7$
$r=\frac{15\times7}{10\times44}=\frac{105}{440}=0.2386cm$
$\text{Radius }=0.2386cm$
$\text{Height}=25cm$
$\text{Curved surface area of one pencil}=2\pi r h$
$=2\times\frac{22}{7}\times0.2386\times25=\frac{262.46}{7}$
$=37.49cm^2$
$1cm=\frac{1}{10}dm$
$\Rightarrow \frac{37.49}{100}dm^2=0.3749dm^2$
$=0.375dm^2$$(approximately)$
$\text{Curved surface area of 120000 pencils}=0.375\times120000=45000dm^2$
Cost of coloring 1dm2 in one day =0.05 Rs
Cost of coloring 45000 dm2=0.05 x 45000
= 2250 Rs.

Question:8

Water is flowing at the rate of 15 km/h 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?

Answer 2 hours
Solution
Length of cuboidal pond = 50m
Breadth = 44 m
$Height=21 cm =\frac{21}{100}m$
$Volume=l \times b \times h$
$=50 \times 44 \times \frac{21}{100}$
Speed of water = 15 km per hour = 15000 m
$\text{ The diameter of the cylindrical pipe }=14 cm =\frac{14}{100}m$
$Radius =\frac{7}{100}m$
$Volume=\pi r^2 h$
$=\frac{22}{7}\times \frac{7}{100}\times \frac{7}{100}\times 15000 = 213m^3$
$\text{ Time required}=\frac{\text{volume of cuboid pond}}{\text{volume of cylindrical pipe}}$
$=\frac{462}{231}=2hours$

Question:9

A solid iron cuboidal block of dimensions 4.4 m × 2.6 m × 1m is recast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

Answer:

Answer 112m
Solution
Length of cuboidal block = 4.4m
Breadth = 2.6 m
Height = 1 m
Volume $=l \times b \times h$
$= 4.4 \times 2.6 \times 1$
$= 4.4 \times 2.6 \times m^3$
Radius of cylindrical pipe r1 = 30cm = 0.3m
r2 = 30+5 = 35cm = 0.35 m
Let, Height h1
Volume $=\pi r ^2 h$
$=\frac{22}{7}\times h_{1} \left ( (0.35)^2-(0.3)^2 \right )$
The volume of cuboid = volume of the cylindrical pipe
$=4.4 \times 2.6\times \frac{22}{7}\times h_{1}\left [ \left ( \frac{35}{100} \right )^2 -\left ( \frac{30}{100} \right )^2 \right ]$
$=\frac{44}{10} \times \frac{26}{10}\times \frac{22}{7}\times h_{1}\left [\frac{(35-30) (35+30)}{(100)^2} \right ]$ $\left (a^2-b^2)=(a-b)(a+b) \right )$
$h_{1}=\frac{44}{10} \times \frac{26}{10}\times \frac{7}{22}\times \frac{100 \times 100}{5 \times 65}$
$h_{1}=112m$

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 0.04m3?

Answer:

Length of cuboidal pond = 80m
Breadth = 50 m
Let, Height = h
$\text{ Volume}\ l \times b \times h$
$80 \times 50 \times h=400 h m^3$
Average water of one person =0.04m3
Average water of 500 persons = $0.04 \times 500$
According to question
The volume of cuboidal pond = Average water of 500 persons
$400 h =0.04 \times 500$
$h = 4 \times \frac{500 }{100\times 400}=\frac{5}{100}=0.05m$

Question:11

16 glass spheres each of radius 2 cm are packed into a cuboidal box of internal dimensions $16 cm \times 8 cm \times 8 cm$ and then the box is filled with water. Find the volume of water filled in the box.

Answer 487.6 cm3
Solution
Length of cuboidal box = 16 cm
Breadth = 8cm
Height = 8 cm
$\text{ Volume}=16 \times 8 \times 8$
Radius of sphere = 2 cm
$\text{ Volume of sphere }=\frac{4}{3}\pi r^3$
$=\frac{4}{3}\times \frac{22}{7}\times 2\times 2\times 2 =33.5238cm^3$
$\text{ The volume of 16 sphere}=16 \times 33.5238$
$=536.3808$
The volume of liquid = volume of the cuboidal box – the volume of 16 spheres
$=1024 -536.3808$
$=487.6cm^3$

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 Rs. 22 per litre which the container can hold.

Answer 230.10 Rs.
Solution
Upper radius of frustum of cone (R) = 20cm
Lower radius of frustum of cone r = 8cm
Height (h)= 16cm
Volume $\frac{\pi h}{3}[R^2+r^2+Rr]$
$\frac{22 \times 16}{7 \times 3}(20^2+8^2+20 \times 8)$
$\frac{22 \times 16}{21}(400 + 64+160)$

$\frac{22 \times 16}{21}(624)$
$\frac{22 \times 16\times 208}{7}=10459.428 m^3$
= 10.459 liter
Cost of 1 liter milk = 22 Rs.
Cost of 10.459 liter milk $22 \times 10.459$
= 2301.10 Rs

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.

Answer:

r = 36 cm, $l = 12\sqrt{13} cm$
Height of cylindrical bucket (h1) = 32 cm
Radius (r1) = 18 cm
Volume = $\pi r{_{1}}^{2}h_{1}$
$=\frac{22}{7}\times 18 \times 18 \times 32$
Height of conical heap (h2) = 24 cm
Let radius = r2
Volume $=\frac{1}{3}\pi r_{2}^{2}h_{2}=\frac{1}{3}\times \frac{22}{7} \times r_{2}^{2}\times 24$
According to question
Volume of cylindrical bucket = Volume of conical heap
$\frac{22}{7}\times 18 \times 18 \times 32 = \frac{1}{3}\times \frac{22}{7}\times r_{2}^{2}\times 24$
$=\frac{18 \times 18 \times 32 \times 3}{24}=r_{2}^{2}$
$=18^{2}\times 4 = r_{2}^{2}$
$r_{2}=\sqrt{18 \times 18 \times 2 \times 2}$
$r_{2}= 18 \times 2 = 36\; cm$
Slant height $(l)=\sqrt{r_{2}^{2}+h_{2}^{2}}$
$\sqrt{36^{2}+24^{2}}$
$=\sqrt{1296+576}$
$=\sqrt{1872}$
$\sqrt{12 \times 12 \times 13}$
$l=12\sqrt{13}cm$

Answer:

Diameter of cylinder = 6 cm
Radius = 3 cm
Height = 12 cm
$\text{ Surface are}$ $=2 \pi r h = 2\pi \times 3 \times 12 = 72 \pi cm^2$
Similarly radius of circle = 3 cm
$\text{ Area}$ $=\pi r^2=\pi 3 \times 3 = 9 \pi cm^2$
Slant height of cone (l)=5cm
Radius = 3 cm
$\text{ Surface area}=\pi r l =\pi \times 5 \times 3 = 15\pi cm^2$
Total surface area = area of cylinder + area of circle + area of the cone
$=72\pi + 9 \pi + 15 \pi$
$=96\pi$
$=96\times 3.14=301.7 cm^2$
Slant height (l) = 5cm
We know that,
$h^2+r^2=l^2$
$h^2+3^2=5^2$
$\\h^2= 25 -9 \\ h^2=16 \\\ h = 4$
$\text{ Volume of cylinder= }\pi r^2 h \\ = \pi \times 3 \times 3 \times 12 = 108 \pi cm^3$
$\\\text{ Volume of cone}=\frac{1}{3}\pi r^2 h \\ = \frac {1}{3} \pi \times 3 \times 3 \times 12 = 12 \pi cm^3$
$\text{ Volume of rocket }$$=180 \pi + 12 \pi$
$=120 \pi$
$=120 \times 3.14 =377.14 cm^3$

Answer:

Give: Volume of building
$41\frac{19}{21}=\frac{880}{21}$
Let total height above the floor = h
Hemisphere‘s diameter = h (given)
$Radius =\frac{h}{2}$
$Volume=\frac{2}{3}\pi r^3=\frac{2}{3}\pi \left ( \frac{h}{2} \right )^3$
Height of cylinder = total height – the height of hemisphere
$h-\frac{h}{2}=\frac{h}{2}$
Volume $\pi r^2 h=\pi \times \left (\frac{h}{2} \right )^2 \times \frac{h}{2} =\pi \left (\frac{h}{2} \right )^3$
According to question
The volume of building = Volume of cylinder + volume of the hemisphere
$\frac{880}{21}=\left ( \frac{h}{2} \right )^3\left [ \pi + \frac{2}{3} \pi \right ]$
$\frac{880}{21}=\left ( \frac{h}{2} \right )^3\left [ \frac{3\pi +2\pi }{3} \right ]$
$\frac{880 \times 2^3}{21}=h^3 \left [ \frac{5\pi }{3} \right ]$
$\frac{880 \times 2\times 2\times 2\times 7\times 3}{21\times 5\times 22}=h^3$
$h^3 = {2 \times 2 \times 2\times 2\times 2\times 2$
$h = \sqrt[3]{2 \times 2 \times 2\times 2\times 2\times 2}$
$h = 2 \times 2$
h=4m

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?

Answer:

The radius of hemispherical bowl = 9 cm
$Volume =\frac{2}{3}\pi r^3$ $=\frac{2}{3} \times \frac{22}{7}\times 9\times 9\times 9$
Radius of cylindrical bottle = 1.5 cm
Height = 4cm
$Volume$$=\pi r^2h$
$=\frac{22}{7}\times \frac{15}{10}\times \frac{15}{10}\times 4$
$\text{ Number of bottles needed}=\frac{\text {volume of hemispherical bowl}}{\text {volume of cylindrical bottle}}$
$=\frac{2}{3} \times \frac{22}{7}\times 9\times 9\times 9$
_________________
$=\frac{22}{7}\times \frac{15}{10}\times \frac{15}{10}\times 4$
$=\frac{2 \times 9 \times 9 \times \times 10 \times 10}{3 \times 15 \times 15 \times 4} =54 bottles$

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 of the cone.

Answer:

Height of cone = 120cm
Radius =60cm
$Volume =\frac{1}{3}\pi r^2 h$
$=\frac{1}{3}\pi( 60)^2 120$
$=452571.43cm^3$
Height of cylinder = 180
Radius = 60 cm (given that radius of the cylinder is equal to the radius of the cone)
$Volume=\pi r^2 h$
$=pi( 60)^2 180=2036571.43cm^3$
Volume of water left in the cylinder = volume of cylinder – volume of cone
2036571.43 – 452571.43
= 1584000 cm3
Or 1.584 m3 [Q 1m = 100cm]
2m3 (approximate)

Question:18

Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm/sec 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?

Answer:

Radius of cylindrical pipe = 1 cm
Height = 80 cm
$Volume=\pi r ^2 h$
$=\frac{22}{7} \times 1 \times 1 \times 80 = 251.4285 cm^3 /sec$
In half an hour volume of water is
$=251.4285 \times30 \times60 = 452571.5 cm^3$
Radius of cylindrical tank = 40 cm
Let height = h
Volume $=\pi r ^2 h$
$\frac{22}{7}\times (40)^2 h= 5828.5714 cm^3$
According to question
The volume of cylindrical pipe = volume of the cylindrical tank
$= 452571.5 = 5028.5714h$
$\frac{452571.5}{5028.5714} = h$
h =89.99
h = 90 cm (approximate)

Question:19

The rain water from a roof of dimensions 22 m × 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.

Answer:

The radius of the cylindrical vessel
$=\frac{2}{2}=1cm$
Height = 3.5 cm
$\text{ Volume}$ $=\pi r^2 h=\frac{22}{7}\times 1 \times 1 \times \frac{35}{10}=11m^3$
Let the height of rainfall = x
Length = 22m
Breadth = 20m
$Volume= l \times b \times h$
$= 22 \times 20 \times x$
Rainfall = volume of water = volume of the cylindrical vessel
$22 \times 20 \times x=11$
$x=\frac{11}{22 \times 20}=\frac{1}{40}=0.025 m$
x=0.025 m
Or x= 2.5cm [Q 1 m = 100 cm]
Hence the rainfall is in 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 dimension of the cuboid 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.

Answer:

Radius of conical depression = 0.5 cm
Depth = 2.1 cm
$Volume=\frac{1}{3}\pi r^2 h$
$=\frac{1}{3}\times \frac{22}{7} \times (0.5)^2 \times 2.1$
$=0.55cm^3$
$\therefore$the volume of 4 cones $=4 \times 0.55=2.2cm^3$
Edge of cube = 3
The volume of cube =33 ( Because the volume of cube = a3 )
Length of cuboid = 10 cm
Breadth = 5 cm
Height = 4 cm
$Volume$ $=l \times b \times h$
$=10 \times 5 \times 4=200 cm^3$
Volume of wood = volume of cuboid – volume of cube - volume of 4 cones
=200 - 2.2 -27 = 170.8 cm3

NCERT Exemplar Solutions Class 10 Maths Chapter 12 Important Topics:

NCERT exemplar Class 10 chapter 12 Maths solutions covers the following topics:

• Surface area and Volume up of frustum.
• Area and volume of multiple composite bodies made by known structures.
• NCERT exemplar Class 10 Maths solutions chapter 12 discusses the method to find out curved surface area and total surface area.

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NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

 Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry & Its Equations Chapter 9 Circles Chapter 10 Constructions Chapter 11 Areas related to Circles Chapter 13 Statistics and Probability

Features of NCERT Exemplar Class 10 Maths Solutions Chapter 12:

These Class 10 Maths NCERT exemplar chapter 12 solutions emphasise finding out curved surface area, total surface area, and volume of various three-dimensional objects such as cones, cylinders, and spheres. These calculations and learning will be helpful in Physics and Maths of higher classes and are essential for both medical and engineering aspirants. The expressive nature of these solutions provides the students a perfect environment to attempt and practice Surface Areas and Volumes based practice problems. The Class 10 Maths NCERT exemplar solutions chapter 12 Surface Areas and Volumes consist of plenty of practice problems, making the student prepared for NCERT Class 10 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths et cetera.

NCERT exemplar Class 10 Maths solutions chapter 12 pdf download is a free-to-use feature that provides the students with the facility to study the NCERT exemplar Class 10 Maths chapter 12 in an offline environment.

Check Chapter-Wise Solutions of Book Questions

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

Frequently Asked Question (FAQs)

1. How to find out the volume of an ice cream cone?

Ice cream cones can be seen as a combination of a hemisphere and a cone in simplified form. We know the volume of cone and volume of the hemisphere, so we can find out its complete volume.

2. How to find out the volume of a sharpened pencil?

We can find out the volume of sharpened pencil by assuming it as a combination of cylinder and a cone.

3. Is the chapter Surface Areas and Volumes important for Board examinations?

The chapter Surface Areas and Volumes is vital for Board examinations as it holds around 8-10% weightage of the whole paper.

4. What are the critical topics of Surface Areas and Volumes from a board examination perspective?

It is highly suggested that students practice and study every topic covered in NCERT exemplar Class 10 Maths solutions chapter 12 to score high in Surface Areas and Volumes.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.

Generally, passing 10th CBSE with a minimum percentage (often 50%) is the basic eligibility for diploma courses. Some institutes might have specific subject requirements depending on the diploma specialization.

There is a wide range of diploma courses available in various fields like engineering (e.g., mechanical, civil, computer science), computer applications, animation, fashion design, hospitality management, and many more.

You can pursue diplomas at various institutions like:

Hi,

Yes, you can get admission for a diploma course if you are a 10th passed out student. The basic eligibility criteria to get enrolled into a Diploma course is to secure minimum of 50% in your 10th boards. Although, several institutions can ask for specific subject requirements.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9
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