NCERT Exemplar Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes

In this chapter, we provide NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes pdf, free NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 13
Chapter NameSurface Areas and Volumes
CategoryNCERT Exemplar

NCERT Exemplar Class 9 Maths Solutions Surface Areas and Volumes

Exercise 13.1: Multiple Choice Questions (MCQs)

Question 1:
If the radius of a sphere is 2r, then its volume will be
ncert-exemplar-problems-class-9-maths-volume-surface-area-1
Solution:
(d) Given, radius of a sphere = 2r
Volume of a sphere =4/3  π(Radius)3
= 4/3 π(2r)3 = 4/3 π 8r3  
= (32 πr3)/3 cu units 
Hence the volume of a sphere is  (32 πr3)/3 cu units.

Question 2:
The total surface area of a cube is 96 cm2. The volume of the cube is
(a) 8 cm3
(b) 512 cm3
(c) 64 cm3      
(d) 27 cm3
Solution:
(c) Surface area of a cube = 96 cm2
Surface area of a cube = 6 (Side)2 = 96 => (Side)2 = 16
=> (Side) = 4 cm
[taking positive square root because side is always a positive quantity]
Volume of cube = (Side)3 = (4)3 = 64 cm3
Hence, the volume of the cube is 64 cm3.

Question 3:
A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is
(a) 4.2 cm
(b) 2.1 cm
(c) 2.4 cm
(d) 1.6 cm
Thinking Process

  1. Firstly, determine the volume of a cone and volume of sphere using formula, volume of sphere = 4/3 πr3 and volume of cone = 1/3 πr2h
  2. Further, equate volume of a cone and volume of sphere; so that we get the radius of sphere.

Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-2

Question 4:
In a cylinder, radius is doubled and height is halved, then curved surface area will be
(a) halved
(b) doubled
(c) same
(d) four times
Solution:
(e) Let the radius be r and height be h of a cylinder.,
∴ Curved surface area of cylinder = 2πrh
We have, radius = 2r, height = h/2
New curved surface area = 2π (2r) x (h/2) = 2 πrh
Hence, the curved surface area will be same.

Question 5:
The total surface area of a cone whose radius is r/2  and slant height 2l is
ncert-exemplar-problems-class-9-maths-volume-surface-area-3
Solution:

Question 6:
The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. The ratio of their volumes is
(a) 10:17
(b) 20:27
(c) 17:27
(d) 20:37
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-5

Question 7:
The lateral surface area of a cube is 256 m2. The volume of the cube is
(a) 512 m3
(b) 64 m3                       
(c) 216 m3
(d) 256 m3
Solution:
(a) Given, lateral surface area of a cube = 256 m2
We know that, lateral surface area of a cube = 4 x (Side)2
=> 256 = 4 x (Side)2
=> (Side)2 = 256/4 = 64
=> Side = √64 = 8 m
[taking positive square root because side is always a positive quantity]
Now, volume of a cube = (Side)3 = (8)3 = 8 x 8 x 8 = 512 m3
Hence, the volume of the cube is 512 m3.

Question 8:
The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is
(a) 1900
(b) 1920
(c) 1800
(d) 1840
Thinking Process

  1. Firstly, determine the volume of plank and volume of pit by using the formula = l x b x h
  2. Further, find the ratio of the volume of pit to the volume of the plank, to get the number of planks.

Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-6

Question 9:
The length of the longest pole that can be put in a room of dimensions (10m x 10m x 5m)is
(a) 15 m
(b) 16 m
(c) 10 m
(d) 12 m
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-7

Question 10:
The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratios of the surface areas of the balloon in the two cases is
(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 2 : 1
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-8

Exercise 13.2: Very Short Answer Type Questions

Write whether True or False and justify your answer
Question 1:
The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.
Solution:
True
Let the radius of the sphere and cylinder be r.
Given, height of the cylinder = diameter of the base  => h = 2r
According to the given condition,
Volume of sphere = (2/3) x Volume of cylinder
4/3 πr3 = (4/3) x πr2 x 2r
4/3 πr3 = 4/3 πr3
Hence, the volume of a sphere is equal to two-third of the volume of a cylinder.

Question 2:
If the radius of a right circular cone is halved and height is doubled, then volume will remain unchanged,
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-9

Question 3:
In a right circular cone, height, radius and slant height do not always be sides of a right triangle,
Solution:
True
On rotating a right-angled triangular lamina AOB about OA, it generates a cone. The point A is the vertex of a cone. Its base is a circle with centre O and radius OB. The length OA is the height of the cone and the length AB is called its slant height.
ncert-exemplar-problems-class-9-maths-volume-surface-area-10

Question 4:
If the radius of a cylinder is doubled and its curved surface area is not changed, the height must be halved.
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-11

Question 5:
The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere radius r.
Solution:
True
Given, edge of cube = 2r, then height of cube becomes h = 2r.
Volume of a cone = 1/3 πr2h = 1/3 πr2(2r)= 2/3 πr3
Volume of a hemisphere = 2/3 πr3
Hence, the volume of a cone is equal to the volume of a hemisphere.

Question 6:
A cylinder and a right circular cone are having the same base and same height. The volume of the cylinder is three times the volume of the cone.
Solution:
True
Let the radius of the base of a cylinder and a right circular cone be r and height be h. Then, Volume of a cylinder = πr2h
Volume of a cone = 1/3  πr2h
Volume of a cylinder = 3 x Volume of a cone
Hence, the volume of a cylinder is three times the volume of the right circular cone.

Question 7:
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is 1 : 2 : 3.
Solution:
True
Let radius of hemisphere is r.
Volume of a cone, V1= 1/3 πr2h
V= 1/3 πr2(r)   [∴h = r]
= 1/3 πr3
Volume of a hemisphere, V2=2/3 πr3
volume of cylinder, V3 = πr2h = πr2 x r = πr3 [∴ h=r]
V1 : V2 : V3 =1/2 πr3 :  2/3 πr3 : πr3 = 1 : 2 : 3
Hence, the ratio of their volumes is 1 : 2 : 3.

Question 8:
If the length of the diagonal of a cube is 6√3 cm, then the length of the edge of the cube is 3 cm.
Solution:
False
Given, the length of the diagonal of a cube = 6√3 cm
Let the edge (side) of a cube be a cm.
Then, diagonal of a cube =a√3
=> 6√3 = a√3
=> a = 6 cm
Hence, the edge of a cube is 6 cm.

Question 9:
If a sphere is inscribed in a cube, then the ratio of the volume of the cube to the volume of the sphere will be 6 : π.
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-12

Question 10:
If the radius of a cylinder is doubled and height is halved, the volume will be doubled.
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-13

Exercise 13.3: Short Answer Type Questions

Question 1:
Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. When 16 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. Give your answer to the nearest integer, [use π = 3.14]
Thinking Process

  1. Firstly, determine the volume of metallic sphere by using formula 4/3 πr3 and then multiply by 16.
  2. Secondly, determine the volume of internal rectangular box by using the formula = l x b x h
  3. Finally, substract the volume of 16 metallic sphere from volume of internal rectangular box, to get volume of preservative liquid.

Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-14

Question 2:
A storage tank is in the form of a cube. When it is full of water, the volume of water is 15.625 m3. If the present depth of water is 1.3 m, then find the volume of water already used from the tank.
Solution:
Let side of a cube be = x m
∴ Volume of cubical tank = 15.625 m3 [given]
=> x3 = 15.625 m3
=> x = 2.5 m
and present depth of water in cubical tank = 1.3 m
∴ Height of water used =2.5 – 1.3 m = 1.2 m
Now, volume of water used = 1.2 x 2.5 x 2.5= 7.5 m3
= 7. 5 x 1000 = 7500 L [∴ 1 m3 = 1000 L]
Hence, the volume of water already used from the tank is 7500 L.

Question 3:
Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water.
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-15

Question 4:
How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m?
Thinking Process

  1. Firstly, determine the slant height of conical test by using the formula,
    ncert-exemplar-problems-class-9-maths-volume-surface-area-16
  2. Further, using the formula for curved surface area =πrl which we get the required canvas.

Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-17

Question 5:
Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
Thinking Process
ncert-exemplar-problems-class-9-maths-volume-surface-area-18
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-19
ncert-exemplar-problems-class-9-maths-volume-surface-area-20

Question 6:
A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. If the glass is filled with milk upto an height, of 12 cm, find how many litres of milk is needed to serve 1600 students.
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-21

Question 7:
A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. How many revolutions did it make?
Thinking Process

  1. Firstly, determine the covered surface area of cylinder by using the formula 2πrh and surface area of cylinder covered in one revolution.
  2. Divide total area covered by area covered by area covered by cylindrical roller in one revolution to get number of revolutions.

Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-22

Question 8:
A small village, having a population of 5000, requires 75 L of water per head per day. The village has got an overhead tank of measurement 40 m x 25 m x 15 m. For how many days will the water of this tank last?
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-23

Question 9:
A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made?
Thinking Process

  1. Firstly, determine the volumes of laddoo and small laddoo by using the formula 4/3 πr3
  2. Further, find the quotient of volume of laddoo to the volume of small laddoo, to get the number of laddoos.

Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-24

Question 10:
A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the solid so formed.
Solution:
When a right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm, then solid formed is a cone whose height of a cone, h = 8 cm and radius of a cone, r = 6 cm. Slant height of a cone, l = 10 cm
Volume of a cone = 1/3 πr2h = (1/3) x (22/7) x 6 x 6 x 8
=> 6336/21 = 301.7 cm3
and curved surface of the area of cone = πrl
=> (22/7) x 6 x 10 = 1320/7 = 188.5 cm2
Hence, the volume and surface area of a cone are 301.7 cm3 and 188.5 cm2, respectively.

Exercise 13.4: Long Answer Type Questions

Question 1:
A cylindrical tube opened at both the ends is made of iron sheet which is 2 cm thick. If the outer diameter is 16 cm and its length is 100 cm, find how many cubic centimetres of iron has been used in making the tube?
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-25

Question 2:
A semi-circular sheet of metal of diameter 28 cm is bent to form an open conical cup. Find the capacity of the cup.
Thinking Process

  1. Firstly, determine the circumference of cone by using formula 2πR and circumference of semi-circle by using formula πr and equating them to get the radius of cone.
  2. Secondly, determine the height of cone by using the formula
    ncert-exemplar-problems-class-9-maths-volume-surface-area-26
  3. Further, determine the capacity of conical cup by using the formula,
    Capacity of cup = Volume of cup=1/3 πR2h

Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-27
ncert-exemplar-problems-class-9-maths-volume-surface-area-28

Question 3:
A cloth having an area of 165 m2 is shaped into the form of a conical tent of radius 5 m.

(i) How many students can sit in the tent, if a student on an average occupies 5/7 m2 on the ground?
(ii) Find the volume of the cone.

Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-29

Question 4:
The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kL of water. Water is pumped into the tank to fill to its capacity. Calculate the volume of water pumped into the tank.
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-30

Question 5:
The volumes of the two spheres are in the ratio 64 : 27. Find the ratio of their surface areas.
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-31

Question 6:
A cube of side 4 cm contains a sphere touching its sides. Find the volume of the gap in between.
Solution:
Given, side of a cubp = 4 cm
Side of cube = Diameter of sphere
4 = Diameter of sphere
ncert-exemplar-problems-class-9-maths-volume-surface-area-32

Question 7:
A sphere and a right circular cylinder of the same radius have equal volumes. By what percentage does the diameter of the cylinder exceed its height?
Solution:
ncert-exemplar-problems-class-9-maths-volume-surface-area-33

Question 8:
30 circular plates, each of radius 14 cm and thickness 3 cm are placed one above the another to form a cylindrical solid.
Find

  1. the total surface area.
  2. volume of the cylinder so formed.

Solution:
Given, radius of a circular plate, r = 14 cm
Thickness of a circular plate = 3 cm
Thickness of 30 circular plates = 30 x 3 = 90 cm
Since, 30 circular plates are placed one above the another to form a cylindrical solid. Then, Height of the cylindrical solid, h = Thickness of 30 circular plates = 90 cm

  1. Total surface area of the cylindrical solid so formed
    = 2πr(h+ r)=2 x (22/7) x 14(90+ 14)
    = 44 x 2 x 104 = 9152 cm2
    Hence, the total surface area of the cylindrical solid is 9152 cm2,
  2. Volume of the cylinder so formed = πr2h
    = (22/7) (14)2 x 90 =(22/7) x 14 x 14 x 90
    = 22 x 28 x 90 = 55440 cm3
    Hence, the volume of the cylinder so formed is 55440 cm3.

All Chapter NCERT Exemplar Problems Solutions For Class 9 maths

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All Subject NCERT Exemplar Problems Solutions For Class 9

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