## KSEEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Students can Download Kannada Chapter 3 Atoms and Molecules Questions and Answers, Summary, Notes Pdf, Siri KSEEB Solutions for Class 9 Science, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka Board Class 9 Science Chapter 3 Atoms and Molecules

### KSEEB Solutions for Class 9 Science Chapter 3 Intext Questions

Question 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid ➝ sodium ethanoate + carbon dioxide + water
Sodium carbonate + ethanoic acid ➝ sodium ethanoate + carbon dioxide + water
5.3 + 6 ➝ 8.2 + 2.2 + 0.9
= 11.3 g = 11.3 g
Weight of reactants is equal to weight of products. This observation is in agreement with the law of conservation of mass.

Question 2.
Hydrogen and Oxygen combine in the ratio or 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas. Because in water the ratio of the mass of hydrogen to the mass of oxygen is always 1 : 8.

Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
The relative number and kinds of atoms are constant in a given compound. This postulate is the result of the law of conservation of mass.

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Atoms combine in the ratio of small whole numbers to form compounds. This postulate explains the law of definite proportions.

Question 1.
Define the atomic mass unit.
One atomic mass unit is a mass unit equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12.

Question 2.
Why is it not possible to see an atom with naked eyes?
Atoms are very small, they are smaller than anything that we can imagine or compare with. Therefore it is not possible to see an atom with naked eyes.

Question 1.
Write down the formulae of:

1. sodium oxide
2. alluminium chloride
3. sodium sulphide
4. magnesium hydroxide

1. sodium oxide: Na2O
2. aluminium chloride: Al2Cl3
3. sodium sulphide : NaS
4. magnesium hydroxide: Mg(OH)2

Question 2.
Write down the names of compounds represented by the following formulae:

1. Al2(SO4)3
2. CaCl2
3. K2SO4
4. KNO3
5. CaCO3

1. Al2(SO4)3: Aluminium sulphate
2. CaCl2: Calcium chloride
3. K2SO4: potassium sulphate
4. KNO3: potassium nitrate
5. CaCO3: calcium carbonate

Question 3.
what is meant by the term chemical formula?
The chemical formula of a compound is a symbolic representation of its composition.

Question 4.
How many atoms are present in a
i) H2S molecule and
ii) PO43- ion?
i) H2S molecule:

ii) PO43- ion

Question 1.
Calculate the molecular masses of
H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
i) H2
= 2 × H
= 2 × 1
= 2u

ii) O2
= 2 × O
= 2 × 16
= 32u

iii) Cl2
= 2 × Cl
= 2 × 35.5
= 7lu

iv) CO2
= 1 × 12 + 2 × O
= 1 × 12 + 2 × 16
= 12 + 32
= 44u

v) CH4
= 1 × 12 + 4 × 1
= 12 + 4
= 16u

vi) C2H6
= 2 × 12 + 6 × 1
= 24 + 6
= 30u

vii) C2H4
= 2 × 12 + 4 × 1
= 24 + 4
= 28u

viii) NH3
= 14 × 1 + 3 × 1
= 14 + 3
= 17u

ix) CH3OH
= 12 × 1 + 3 × 1 + 16 × 1 + 1 × 1
= 12 + 3 + 16 + 1
= 32u

Question 2.
Calculate the formula unit masses of ZnO, Na2O, K2CO3 given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u and O = 16u.
i) ZnO
(Automic mass of Zn) + (Automic mass of O)
= 65 + 16 = 81u.

ii) Na2O
(2 × Automic mass of Na) + (Automic mass of O)
= 2 × 23 + 1 × 16
= 46 + 16
= 62u.

iii) K2CO3
(2 × Atomic mass of K)+ (Atomic mass of C) + (3 × Atomic mass of oxygen)
= 2 × 39 + 12 × 1 + 16 × 3
= 78 + 12 + 48
= 138u.

Question 1.
If one mole of carbon atoms weighs 12 gms, what is the mass (in gms) of 1 atom of carbon?
Number of moles = n
Given mass = m
molar mass = M
Given the number of particles = N
Avogadro number of particles = N0
i) Atomic mass of carbon = 12u.
Atomic mass of one mole of carbon=12g
But Atomic mass of carbon = 12 g mass of 6.022 × 1023 carbon atoms = 12g.

∴ Mass of carbon atoms
= 1.9926 x 10-23 gm.

Question 2.
Which has more number of atoms, 100 g, 100 gms of sodium or 100 gms of sodium or 100 gms of iron (given, the atomic mass of Na=23u, Fe = 56u).
Atomic mass of sodium = 23u (data)
It means the gram atomic mass of sodium = 23 gm
Now atoms present in 23 gm sodium = 6.022 × 1023

It means Number of atoms in 100 gm sodium = 1.6753 × 1024
∴ 100 gm sodium has more number of atoms rather than 100 gms of iron.

### KSEEB Solutions for Class 9 Science Chapter 3 Textbook Exercises

Question 1.
A 0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Mass of Boron = 0.096 g (Data)
Mass of Oxygen = 0.144g (Data)
Given mass = 0.24 g (Data)
∴ The percentage composition of Boron

∴ The percentage of O2 = $$\frac{0.144}{0.24}$$ × 100
= 60%.

Question 2.
When 3.00 g of carbon is burnt in 8.009 of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50,000 g of oxygen? Which law of Chemical combination will govern your answer?
3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.
But when 3 g of carbon is burnt in 50 g of oxygen, only 3 g of carbon reacts with 8 g of oxygen.
Remaining 42 gm of oxygen will not react.
11 gm of carbon dioxide is produced.
∴ Our answer obeys law of constant proportion.

Question 3.
What are polyatomic ions? Give examples.
A group of atoms carrying a charge is known as a polyatomic ion.

Question 4.
Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
(a) Magnesium chloride : MgCl2
(b) Calcium oxide : CaO
(c) Copper nitrate : Cu(NO3)2
(d) Aluminium chloride : AlCl3
(e) Calcium carbonate : CaCO3.

Question 5.
Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
(a) Quick lime: Calcium, carbon, oxygen
(b) Hydrogen bromide: Hydrogen, Bromine
(c) Baking powder: Sodium, Bicarbonate
(d) Potassium sulphate: Potassium, Sulphur, Oxygen.

Question 6.
Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus Molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid HNO3.
(a) Ethyne C2H2
Molar Mass = 2 × 12 + 2 × 1
= 24 + 2
= 26 g.

(b) Molar mass of Sulphur molecule
= 8 × 32
= 256 g.

(c) Molar Mass of Phosphorus molecule = 4 × 31
(Atomic mass of Phosphorus) = 124 g.

(d) Molar mass of Hydrochloric acid = HCl
= 1 + 35.5
= 36.5 g.

(e) Molar mass of HNO3
= 1 + 14 + 3 × 16
= 15 + 48
= 63 gm.

Question 7.
What is the mass of –
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of Aluminium = 27)
(c) 10 moles of Sodium Sulphite (Na2SO3)
(a) Mass of 1 mole of nitrogen= 14g.

(b) Mass of 4 moles of Aluminium
= 4 × 27
= 108 g.

(c) Mass of 10 moles of Sodium sulphite
= 10 × [2 × 23 + 32 + 3 × 16]
= 10 × 126
= 1260 gm.

Question 8.
Convert into mole:
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
(a) 32gm of oxygen means 1 mole
12gm oxygen means $$\frac{12}{32}$$ mole
∴ 12g of oxygen gas = 0.375 mole
(b) 18 gm of water means = 1 mole
20 gm of water means $$\frac{20}{18}$$ mole = 1.11 mole
(c) 22 g of carbon dioxide means
$$\frac{22}{44}$$ = 0.5 mole.

Question 9.
What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
(a) Mass of 1 mole of oxygen = 16 g
Mass of 0.2 mole of oxygen = 0.2 × 16
= 3.2 g.
(b) Mass of 1 molecule of water= 18 gm.
Mass of 0.5 mole of water = 0.5 × 18
= 9 gm.

Question 10.
Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
One mole of Sulphur (S) = 8 × 32
= 256 gm.
256 g of solid sulphur = 6.022 × 1023 molecules.

= 3.76 × 1022
(Approximate)

Question 11.
Calculate the number of aluminium ions present in 0.0519 of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u).
One mole of Aluminium oxide
= 2 × 27+ 3 × 16
= 54 + 48
= 102 g.
102 g of Al2O3 = 6.022 × 1023 moles (aluminium oxide)

It means Aluminium present in 0.051 gm
= 3.011 × 1020 Aluminium oxide molecules
Number of Al ions in one mole of Al2O3 = 2
∴ Number of Al Ions In 3.011 × 1026 molecules
0.051 Al2O3
= 2 × 3.011 × Number of Al ions In 3.011 × 1020
= 6.022 × 1020.

### KSEEB Solutions for Class 9 Science Chapter 3 Additional Questions

Question 1.
Write the symbols of the following:
a) Iron
c) Zinc
d) Oxygen
e) Chlorine
a) Iron = Fe
c) Zinc = Zn
d) Oxygen = O
e) Chlorine = Cl

Question 2.
What is a molecule?
The smallest particle of an element or a compound that is capable of independent existence and shows all the properties of that substance.

Question 3.
What is an ion?
An ion is a charged species present in metals and non-metals.

Question 4.
What is molecular mass?
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance.

Question 5.
What is a mole?
One mole of any species (atoms, molecules, ions, or particles) is that quantity in number having a mass equal to its atomic or molecular mass in grams.

### KSEEB Class 9 Science Atoms and Molecules Additional Questions and Answers

Question 1.
An element Z forms an oxide with formula Z2O3. What is its valency?
Valency is 3+

Question 2.
Mention the elements present in (1) quick line (2) sodium hydrogen carbonate.,
(1) Quick line (calcium oxide) (CaO) element present are calcium and oxygen.
(2) Sodium hydrogen carbonate (NaHCO3) elements are sodium, hydrogen, hydrogen, carbon and oxygen.

Question 3.
Calculate the total number of ions in 0.585 g of sodium chloride.
Gram formula mass of NaCl = 23 + 35-5 = 58.5 g
58.5 g of NaCl have ions = 2 × NA
58.5 g of NaCl have ions
$$=2 \times \mathrm{N}_{\mathrm{A}} \times \frac{0.585}{58 \cdot 5}=0 \cdot 02 \times \mathrm{N}_{\mathrm{A}}$$
= 0.02 × 6.022 x 1023
= 1.20 × 1022 ions

## KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.4.

## Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.4

Question 1.
The following number of goals were scored by a team in a series of 10 matches :
2, 3, 4, 5, 0, 1. 3, 3, 4, 3
Find the mean, median and mode of these scores.
Solution:
Number of goals scored by a team:
(i) 2, 3, 4, 5, 0, 1, 3, 3, 4, 3

= 2.8

(ii) If number of goals are arranged in ascending order,
0, 1, 2, 3, 3. 3, 3, 4, 4. 5
No. of observations =10 (even number)

= 5th observation.
= 3

= 5 + 1
= 6th observation ➝ 3

= 3

(iii) Mode is the value of the observation which occurs most frequently.
Here 3 is repeated 4 times.
∴ Mode = 3.

Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded : 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Solution:
In a mathematics test given to 15 students, the following marks (out of 100) are recorded :
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60

Mode= 54.8

(ii) When marks are arranged in ascending order,
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Number of observations = 15 (odd number)

∴ Median = 52.

(iii) Mode: It is the repeated value.
Here 52 is repeated three times.
∴ Mode = 52.

Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63. find the value of x,
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Median = 63.
Number of observations = 10 (even number)

∴ x + 1 = 63 ∴ x = 63 – 1
∴ x = 62.

Question 4.
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Solution:
14, 25, 14, 28, 18, 17, 18, 14, 23, 22; 14, 18
Mode = ?
Arranging the scores in ascending order,
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Mode is the repeated value.
∴ Here 14 is repeated 4 times.
∴ Mode = 14.

Question 5.
Find the mean salary of 60 workers of a factory from the following table :

Solution:

∴ The men salary of workers is Rs. 5083.33.

Question 6.
Give one example of a situation in which
(i) The mean is an appropriate measure of central tendency.
(ii) The mean is not an appropriate measure of central tendency but the median is appropriate of central tendency.
Solution:
(i) Each score is in lesser difference, it is easy to calculate the average than the median.
Ex: monthly salary of 5 persons, 10000, 10100, 10200, 10300, 10400
Mean of the scores is 10200.

(ii) If scores have more differences, it is easy to calculate the median than the mean.
Ex.: Marks obtained by 7 students in Mathematics :
2, 10, 20, 15, 4, 23, 3
Ascending Order : 2, 3, 4, 10, 15, 20, 23
Median =10 but the mean is 11.

We hope the KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.4, drop a comment below and we will get back to you at the earliest.

## KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.3.

## Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.3

Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 – 44 (in years) worldwide, found the following figures (in %).

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Solution:
(i) Scale: x axis ➝ 1 horizontal axis = 1 cm.
y axis ➝ 5% = 1 cm.

(ii) Reproductive health conditions.
(iii) Nerve and skin.

Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys In different sections of Indian society is given below.

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Solution:
(i) Scale: x axis ➝ 1 horizontal = 1 cm.
y axis ➝ 100 girls = 1 cm.

(ii) After verification we came to know that Strength of Girls is more in ST section and it is less in Urban section.

Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections :

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats ?
(ii) A Political Party won the maximum number of seats.
Solution:
(i) Scale: x axis ➝ 1 horizontal(Party) = 1 cm.
y axis ➝ 10 Parties = 1 cm.

(ii) ‘A’ political Party won the maximum number of seats.

Question 4.
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is there any other suitable graphical representation for the same data ?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long ? Why ?
Solution:
(i) Scale: x axis ➝ 1 Class interval = 1 cm.
y axis ➝ 2 leaves = 1 cm.

(ii) Frequency Polygon.
(iii) No. Because from 145 to 153, length of leaves is 153.

Question 5.
The following table gives the life times of 400 neon lamps:

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours ?
Solution:
(i) Scale: x axis ➝ 1 Class interval = 1 cm.
y axis ➝ 10 Bulbs = 1 cm.

(ii) 184 lamps have a life time of more than 700 hours.

Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them :

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Solution:
(i) Scale: x axis ➝ 5 marks = 1 cm.
y axis ➝ 2 frequency = 1 cm.

Section B students scored less marks in great numbers.

Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below :

Represent the data of both the teams on the same graph by frequency polygons.
(Hint: First make the class intervals continuous.)
Solution:
Class interval is continuous,
1 – 6, 7 – 12 Here the difference is 1.
∴ $$\frac{1}{2}$$ = 0.5 is lower limit is subtracted,
0.5 is taken upper limit, it becomes
0.5 – 6.5
6.5 – 12.5.

Scale: x-axis ➝ 3.5 mark = 1 cm.
y – axis ➝ 1 frequency = 1 cm

Question 8.
A random survey of the number of children of various age groups playing in a park was found as follows :

Draw a histogram to represent the data above.
Solution:
Scale: x-axis ➝ 1 year = 1 cm.
y – axis ➝ 1 child = 1 cm.

Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows :

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie
Solution:
(i) Scale: x-axis ➝ 1 letter = 1 cm.
y – axis ➝ 4 surnames = 1 cm.

(ii) Class interval which has maximum number of surnames is : 6 – 8.

We hope the KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.3, drop a comment below and we will get back to you at the earliest.

## Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.6

(Assume π = $$\frac{22}{7}$$ unless stated otherwise)

Question 1.
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold ? (1000 cm3 = 1 l)
Solution:
The circumference of the base of cylindrical vessel, C = 132 cm.
height, h = 25 cm
C = 2πr = 132

r = 21 cm
∴ Volume of Cylinder, V = πr2h

= 34650 cm3
1000 cubic cm. = 1 litre
∴ 34650 cm3 = … ? …

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Solution:
Let the inner diameter of a cylindrical vessel be d1 and outer diameter be d2.

height of pipe, h = 35 cm.
∴ volume of pipe, V = $$\pi r_{2}^{2} h-\pi r_{1}^{2} h$$

= 110 × 52
∴ V = 5720 cm3
Mass of pipe of 1 ccm is 0.6 gm.
∴ Mass of 5720 cm3. … ? …
= 5720 × 0.6
= 3432 Kg.

Question 3.
A soft drink is available in two packs –
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.
Which container has greater capacity and by how much ?
Solution:
Length of rectangular tin, l = 5 cm,
height, h = 15 cm.
∴ Volume of rectangular tin, V = l × b × h
V = 5 × 4 × 15
= 300 cm3
Diameter of plastic cylinder, d = 7 cm,
∴ r = $$\frac{7}{2}$$
Volume of cylinder, v = $$\pi \mathrm{r}^{2} \mathrm{h}$$

= 11 × 7 × 5
= 385 cm3.
∴ Plastic cylinder’s volume is great.
= Volume of cylindrical tin – Volume of rectangular tin.
= 385 – 300
= 85 cm3.
∴ Volume of plastic tin is greater than rectangular tin by 85 cm3.

Question 4.
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find
(ii) its volume. (Use π = 3.14)
Solution:
Curved surface area of cylinder = 94.2 cm2
height, h = 5 cm.
r = ?
V = ?
(i) Curved Surface area of cylinder, L.S.A.
A= 2πrh
94.2 = 2 × 3.14 × 5 × r

= 3.01 cm.
= 3 cm.
(ii) Volume of cylinder, V = πr2h
= 3.14 × (3)2 × 5
= 3.14 × 9 × 5
= 141.3 cm3.

Question 5.
It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs. 20 per m2, find
(i) inner curved surface area of the vessel.
(iii) capacity of the vessel.
Solution:
(i) Cost of painting for cylindrical vessel = Rs. 2200
Cost of painting is at the rate of Rs. 20 per m2.
∴ Curved Surface area of vessel = $$\frac{2200}{20}$$
= 110 m2

(ii) Depth of vessel, h = 10 m
r = ?
Curved surface area of vessel, V = 2πrh

(iii) Capacity of the Vessel = Volume of Vessel
V = πr2h

V = 96.25 m3.

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it ?
Solution:
Volume of a cylindrical vessel = 15.4 litres
1000 c.cm. = 1 litre.

height, h = 1 m,
Curved surface area =?
Volume of cylinder, V = πr2h
0. 0154 = $$\frac{22}{7}$$ × r2 × 1

r = 0.7 m
Now, curved surface area of cylinder vessel,

= 0.44 × 1.07
A = 0.4708 m2

Question 7.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Solution:
Diameter of pencil, d = 7 mm = 0.7 m.

length of pencil, h = 14 cm
diameter of pencil, d = 1 mm = 0.1 cm

Volume of wood in pencil, V

Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:
Diameter of pencil, d = 7 cm

height of soup, h = 4 cm
Volume of soup given for 1 patient is 154 cm3
Volume of soup given for 250 patients .. ? ….
= 154 × 250 = 38500 cm3
For 1000 cm3, 1 litre
For 38500 cm3 ….. ? …
= $$\frac{38500}{1000}$$
= 38.5 litres soup.

## KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.2.

## Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The blood groups of 30 students of Class VIII are recorded as follows :
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B. O
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?
The blood groups of 30 students of class VIII are as follows

i) Out of 30 students only three students have blood group AB. This is rarest blood group.
ii) Most common blood group students are 12. Hence ‘O’ is the maximum blood group.

Question 2.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows :

Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0 – 5 (5 not included). What main features do you observe from this tabular representation ?
Solution:
Size of the class interval = 5
First class interval is 0 – 5 (5 is not included).

More number of people, i.e. 5 to 15 Km come.

Question 3.
The relative humidity (in %) of a certain city for a month of 30 days was as follows:

(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.
(ii) Which month or season do you think this data is about ?
(iii) What is the range of this data ?
Solution:
Size of class interval = 2
(i) Frequency distribution table :

(ii) Relative humidity is more, these data are taken during rainy season.
(iii) Maximum humidity = 99.2
Minimum humidity = 84.0
∴ Range = Maximum – Minimum
= 99.2 – 84.9
= 14.3

Question 4.
The heights of 50 students, measured to the nearest centimetres, have been found to be as follows :

(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170, etc.
(ii) What can you conclude about their heights from the table ?
Solution:
Maximum height : 150 cm.
Minimum height : 173 cm.
Size of class interval = 5
(i) Frequence distribution table for grouped data:

(ii) Referring to above table, 50% students have height less than 165 cm.
Height of majority students is 160 – 165 cm.

Question 5.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per millon ?
Solution:
(i) Grouped frequency distribution table:

(ii) In 8 days the concentration of sulphur dioxide is more than 0.11 parts per million.

Question 6.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows :

Prepare a frequency distribution table for the data given above.
Solution:
Frequency distribution table :

Question 7.
The value of n upto 50 decimal places is given below :
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits ?
Solution:
Frequency distribution table :

(ii) Digits which have more frequency are 3 and 9.
‘0’ is the digit which has less frequency.

Question 8.
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows :

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 – 10.
(ii) How many children watched television for 15 or more hours a week ?
Solution:
(i) Size of Class Interval = 5

(ii) Number of children watched TV for 15 days or more hours a week is 2.

Question 9.
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows :

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Solution:
Size of the class interval = 0.5
Class interval : 2 – 2.5, 2.5 – 3 …….
Grouped Frequency distribution table:

We hope the KSEEB Solutions for Class 9 Maths Chapter 15 Statistics Ex 14.2 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 15 Statistics Exercise 14.2, drop a comment below and we will get back to you at the earliest.

## Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.5

Question 1.
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?
Solution:
Match box is in the form of cuboid.
Its length, l = 4 cm.
height, h = 1.5 cm.
∴ Volume of Cuboid, V = l × b × h
= 4 × 2.5 × 1.5
V = 15 cm3.
Volume of 1 match box is 15 cm3.
Volume of 12 match boxes …?… .
= 15 × 12
= 180 cm3.

Question 2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold ? (1 m3 = 1000 l)
Solution:
Length of cuboidal water tank, l = 6 m
height, h =4.5 m
∴ Volume of cuboidal water tank, V= l × b × h
= 6 × 5 × 4.5
V= 135 m3.
1 m3 = 1000 l.
135 m3 = ? = 135 × 1000
= 135000 lit.
∴ Number of litres of water tank = 1350000 litres.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid ?
Solution:
Volume of cuboidal vessel, V = 380 m3.
length, l = 10 m
breadth, b = 8 m height, h = ?
Volume of vessel, V= l × b × h
380 = 10 × 8 × h
380 = 80 h

∴ h = 4.75 m.

Question 4.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m3.
Solution:
Length of cuboidal pit, l = 8 m
breadth, b = 6 m height, h = 3 m
∴ Volume of pit, V = l × b × h
= 8 × 6 × 3
= 144 m3.
∴ Cost of digging a cuboidal 1 m3 is Rs. 30
Cost of digging a cuboidal pit for 144 m3 … ? …
= 144 × 30
= Rs. 4320.

Question 5.
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Solution:
Volume of cuboidal tank, V = 50000 lit.
1 c.c = 1000 litres
∴ Volume of tank = $$\frac{50000}{1000}$$ = 50 c.c
length of tank, l = 2.5 m
height, h = 10m
Volume of cuboid, V = l × b × h

∴ b = 2m.

Question 6.
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last ?
Solution:
Length of cuboid water tank, l = 20 m
height, h = 6 m
Volume, V = ?
Volume of water tank, V = l × b × h
= 20 × 15 × 6
= 1800 m3.
1m3 = 1000 litres
∴ 1800m3 = 1800 x 1000
= 1800000 litres.
Water required daily per man= 150 litres
∴ Quantity of water required for 4000 men?
= 4000 × 150
= 600000 litres.
Quantity of water for 1 day = 600000
Number of days for consuming 1800000 litres … ? …

= 3 Days.

Question 7.
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown
Solution:
Length of rectangular godown, l1 = 40 m
height, h1 = 10 m
∴ Volume of godown, V= l1 × b1 × h1
= 40 × 25 × 10
= 10000 m3.
Length of wooden crate, l2 = 1.5 m
height, h2 = 0.5 m
∴ Volume of wooden crate. V = l2 × b2 x h2
= 1.5 × 1.25 × 0.5
= 0.9375 m3.
For 0.9375 m3, 1 crate

= 10666.66

Question 8.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Solution:
Each side of a cube, a = 12 cm.
∴ Volume of cube, V = a3
= (12)3
= 1728 cm3.
Solid cube is cut into 8 cubes of equal volume.
Volume of each small cube,

∴ l3 = 216
∴ l = 6 cm.
∴ Each side of small cube is 6 cm.
Ratio of outer area :

Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Solution:
Depth of river, h = 3m
1 Km. = 1000 m
∴ 2 Km. = 2000 m.
Speed of the river per hour is 2 Km.
For 60 minutes, 2000m
For 1 minute …?… = $$\frac{2000}{60}$$

∴ Depth of river in 1 minute, l = $$\frac{100}{3}$$ m.
Volume of water flows in 1 minute,
V = l × b × h
= $$\frac{100}{3}$$ × 40 × 3
= 4000 m3
∴ In One minute, 4000 m3 water reaches the sea.

## Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.4

(Assume π = $$\frac{22}{7}$$, unless stated otherwise)

Question 1.
Find the surface area of a sphere of radius :
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm.
Solution:

Surface area of the sphere = $$4 \pi r^{2}$$

= 22 × 3 × 21
= 1386 cm2.

(ii) r = 5.6 cm. = $$\frac{56}{10}$$cm.
Surface area of the sphere = $$4 \pi r^{2}$$

= 394.24 cm2.

(iii) r = 14 cm.
Surface area of the sphere = $$4 \pi r^{2}$$

= 88 × 28
= 2464 cm2.

Question 2.
Find the surface area of a sphere of diameter :
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m.
Solution:
(i) Diameter, d = 14 cm.

Surface area of a sphere = $$4 \pi r^{2}$$

= 616 cm2.
(ii) Diameter, d = 21 cm.

Surface area of a sphere = $$4 \pi r^{2}$$

= 1386 cm2.
(iii) Diameter, d = 3.5 m. = $$\frac{7}{2}$$

Surface area of a sphere = $$4 \pi r^{2}$$

= 38.5 m2.

Question 3.
Find the total surface area of a hemisphere of radius 10 cm.
(Use π = 3.14 )
Solution:
r = 10 cm.
Total surface area of a hemisphere = $$3 \pi r^{2}$$

= 942 cm2.

Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
Radius of spherical balloons be r1 and r2, r1 = 7 cm. and r2 = 14 cm.
Ratio of surface area

∴ Ratio of surface areas of the balloons = 1 : 4.

Question 5.
A hemispherical bowl made of brass has an inner diameter of 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm2.
Solution:
Inner diameter of hemispherical bowl = 10.5 cm.

Curved surface area of the bowl = $$2 \pi r^{2}$$

= 173.25 cm2.
Cost of tin-plating for 100 cm2 is Rs. 16.
Cost of tin-plating for 173.25 sq.cm. …?…

= Rs. 27.72.

Question 6.
Find the radius of a sphere whose surface area is 154 cm2.
Solution:
Total surface area of a sphere= 154 cm2

∴ r = 3.5 cm

Question 7.
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution:
Let the diameter of earth be d’ unit
∴ Diameter of Moon = $$\frac{d}{4}$$
radius of the earth = $$\frac{\mathrm{d}}{2}$$
radius of the moon = $$\frac{1}{2} \times \frac{\mathrm{d}}{4}=\frac{\mathrm{d}}{8}$$
Ratio of surface area,

= 1 : 16

Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
Thickness of the hemispherical bowl = 0.25 cm.
inner radius, r = 5 cm.
∴ Outer Surfce area= 5 + 0.25 = 5.25 cm.
Outer Curved Surface area = 2πr2

= 173.25 cm2.

Question 9.
A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder
(iii) ratio of the areas obtained in (i) & (ii).

Solution:
(i) If radius of a sphere is ‘r’ cm, its Surface area = 4πr2
(ii) height of cylinder, h = diameter of sphere
h = r + r
∴ h = 2r
∴ Curved surface area of cylinder = 2πrh
= 2πr × h
= 2πr × 2r
= 4πr2

= 1 : 1

## KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1

KSEEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 14 Statistics Exercise 14.1.

## Karnataka Board Class 9 Maths Chapter 14 Statistics Ex 14.1

Question 1.
Give five examples of data that you can collect from your day-to-day life.
Some of the examples of data that we can collect from our day-to-day life are as follows :

1. Number of TV viewers in the city.
2. Number of Colleges in the city.
3. Number of sugar factories in the city.
4. Measuring the height of students in the classroom.
5. Number of children below 15 years in India.

Question 2.
classify the data in Q.1 above as primary or secondary data.
i) Secondary Data
Because in example (iv) to find our heights of students, the investigator is in contact with the student.
ii) E.g., (i), (ii), (iii) and (v) are primary data. Because these are information collected from a source.

We hope the KSEEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 15 Probability Exercise 15.1, drop a comment below and we will get back to you at the earliest.

## KSEEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1

KSEEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 15 Probability Exercise 15.1.

## Karnataka Board Class 9 Maths Chapter 15 Probability Ex 15.1.

Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Total number of hittings of ball = 30
∴ n(S) = 30.
Total number of boundary = 6
Total number of No boundaries = 30 – 6 = 24
∴ n(E) = 24
Probability of No boundary hits =

Question 2.
1500 families with 2 children were selected randomly, and the following data were recorded :

 Number of girls in a family 2 1 0 Number of families 475 814 211

Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl,
(iii) No girl
Also check whether the sum of these probabilities is 1.
Solution:
Total number of families =1500
(i) Family with 2 girls = 475
∴ P1(Family with 2 girls)

(ii) Family with 1 girl = 814
∴ P2 (Family with 1 girl)

(iii) Family without girls = 211
∴ P3 (Family without girls)

Sum of probabilities

∴ Sum of probabilities is 1.

Question 3.
Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.
Solution:
Total number of students in IX class = 40
∴ n(S) = 40
Number of students born in August = 6
∴ n(E) = 6

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes :

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Solution:
Total number of events tossing 3 coins simultaneously = 200
∴ n(S) = 200
Frequency in which two heds up = 72
∴ n(E) = 72

Question 5.
An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below :

 Monthly Income (in Rs.) Vehicles per family 0 1 2 Above 2 Less than 7000 10 160 25 0 7,000 – 10,000 0 305 27 2 10,000- 13,000 1 535 29 1 13,000- 16,000 2 469 59 25 16,000 or more 1 579 82 88

Suppose a family is chosen. Find the probability that the family chosen is
(i) earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs. 7000 per month and does not own any vehicle.
(iv) earning Rs. 13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
Total number of families surveyed = 2400
∴ n(S) = 2400
(i) earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles = 29
∴ n(A) = 29

(ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579
∴ n(B) = 579

(iii) earning less than Rs. 7000 per month and does not own any vehicle =10
∴ n(C) = 10

(iv) earning Rs. 13000 – 16000 per month and owning more than 2 vehicles = 25
∴ n(D) = 25

(v) owning not more than 1 vehicle 10 + 0 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579 = 2062
∴ n(E) = 2062

Question 6.
Refer to Table 14.7, Chapter 14.
(i) Find the probability that a student obtained less than 20% in the Mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Solution:
(i) Number of students obtained less than 20% in the Mathematics test = 7
n(A) = 7
Total number of students = 90
n(S) = 90

(ii) Number of students obtained marks 60 or above
= 15 + 8 = 23 ∴ n(B) = 23
Total Number of students = 90
n(S) = 90

Question 7.
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table:

 Opinion No. of students like dislike 135 65

Find the probability that a student chosen at random.
(i) likes statistics,
(ii) does not like it.
Solution:
Total number of students = 135 + 65
= 200
n(S) = 200
(i) Number of students who like Statistics
= 135
n(A) = 135

(ii) Number of students who does not like Statistics = 65
n(B) = 65

Question 8.
Refer to Q. 2, Exercise 14.2. What is the empirical probability that an engineer lives :
(i) less than 7 km from her place of work ?
(ii) more than or equal to 7 km from her place of work ?
(iii) within $$\frac{1}{2}$$ km from her place of work.
Solution:
Referring to Q.2 Exercise 14.2
(i) Total Number of engineers = 40
n(S) = 40
Number of engineers living less than 7 km from her place of work = 9
n(A) = 9

(ii) Number of engineers living more than or equal to 7 km from her place of work.
= 40 – 9 = 31
n(B) = 31

(iii) Number of engineers living within $$\frac{1}{2}$$ Km from her place of work = 0
n(C) = 0

Question 9.
Activity: Note the frequency of two-wheelers, three-wheelers and four- wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.

Question 10.
Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

Question 11.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg.)
4.97 5.05 5.08 5.03 5.00 5.06
5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:
Total number of bags of wheat flour = 11
n(S) = 11
Number of bags which have flour more than 5 kg. = 7
n(A) = 0

Question 12.
In Q. 5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12- 0.16 on any of these days.
Solution:
In Q.5, Exercixe 14.2 of the Text Book,
Number of days recorded = 30
n(S) = 30
Concentration of Sulphur Dioxide (SO2) in the interval 0.12 – 0.16 = 2
n(A) = 2

Question 13.
In Q.1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Solution:
In Q. 1, Exercise 14.2 of the Text Book,
Total Number of students = 30
n(S) = 30
Number of students having Blood group AB = 3
n(A) = 3

We hope the KSEEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 15 Probability Exercise 15.1, drop a comment below and we will get back to you at the earliest.

## Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3

(If values are not given for ‘n’ Assume n = $$\frac{22}{7}$$.)

Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
Diameter of the base of a cone, d = 10.5 cm.
height h = 10 cm.
Curved Surface Area, C.S.A. = ?
d = 10.5 cm, $$=10 \frac{1}{2}=\frac{21}{2}$$ cm.
h = l = 10 cm.
∴ Curved Surface Area of a cone = πrl

Question 2.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Diameter of base of a cone, d = 24 m.
height, h = 21 m.
Total Surface Area, T.S.A. = ?
d = 24 m, ∴ r = $$\frac{\mathrm{d}}{2}=\frac{24}{2}=12$$
h = l = 21 m.
T.S.A of a cone = πr(r + l)

= 1244.57 m2

Question 3.
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base, and
(ii) total surface area of the cone.
Solution:
If Curved surface area of a cone, πrl = 308 cm2.
height, h = l = 14 cm.
Then, (i) r = ?, (ii) T.S.A. = ?
(i) CSA of a Cone = πrl

∴ r = 7 cm
(ii) TSA of a cone= πr(r + l)

= 22 × 21
= 462 cm2

Question 4.
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs. 70.
Solution:
(i) ABC is a conical tent,
height, h = 10 m.
Slant height of the tent, l = ?

In ⊥∆AOC, ∠O = 90°
As per Pythagoras theorem,
AC2 = AO2 + OC2
l2 = (01)2 +(24)2
= 100 + 576
l2 = 676
l = $$\sqrt{676}$$
Area of canvas required to prepare tent, C.S.A. = ?
C.S.A.= πrl

Cost of 1 m2 canvas is Rs. 70,

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14.)
Solution:
ABC is a conical tent.
Height, AO = h = 8 m.
base radius, OC = r = 6 cm
In ⊥∆AOC,
AC2 = AO2 + OC2
l2 = (8)2 +(6)2
l2 = 64 + 36
l2 = 100
∴ l = $$\sqrt{100}$$
l = 10 m
Curved surface Area of cone C.S.A. = πrl
= 3.14 × 6 × 10
= 188.4 m2
Let the length of tarpaulin required be 1 m.
In that 20 cm. (0.2 m) is wastage means remaining tarpaulin is (1 – 0.2 m)
∴ Area of tarpaulin = curved surface area of tent.
(l – 0.2 m) × 3 = 188.4 m2

l – 0.2 = 62.8
∴ l = 62.8 + 0.2
∴ l = 63 m.

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2.
Solution:
Slant height of a cone, l = 25 m
Diameter, 2r = 14m ∴ r = 7 m
Curved Surface Area = πrl

= 22 × 25
= 550 m2.
Cost of white-washing is Rs. 210 per m2.
Cost of white-washing for 550 m2 … ? …

= Rs. 1155.

Question 7.
A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:
A joker’s cap is in the form of a right circular cone.
radius of base, r = 7 cm.
height, h = 24 cm.
slant height, 1 = ?
Curved surface area, C.S.A.= ?

In ⊥∆AOC, ∠AOC = 90°
∴ AC2 = AO2 + OC2

∴ Curved surface area= πrl

= 550 m2
Area of 1 cap is 550 m2
Area of 10 caps…,v … ? …
= 550 × 10
= 5500 m2.

Question 8.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m2, what will be the cost of painting all these cones ?
(Use π = 3. 14 and take $$\sqrt{1.04}$$ = 1.02)
Solution:
Diameter of a cone, d = 40 cm = 0.40 m.

height, h = l m.

Curved surface area of cone = πrl

= 0.64056 m2
Area of 1 cone = 0.64056 m2
Area of 50 cone = ?
= 50 × 0.64056
= 32.028 m2
Cost of painting for 1 sq.m, is Rs. 12.
Cost of painting for 32.028 sq.m. …? …
= 32.028 × 12
= Rs.384.37.

##### KSEEB Solutions for Class 9 Maths
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