Class 9

MP Board Class 9th Maths Solutions Chapter 12 Heron’s Formula Ex 12.2

Question 1.
A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD, = 5 m and AD = 8 m. How much area does it. occupy?
Solution:
In ∆DCB
DB² = DC² + BC²

In ∆DBA P = (8 + 9 + 13)m = 30m
s = \(\frac{P}{2}\) = 15 m
s – a = 15 – 13 = 2
s – b = 15 – 9 = 6
s – c = 15 – 8 = 7
Area of ∆DBA = \(\sqrt{15x 2x6x7}\)
\(\sqrt{3x5x2x2x3x7}\)
= 2 x 3\(\sqrt{5×7}\) x 7 = 6\(\sqrt{35}\)m²
Area of quadrilateral ABCD = (30 + 6\(\sqrt{35}\)) m²
= 30 + 6 x 5.91
= 30 + 35.46
= 65.46 m2

Question 2.
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution:
In ∆ABC

P = 3 + 4 + 5 = 12cm
s = \(\frac{12}{2}\) = 6 cm
s – a = 6 – 5 = 1 cm
s – b = 6 – 4 = 2cm
s – c = 6 – 3 = 3 cm
Area of ∆ABC = \(\sqrt{6x1x2x3}\)
= \(\sqrt{2x3x2x3}\)
= 2 x 3 = 6 cm²
In ∆ADC P = 5 + 5 + 4 = 14 cm
s = \(\frac{P}{2}\) = 7 cm
s – a = 7 – 5 = 2cm
s – b = 7 – 5 = 2cm
s – c = 7 – 4 = 3cm
Area of ∆ADC = \(\sqrt{7x2x2x3}\) = 2\(\sqrt{21}\) cm²
Area of ∆BCD = Area of ∆ABC + Area of ∆ADC
= (6 + 2\(\sqrt{21}\)) cm²
= (6 + 2 x 4.58)
= 15.16 cm²

Question 3.
Radha made a picture of an aeroplane with coloured paper is shown in Fig. Find the total area of the paper used.
Solution:
The figure is divided into five parts as shown in Fig.

• Part I – Triangle having sides 5 cm, 5 cm and 1 cm
• Part II – Rectangle having sides 1 cm and 6.5 cm
• Part III – Trapezium having sides 2,1,1,1.
• Part IV and V. Right angled triangles having sides 6 cm and 1.5 cm.

For Part I:
a = 5 cm
b = 5 cm
c = 1 cm
s = \(\frac{P}{2}\) = \(\frac{5+5+1}{2}\) = \(\frac{11}{2}\) = 5.5 cm
s – a = 5.5 – 5 = 0.5
s – b = 5.5 – 5 = 0.5
s – c = 5.5 – 1 = 4.5

Area of triangle IV and V = \(\frac{1}{2}\) x 1.5 x 6 = 4.5 cm²
∴ Area of paper required = Area of part I + Area of part II + Area of part III + Area of part IV + Area of part V

= 2.49 + 1.27 + 15.5
= 19.26 cm²

Question 4.
A triangle and a parallelogram have the same base and the &me area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:

In ∆ABE
a = 30 cm
b = 28 cm
c = 26 cm
s = \(\frac{P}{2}\) = \(\frac{30+28+26}{2}\) = 42 cm
s – a = 42 – 30 = 12
s – b = 42 – 28 = 14
s – c = 42 – 26 = 16
Area of ∆ABE = \(\sqrt{42x12x14x16}\)
= \(\sqrt{2x3x7x2x2x3x2x7x4x4}\)
= 2 x 3 x 7 x 2 x 4
= 336 cm²

Area of parallelogram ABCD = area of ∆ABE = 336 cm² (given)
Area of parallelogram = b x h
336 = 28 x h
⇒ \(\frac{336}{28}\)
h= 12 cm.

Question 5.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Solution:
We know that the diagonal of a rhombus divide it into two triangles of equal area.

a = 48m
b = 30m
c = 30m
s = \(\frac{P}{2}\) = \(\frac{48+30+30}{2}\) = \(\frac{108}{2}\) = 54
s – a = 54 – 48 = 6 cm
s – b = 54 – 30 = 24 cm
s – c = 54 – 30 = 24 cm
Area of ∆ABD
= \(\sqrt{54x6x24x24}\)
= \(\sqrt{2x3x3x3x2x3x2x2x2x3x2x2x2x3}\)
= 2 x 3 x 3 x 2 x 2 x 3 x 2 = 432 cm²
Area of rhombus = 2 x 432 = 864 cm2
Area of grass field for each cow = \(\frac{864}{18}\) = 48 cm²

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.), each piece measurig 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umberella?
Solution:

a = 50 cm
b = 50cm
c = 20 cm
s = \(\frac{P}{2}\) = \(\frac{50+50+20}{18}\) = 60 cm
s – a = 60 – 50 = 10
s – b = 60 – 50 = 10
s – c = 60 – 20 = 40
Area of a ∆ = \(\sqrt{60x10x10x40}\)
= \(\sqrt{2x3x10x10x10x2x2x10}\)
= 10 x 10 x 52\(\sqrt{6}\)
= 200\(\sqrt{6}\) cm²
Area of cloth of each type = 200\(\sqrt{6}\) x 5 = 1000\(\sqrt{6}\) cm²

Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. How much paper of each shade has been used in it?
Solution:

∠AOB = 90°
BD = AC = 32 cm
OA = OC = 16 cm
Area (ABD) = area (DBC) = \(\frac{1}{2}\) x 32 x 16 = 256 cm²
In ∆CEF
a = 6 cm
b = 6 cm
c = 8 cm
s = \(\frac{P}{2}\) = \(\frac{6+6+8}{2}\) = 10
s – a = 10 – 6 = 4
s – b = 10 – 6 = 4
s – c = 10 – 8 = 2

Area of ∆CEF = \(\sqrt{10x4x4x2}\)
= \(\sqrt{5x2x4x4x2}\)
= 8\(\sqrt{5}\) cm²
= 8 x 2.24 = 17.92 cm²

Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig.). Find the cost of polishing the tiles at the rate of the field.
Solution:

a = 35 cm
b = 28 cm
c = 9 cm
s = \(\frac{P}{2}\) = \(\frac{35+28+9}{2}\)
s = a = 36 – 35 = 1
5 = b = 36 – 28 = 8
c = c = 36 – 9 = 27
Area of a tile = \(\sqrt{36x1x8x27}\)
= \(\sqrt{2x2x3x3x2x2x2x3x3x3}\)
= 2 x 3 x 2 x 3\(\sqrt{6}\) = 36\(\sqrt{6}\) cm²
= 36 x 2.45 = 88.2 cm²
Total area of tiles = 16 x 88.2 = 1411.2 cm²
Cost of polishing the tiles per cm² = 50 P
Total cost of polishing the tiles = 1411.2 x 50 P
= ₹ \(\frac{1411.2×50}{100}\) = ₹ 705.60

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
ABCD is a trapezium. Draw BE parallel to AD. Draw BF ⊥ DC. ABED is a parallelogram
AB = DE = 10 m and AD = BE = 14 m
EC = DC – DE = 25 – 10 = 15 m
In ∆BEC
a = 15m
b = 14m
c = 13m
s = \(\frac{15+14+13}{2}\) = 21

s – a = 21 – 15 = 6
s – b = 21 – 14 = 7
s – c = 21 x 13 = 8

MP Board Class 9th Maths Solutions

MP Board Class 9th General English Composition: Verbal/Visual Stimulus

You would be given some points (verbal stimulus) or a picture (visual stimulus) and would be asked to compose a short report, an article, a speech or description of about 80 words.

(A) Verbal Stimulus

On the basis of the points given write para-graphs of about 80 words on the given topics.

1. Effect of Music
Effect of music on humans, soothing effect on heart, cure even madness, even animals are affected, tigers, deers, snakes sensitive to music, one who has no delight is incapable of anything.
Answer:
The effect of music on the human mind and heart is great. Music has the power of soothing the oppressed heart and of infusing courage and cheerfulness into it. So great is its power that under its influence even madness is found to be cured. Even wild and savage animals cannot resist the charm of music. Tigers, deer and snakes are particularly sensitive to music. The man who has no ear for music and finds no delight in it is incapable of anything.

2. Alice in Wonderland
Alice, a little girl with imagination, falls asleep sitting near her sister, dreams of running in a field behind rabbit, falls down and enters a strange land with many adventures, tells her sister on waking up, sister finds her dreams illogical.
Answer:
Alice was a little girl with an imaginative mind. One day while sitting by her sister on the bank of the river she fell asleep. She dreamt that she was running across the field after a white beautiful rabbit. She fell down a large rabbit-hole and entered a strange land where she came across many odd characters and had many strange adventures. When she woke up she found herself lying on the bank with her head on her sister’s lap. She told her sister about her curious and strange dream and her extraordinary adventures in that wonderland. The sister found no sense in her dream. It was futile to expect any sense or logic in the dreams of a little girl.

3. TV Watching
Greatest source of entertainment attracts all, educative and entertaining, music, bad effects.
Answer:
These days TV is the greatest source of entertainment. It offers every kind of entertainment to all of the different ages. Cable TV has attracted all alike to it. TV watching is both educative and entertaining. It gives us good morals through various programmes, like serial and films. Its news channels let us know what is happening all around the world. Its music channels provide good music. Educational programmes add to our knowledge. But TV watching has bad effects also. It harms children’s eye-sight. They lag in their studies due to it. Thus, TV watching is both useful and harmful.

(B) Visual Stimulus

1. See the following picture and write a paragraph on it,

Answer:
It is a special campaign of girls education. The teacher is teaching them. The girls are listening to her with great interest. Education of girls is very
important as an educated girl is the backbone of a family. If she is educated then only she would be able to play the roles of a sister, mother and wife efficiently and become aware about the world and her rights. Hence, it is very necessary to educate girls.

2. Read the following flowchart describing the preparation of rice. Write a suitable paragraph with its help.

Answer:
Preparation of Rice : Take some rice in a bowl. Wash it with plenty of water three to four times. Keep the rice soaked in water for some time. Put in pressure cooker and add one and a half times of water and salt to taste. Switch on the gas and close the pressure cooker lid. Give one or two whistles and switch off the gas. Take out in the rice bowl and serve.

3. Write a story with the help of the given picture.

Answer:
A farmer once needed money because his crops were ruined due to snow. He went to a money lender with a pot containing jewellery of his wife. The money lender was a greedy man. He wished to take his jewellery and give him a small amount of money at high rate of interest so that the farmer may not be able to take it back. Money-lender’s wife also liked the farmer’s jewellery and she wished that her husband took it from the farmer any how. But farmer understood money lender’s greed and took his jewellery back from him.

MP Board Class 9th English Solutions

MP Board Class 9th General English Note Making and Summarising

Q. 1. Read the following passage carefully, supply a title, make notes and prepare a summary of it.

In democratic countries, any effort to restrict the freedom of the press are rightly criticized. However, this freedom can be easily abused. Stories about people often attract far more public attention than political events. Though we may enjoy reading about the lives of others, it is extremely doubtful whether we would equally enjoy reading about ourselves. Acting on the basis that facts are sacred, reporters can cause untold sufferings to individuals by publishing details about their private lives. Newspapers have such a great influence that not only can they bring about major changes to the lives of ordinary people but can even overthrow the government.

Answer:
(a) Title—Power of the Press.
(b) Notes—Freedom of the Press—sacred in Democracy.
(i) easily abused.
(ii) people more fascinating than politics.
(iii) misreport causes sufferings; newspaper have great influence on private lives.
(c) Summary—Freedom of the press can’t be restricted in democracies. False reporting may cause untold sufferings to individuals. Newspaper can bring about major changes to the lives of individuals.

Q. 2. Read the following passage carefully, supply a title, make notes and prepare a summary.

On one hand, the cinema is a source of enjoyment and entertainment and on the other hand it is also a source of knowledge and information. It also informs us of the happenings around us. There are cinemas for all tastes and for all types of people.

There are religious movies. These movies attract more people in towns and villages, and are most popular among ladies. Historical cinemas have their own style. They tell us about the past society, culture and life style. Social films awaken the public about illiteracy, dowry system and casteism. Some realistic films expose rich people and politicians who exploit and cheat poor people. The popular commercial cinema is meant for entertainment of all people. Cinema also helps in awakening the public on social issues and offers solution of various social issues.
Answer:
(a) Title—Cinema For All.
(b) Notes:
1. Cinema—
(a) source of entertainment.
(b) also source of knowledge and information.
(c) caters to tastes of all types of people.

2. Types of movies—
(a) Religious Movies—More popular in villages and among ladies.
(b) Historical Cinema—reveals past society, culture and life style.
(c) Social Films—awaken public about illiteracy, dowry system and casteism.
(d) Realistic Cinema—exposes rich people and politicians
(e) Popular Commercial Cinema—entertains all.
3. Cinema—awakens the public.

(c) Summary—The cinema is the cheapest source of entertainment. It is also a source of knowledge and information. It informs us of the happenings around the world. Various types of movies awaken the public and make them conscious of many concepts.

Q. 3. Read the following passage carefully, supply a title, make notes and prepare a summary.

All vehicles should keep to the left and leave the right half of the road free for those coming from the opposite direction. This is the traffic rule in all parts of India. In some countries in the west, however, vehicles have to keep to the right and not to the left. It does not matter whether it is right or left, but everyone should obey the rule. Cyclists should always keep to the edge of the road and not get in the way of other vehicles or of pedestrians We often see two or more cyclists riding together side by side right in the middle of the road. Traffic rules do not allow this.
Answer:
(a) Title—Traffic Rules
(b) Notes—
(i) In India vehicles should keep to the left.
(ii) In some countries vehicle keep to the right.
(iii) Traffic rules should be followed.
(iv) Cyclists should move on the edge of road and not in the middle.

(c) Summary—Everyone should obey traffic rules. Vehicles should be kept on the left side of the road. Cyclists should move on the edge of the road. They should not move in the middle of the road in groups.

Q. 4. Read the following passage carefully supply a title, make notes and prepare a summary.
Travelling is a means of education. The real aim of education is character building. When we travel, we have to put things in order. We have to buy our tickets and catch the train at the right time. The rich men, can indeed, get all this done by their servants, but India is a country of the poor. In journey we have to help ourselves. We learn many new things by seeing different places and by conversing with people of all kinds. In Europe education without travelling is considered incomplete. In ancient India great importance was given to pilgrimage. Travelling in this country is a great pleasure.
Answer:
(a) Title—Travelling, a means of Education.
(b) Notes—
(i) Travelling educates a person.
(ii) Education is character building.
(iii) Travelling makes one punctual, disciplined and self depended.

(c) Summary—Travelling is a means of education and education without travelling is incomplete. By travelling one learns new things and also becomes punctual, disciplined and self dependend. Travelling is also a great pleasure.

Q. 5. Read the following passage carefully, supply a title, make notes and prepare a summary.

Socrates was a simple man. He walked about barefooted, always wearing an old coat. He was a philosopher who believed that everyone should learn to think for himself. He spent his life in search of truth. He liked discussing questions relating to politics, religion, how life ought to be lived. Many people liked and respected him, but a few hated him. They hated him because he told them that they were wrong. His enemies were powerful. They arrested and condemned him to death. Socrates faced death calmly and cheerfully.
Answer:
(a) Title—Socrates
(b) Note—
(i) Socrates led a simple and austere life.
(ii) He was a philosopher and thinker.
(iii) He had both followers and enemies.
(iv) His enemies led him to death.
(v) He faced death boldly.

(c) Summary—Socrates, a philosopher and thinker led an austere life. He taught people to think for themselves. He pointed out people’s wrong doing which made people his enemies. They led him to death. He faced death calmly.

MP Board Class 9th English Solutions

MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:

1. The area of the sheet required for making the box.
2. The cost of sheet for it, if a sheet measuring 1m² costs ₹ 20.

Solution:
Given
l = 1.5 m = 150 cm
b = 1.25 m = 125 cm
h = 65 cm

1. Total area of plastic sheet required = LSA + Area of base
= 2h(l + b) + l x b
= 2 x 65 (150 + 125) + 150 x 125
= 130 (275) + 18750
= 35750 + 18750
= 54500 cm²
= 5.45 m²

2. Cost of sheet = area of plastic sheet x rate = 5.45 x 20
= 109.00
= ₹ 109.

Question 2.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m².
Solution:
Given
l = 5m
b = 4m
h = 3 m
Area of the room to be white washed = LSA + Area of ceiling
= 2h(l + b) + l x b
= 2 x 3 (5 + 4) + 5 x 4
= 6(9) + 20
= 54 + 20 = 74 m²
Cost of painting = 7.50 x 74 m²
= ₹ 555

Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹ 10 per m² is ₹ 15000, find the height of the hall.
Solution:
Given
P = 250 m
Rate of painting = ₹ 10/ m²
Cost of painting = ₹ 15000
Area of walls to be painted = \(\frac{15000}{10}\) = 1500 m²
Perimeter = 2 (l + b) = 250
∴ l + b = 250/2 = 125m
Area of walls = LSA = 2h (l + b) = 1500
= h (l + b) = \(\frac{1500}{10}\) = 750 m² …..(i)
Putting the value of (l + b) in (i), we get
h(125) = 750
h = \(\frac{750}{125}\) = 6 m.

Question 4.
The paint in a certain container is sufficient to paint an area equal to 9315 m². How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?
Solution:
Given
l = 22.5 cm
h = 7.5 cm = 0.075 m
b = 10 cm = 0.1 m
Area to be painted = 9.375 m² = 93750 cm²
Area of one brick = 2(lb + bh + hl)
= 2(22.5 x 10 + 10 x 7.5 + 7.5 x 22.5)
= 2(225 + 75 + 168.75)
= 2 x 468.75
= 937.5 cm²
No. of bricks which can be painted = \(\frac{93750}{937.5}\) = 100

Question 5.
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.

1. Which box has the greater lateral surface area and by how much?
2. Which box has the smaller total surface area and by how much? Sol.

Solution:
1. LSA of cubical box = 4a²
= 4 x 10 x 10 = 400 cm²
LSA of cuboidal box = 2h (l + b)
= 2 x 8 (12.5 x 10)
= 16 x 22.5
= 360 cm²
LSA of cubical box is more than cuboidal box by 40 cm².

2. TSA of cubical box = 6a² = 6 x 10 x 10
= 600 cm²
TSA of cuboidal box = 2 (lb + bh + hl)
= 2(12.5 x 10 + 10 x 8 + 12.5 x 8)
= 2(125 + 80 + 100)
= 2 x 305 = 610 cm².
∴ TSA of cuboidal box is more than cubical box by 10 cm².

Question 6.
A small indoor green house (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

1. What is the area of the glass?
2. How much of tape is needed for all the 12 edges?

Solution:

Given
l = 30 cm
b = 25 cm and
h = 25 cm
Area of glass required = 2(30 x 25 + 25 x 25 x 30)
= 2(750 + 625 + 750)
= 2 x 2125 = 4250 cm²
Length of tape required = 4(l + b + h)
= 4(30 + 25 + 25)
= 4 x 80 = 320 cm

Question 7.
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹ 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
Big Box :
l = 25 cm
b = 20 cm
h = 5 cm
Small Box:
l = 15 cm
b = 12 cm
h = 5 cm
Number of boxes required = 250
Rate = ₹ 4 per 1000 cm²
TSA of bigger box = 2 (25 x 20 + 20 x 5 + 5 x 25)
= 2(500 + 100 + 125)
= 2 x 725 = 1450 cm²
TSA of smaller box = 2 (15 x 12 +12 x 5 + 5 x 15)
= 2 (180 + 60 + 75)
= 2 x 315 = 630 cm²
Area of cardboard required one bigger = TSA + 5% of TSA
= 1450 + \(\frac{5}{100}\) x 1450
= 1522.5 cm²
Area of cardboard required for 250 boxes of bigger size = 250 x 1522.5
= 380625 cm²
Area of cardboard required for one small box = 630 + \(\frac{5}{100}\) x 630
= 661.5 cm²
Total area of cardboard required for 250 boxes of smaller size = 250 x 661.5
= 165375 cm²
Total area = (165375 + 380625) cm²
= 546000 cm²
Total cost of each kind of cardboard = \(\frac{4}{1000}\) x 546000
= ₹ 2184/-

Question 8.
Parveen wanted to make a temporary shelter for her car, by making a box – like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rollpd up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3 m?
Solution:
Given
l = 4m
b = 3 m and
h = 2.5 m
Area of the tarpaulin required = 2h(l + b) + l x b
= 2 x 2.5(4 + 3) + 4 x 3
= 5 x 7 + 12
= 35 + 12= 47m

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 12 Sound

Sound Intext Questions

Sound Intext Questions Page No. 162

Question 1.
How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
When an object vibrates, it allows the particles of the medium around it to vibrate. It exerts force on the adjacent particles and continue oscillating in all directions and one of it, hit our ear’s medium which creates sound. This process continues in the medium till the sound reaches your ear.

Sound Intext Questions Page No. 163

Question 1.
Explain how sound is produced by your school bell.
Answer:
When the bell rings, it continues to move forward and backward which creates vibration and simultaneously a series of compressions and rarefactions which produce a very loud sound.

Question 2.
Why are sound waves called mechanical waves?
Answer:
Sound waves needs medium to propagate therefore, they are called mechanical waves. Sound cannot travel in the absence of a medium. Sound waves are propagated through a medium because of the interaction of the particles present in that medium.

Question 3.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
No, because sound waves needs a medium through which they can propagate. And since there is no medium on the moon due to absence of atmosphere, one cannot hear any sound produced by his / her friend on the moon.

Sound Intext Questions Page No. 166

Question 1.
Which wave property determines.

1. loudness
2. pitch?

Answer:

1. Amplitude determines loudness of a sound wave.
2. Frequency determines pitch of a sound wave.

Question 2.
Guess which sound has a higher pitch: guitar or car horn?
Answer:
Guitar has a higher pitch than car horn, because sound produced by the strings of guitar has higher frequency and since pitch is proportional to frequency, pitch of guitar will be higher.

Question 3.
What are wavelength, frequency, time period and amplitude of a sound wave?
Answer:

1. Wavelength: Wave length is the length between two consecutive peaks it is represented by Greek letter λ (lambda). Louder sound has shorter wavelength and softer sound has longer wavelength.
2. Frequency: The number of sound waves produced in unit time is called the frequency of sound waves. Frequency is measured in seconds. Frequency is denoted by Greek letter v (nu). The SI unit of frequency is ‘hertz’.
3. Amplitude: Amplitude of a wave is magnitude of maximum disturbance on either side of the normal position or mean value in a medium.

Question 4.
How are the wavelength and frequency of a sound wave related to its speed?
Answer:
Speed, wavelength, and frequency of a sound wave are related as follows:
Speed (ν) = Wavelength [λ] × Frequency (v)
⇒ ν = λ × v

Question 5.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer:
Given,
Frequency of the sound wave, v = 220 Hz
Speed of the sound wave, u = 440 ms^-1
Putting the equations,
Speed = Wavelength (λ) × Frequency (v)
ν = λ × v
∴ λ = \(\frac { ν }{ v  }\) = \(\frac { 440 }{ 220 }\) = 2 m
Hence, Wavelength = 2 m.

Question 6.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Answer:
Given,
Frequency = 500 Hz
Distance = 450 m
Since, T = \(\frac { 1 }{ Frequency }\)
= \(\frac { 1 }{ 500 }\) = 0.002 sec.
Hence, person will listen the sound after every 0.002 sec.

Question 7.
Distinguish between loudness and intensity of sound.
Answer:
Intensity drives loudness of a sound. These both qualities of sound are proportional to each other. The amount of sound passing through a unit area per second represents intensity of a sound wave. While loudness is the response of the ear to the sound (amount received to pinna). The loudness of a sound is defined by its amplitude. The amplitude of a sound decides its intensity, which in turn is perceived by the ear as loudness.

Sound Intext Questions Page No. 167

Question 1.
In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer:
Sound travels the fastest in solids medium (here Iron) then in liquids (water) and it is the slowest in gases (air). Therefore, for a given temperature, sound travels as follows (decreasing order): ⇒ Iron > water >air.

Sound Intext Questions Page No. 168

Question 1.
An echo returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms^-1?
Answer:
Given,
Speed, v = 342 ms^-1
Time, t = 3 s
Since, Distance = v × t
= 342 × 3 = 1026 m
Condition,
In fixed time interval, sound has to travel a distance that is twice the distance of the reflecting surface and the source.
Hence, the distance of the reflecting surface from the source = \(\frac { 1026 }{ 2 }\) m = 513 m.

Sound Intext Questions Page No. 169

Question 1.
Why are the ceilings of concert halls curved?
Answer:
Ceilings of concert halls are curved to:

1. Enhance loudness and echo of sound created.
2. Sound after reflection (from the walls) spreads uniformly in all directions.

Sound Intext Questions Page No. 170

Question 1.
What is the audible range of the average human ear?
Answer:
The audible range of an average human ear is between 20 Hz to 20,000 Hz.

Question 2.
What is the range of frequencies associated with

1. Infrasound?
2. Ultrasound?

Answer:

1. Infrasound: frequencies less than 20 Hz.
2. Ultrasound: frequencies more than 20,000 Hz.

Sound Intext Questions Page No. 172

Question 1.
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?
Answer:
Given,
Time taken by the sonar pulse to return, t = 1.02s
Speed of sound in salt water, v = 1531 ms^-1
Distance of the cliff from the submarine = Speed of sound × Time taken
= 1531 × 1.02 m = 1561.62 m
Distance travelled by the sonar pulse during its transmission and reception in water
= 2 × Actual distance = 2d
Actual distance,
d = Distance of the cliff from the submarine / 2 = \(\frac { 1561 }{ 2 }\) = 780.31 m.

Sound NCERT Textbook Exercises

Question 1.
What is sound and how is it produced?
Answer:
Sound is a form of energy which is received at our ear pinna and gives the sensation of hearing. It is a vibration which propagates in air and developed by vibrating objects.

Question 2.
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer:
When vibration is produced by a body, it moves in backward and forward direction till the energy lasts. During forward movement it creates a region of high pressure. This region of high pressure is known as compressions. When it moves backward, it creates a region of low pressure. This region is known as a rarefaction. This is shown in figure given below.

Here,
C = compressions
R = rarefaction.

Question 3.
Cite an experiment to show that sound needs a material medium for its propagation.
Answer:
Arrange the instrument according to the picture given below:

• Take an electric bell and an air tight jar with glass bell and connect it to a vacuum pump.
• Suspend the bell inside the jar, and press the switch of the bell.

Method: Now pump out the air from the glass jar. The sound of the bell will become fainter and after some time, the sound will not be heard. This shows that sound needs a material medium to travel.

Question 4.
Why is sound wave called a ‘longitudinal wave’?
Answer:
Sound wave is called longitudinal wave because the air particles vibrates parallel to the direction of propagation of sound wave,it is produced by compressions and rarefactions in the air.

Question 5.
Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:
The quality of pitch and loudness of sound enables us to identify our friend by his voice.

Question 6.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer:
Speed difference is the main reason of this happening. The speed of sound (344 m/s) is very less than the speed of light (3 × 10⁸ m/s). A flash is seen before we hear a thunder because sound of thunder takes more time to reach the Earth as compared to light.

Question 7.
A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 ms^-1.
Answer:
We know,
Speed (ν) = Wavelength × Frequency v
ν = λ × v
Given,
Speed of sound in air ν = 344 m/s (Given)

(i) For, v = 20 Hz
λ₁ = \(\frac { ν }{ v }\)
= \(\frac { 344 }{ 20 }\) = 17.2 m

(ii) For, v = 20000 Hz
12 = \(\frac { ν }{ v }\)
= \(\frac { 344 }{ 20000 }\) = 0.172 m
Hence, for humans, the wavelength range for hearing is 0. 0172 m to 17.2 m.

Question 8.
Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of time taken by the sound wave in air and in aluminium to reach the second child.
Answer:
Velocity of sound in air= 346 m/s
Velocity of sound waves in aluminium 6420 m/s
Let length of rod be 1.
Time taken for sound wave in air, t₁ = \(\frac { 1 }{ Velocity }\) in air.
Time taken for sound wave in aluminium, t₂ = \(\frac { 1 }{ Velocity }\) in aluminium.
Therefore,

= \(\frac { 6420 }{ 346 }\) = 18.55.

Question 9.
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
Frequency =100 Hz. (Given)
This means the source of sound vibrates 100 times in one second.
Therefore, number of vibrations in 1 minute, i.e. in 60 seconds = 100 × 60 = 6000 times.

Question 10.
Does sound follow the same laws of reflection as light does? Explain.
Answer:
Yes, sound waves also follow the same laws of reflection as light wave do because:

1. Angle of incidence of sound is always equal to that of angle of reflection of sound waves.
2. The direction in which sound is incident, the direction in which it is reflected and normal all lie in the some plane.

Question 11.
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of
sound production remains the same. Do you hear echo sound on a hotter day?
Answer:
An echo is heard when the time for the reflected sound is just 0.1 s
Time taken = \(\frac { Total Distance }{ Velocity }\)
On a hotter day, due to lighter medium the velocity of sound is more then a colder day. Hence, sound wave will move faster and if time taken by echo is less than 0.1 sec it will not be heard.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:
Two practical applications of reflection of sound waves are:

1. Reflection of sound is used to measure the distance and speed of underwater objects. This method is known as SONAR.
2. Working of a stethoscope is also based on reflection of sound. In a stethoscope, the sound of the patient’s heartbeat reaches the doctor’s ear by multiple reflection of sound.

Question 13.
A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 ms^-2 and speed of sound = 340 ms^-1.
Answer:
Height of the tower, s = 500 m
Velocity of sound, v = 340 ms^-1
Acceleration due to gravity, g = 10 ms^-2
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower t₁,
According to the second equation of motion:
s = ut₁ + \(\frac { 1 }{ 2 }\) gt₁²
500 = 0 × t₁ + \(\frac { 1 }{ 2 }\) × 10 t₁²
t₁² = 500
t₁ = 10 s
Now, time taken by the sound to reach the top from the base of the tower,  t₂ = \(\frac { 500 }{ 340 }\) = 1.47 s.
Therefore, the splash is heard at the top after time, t Where, t = t₁ + t₂ = 10 + 1.47 = 11.47 s.

Question 14.
A sound wave travels at a speed of 339 ms^-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Answer:
Speed of sound, v = 339 ms^-1
Wavelength of sound, λ = 1.5 cm = 0.015 m
Speed of sound = Wavelength × Frequency
ν = λ × v
∴ v = \(\frac { ν }{ λ }\)
= \(\frac { 339 }{ 0.015 }\)
= 22600 Hz.
The frequency range of audible sound for humans lies between 20 Hz to 20,000 Hz.
Since, the frequency of the given sound is more than 20,000 Hz, it is not audible.

Question 15.
What is reverberation? How can it be reduced?
Answer:
The repeated multiple reflections of sound in any big enclosed space is known as reverberation. The reverberation can be reduced by covering the ceiling and walls of the enclosed space with sound absorbing materials, such as fibre board, loose woollens, etc.

Question 16.
What is loudness of sound? What factors does it depend on?
Answer:
The effect produced in the brain by the sound of different frequencies is called loudness of sound. Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

Question 17.
Explain how bats use ultrasound to catch a prey.
Answer:
Bats produce high – pitched ultrasonic squeaks. These high – pitched squeaks are reflected by objects such as preys and returned to the bat’s ear. This allows a bat to know the distance of his prey.

Question 18.
How is ultrasound used for cleaning?
Answer:
Objects to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency of these ultrasound waves detaches the dirt from the objects.

Question 19.
Explain the working and application of a SONAR.
Answer:
Sonar stands for Sound Navigation And Ranging. It is a device used to measure the depth, direction and speed of under – water objects such as submarines and ship wrecks with the help of ultrasounds and is also used to measure the depth of seas and oceans. An ultrasonic sound is produced and transmitted by the transducer (it is a device that produces ultrasonic sound) of the Sonar, which travels through sea water.

The echo produced by the reflection of this ultrasonic sound is detected and recorded by the detector, which is converted into electrical signals. The distance (d) of the under – water object is calculated from the time (t) taken by the echo to return with speed (v) is given by 2d = v × t.

Question 20.
A SONAR device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Answer:
Given,
Time taken to hear the echo, t = 5 s
Distance of the object from the submarine, d = 3625 m
Total distance travelled by the sonar waves during the transmission and reception in water = 2d
Using formula:
Velocity of sound in water,
v = 2 \(\frac { d }{ t }\)
= 2 × \(\frac { 3625 }{ 5 }\) = 1450 ms^-1.

Question 21.
Explain how defects in a metal block can be detected using ultrasound.
Answer:
Defects in metal blocks do not allow ultrasound to pass through them and they are reflected back. This fact is used to detect defects in metal blocks. Ultrasound is passed through one end of a metal block and detectors are placed on the other end. The defective part of the metal block does not allow ultrasound to pass through it. As a result, it will not be detected by the detector. Hence, defects in metal blocks can be detected using ultrasound.

Question 22.
Explain how the human ear works.
Answer:
The human ear consists of three parts – the outer ear, middle ear and inner ear.

1. Outer ear: It is also termed ‘pinna’. It collects the sound from the surrounding and directs it towards auditory canal.
2. Middle ear: It is at the auditory canal where there is a thin membrane called eardrum or tympanic membrane. The sound waves set this membrane to vibrate. These vibrations are amplified by three small bones hammer, anvil and stirrup.
3. Inner ear: When vibration reach the cochlea in the inner ear and are converted into electrical signals which are sent to the brain by the auditory nerve, and the brain interprets them as sound.
   

Sound Additional Questions

Sound Multiple Choice Questions

Question 1.
Which of the following waves have no requirement of any medium to propagate?
(a) sound
(b) radio
(c) light waves
(d) none of above.
Answer:
(c) light waves

Question 2.
What kinds of waves are produced by sound?
(a) longitudinal only
(b) transverse waves only
(c) electromagnetic Waves
(d) both Longitudinal and Transversal.
Answer:
(a) longitudinal only

Question 3.
When a wave propagate, it transfers ___________ .
(a) energy only
(b) matter only
(c) both energy and matter
(d) none of these.
Answer:
(a) energy only

Question 4.
Note is a sound ___________ .
(a) of mixture of several frequencies
(b) of mixture of two frequencies only
(c) of a single frequency
(d) always unpleasant to listen.
Answer:
(a) of mixture of several frequencies

Question 5.
A key of a mechanical piano struck gently and then struck again but much harder this time. In the second case:
(a) sound will be louder but pitch will not be different.
(b) sound will be louder and pitch will also be higher.
(c) sound will be louder but pitch will be lower.
(d) both loudness and pitch will remain unaffected.
Answer:
(d) both loudness and pitch will remain unaffected.

Question 6.
Earthquake produces which kind of sound before the main shock wave begins ___________ .
(a) ultrasound
(b) Infrasound
(c) audible sound
(d) none of the above.
Answer:
(b) Infrasound

Question 7.
Infrasound can be heard by ___________ .
(a) dog
(b) bat
(c) rhinoceros
(d) human beings.
Answer:
(c) rhinoceros

Question 8.
Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting ___________ .
(a) intensity of sound only.
(b) amplitude of sound only.
(c) frequency of the sitar string with the frequency of other musical instruments.
(d) loudness of sound.
Answer:
(c) frequency of the sitar string with the frequency of other musical instruments.

Question 9.
What we term to the number of oscillations completed in one second?
(a) time period
(b) velocity
(c) frequency
(d) wavelength.
Answer:
(c) frequency

Question 10.
Sound waves which can be propagate in solids are ___________ .
(a) Longitudinal only.
(b) Transverse only.
(c) Either longitudinal or transverse.
(d) Non mechanical waves only.
Answer:
(c) Either longitudinal or transverse.

Question 11.
In SONAR, we use ___________ .
(a) ultrasonic waves
(b) infrasonic waves
(c) radio waves.
(d) audible sound waves.
Answer:
(a) ultrasonic waves

Question 12.
Sound travels in air if ___________ .
(a) particles of medium travel from one place to another.
(b) there is no moisture in the atmosphere.
(c) disturbance moves.
(d) both particles as well as disturbance travel from one place to another.
Answer:
(c) disturbance moves.

Question 13.
When we change feeble sound to loud sound we increases ___________ .
(a) Frequency
(b) Amplitude
(c) Velocity
(d) Wavelength.
Answer:
(b) Amplitude

Question 14.
In the curve (Figure) half the wavelength is ___________ .
image
(a) AB
(b) BD
(c) DE
(d) AE
Answer:
(a) AB

Question 15.
The time taken to complete an osoillation ___________ .
(a) time period
(b) velocity
(c) frequency
(d) wavelength.
Answer:
(a) time period

Question 16.
The period of a vibrating body of frequency 100 Hz is ___________ .
(a) 100 seconds
(b) 10 seconds
(c) 0.1 second
(d) 0.01 second.
Answer:
(a) 100 seconds

Question 17.
Which of the following is SI unit of amplitude?
(a) metre
(b) s^-1
(c) metre / second
(d) Hertz.
Answer:
(a) metre

Question 18.
The quantity λ is known as ___________ .
(a) wave velocity
(b) frequency
(c) wavelength
(d) wave number.
Answer:
(c) wavelength

Question 19.
Sound wave of which of the following frequency is an ultrasonic sound?
(a) 30 Hz
(b) 300 Hz
(c) 3000 Hz
(d) 30,000 Hz.
Answer:
(d) 30,000 Hz.

Question 20.
In which of the following, speed of the sound is maximum?
(a) air
(b) water
(c) steel
(d) kerosene.
Answer:
(c) steel

Sound Very Short Answer Type Questions

Question 1.
How does sound travels in gases and liquids – as longitudinal or as transverse waves?
Answer:
As longitudinal waves.

Question 2.
Give examples of longitudinal waves.
Answer:
Sound waves.

Question 3.
What is speed of sound in air?
Answer:
At 0° C, it is 331 m/s. At 20°C, it is 341 m/s.

Question 4.
What is reverberation?
Answer:
The repeated reflection that results in the persistence of sound is called reverberation.

Question 5.
Among solids, liquids and gases sound travels faster in which medium?
Answer:
Sound travels the fastest in solids.

Question 6.
What is one Hz?
Answer:
Hz is the unit of frequency, called as Hertz. One Hertz is equal to one cycle per second.

Question 7.
What is ‘note’ of second?
Answer:
The sound produced due to a mixture of several frequency is called a note, it is pleasant to listen to.

Question 8.
What is pitch?
Answer:
The way our brain interprets the frequency of an emitted sound is called the pitch.

Sound Short Answer Type Questions

Question 1.
What is Sound? Why it is important for us?
Answer:
Sound is a longitudinal mechanical wave. Sound has great importance in our daily life. It gives us a sensation of hearing. It makes it possible to communicate with other persons through speech.

Question 2.
What is a mechanical wave?
Answer:
A mechanical wave is a disturbance that moves through a medium when the particles of the medium set neighbouring particles into motion. The particles vibrating in turn do not move forward but the disturbance is carried forward.

Question 3.
What is a longitudinal wave?
Answer:
If the particles of a medium vibrate in a direction, parallel to or along the direction of propagation of wave, it is called longitudinal wave.

Question 4.
What type of waves can travel in vacuum? Give example(s).
Answer:
Electromagnetic waves can travel in vacuum. Sun light, x – rays are examples of electromagnetic waves.

Question 5.
Suppose you and your friend are on the Moon. Will you be able to hear any sound produced by your friend?
Answer:
No, we will not hear any sound on the Moon. The Moon does not have any atmosphere, since sound is a mechanical wave and requires a medium to travel.

Question 6.
What are the factors, speed of sound wave depends upon?
Answer:
Speed of the sound depends on the following factors:

1. Inertial property of the medium (to store kinetic energy).
2. Elastic property of the medium (to store potential energy).
3. Temperature of the medium.
4. Density of the medium.
5. Humidity present in the medium (in air / gases).

Question 7.
Three persons A, B and C are made to hear a sound travelling through different mediums as given below:

Persons	Medium
A	Iron Rod
B	Air
C	water

Who will hear the sound first ? Why ?
Answer:
Person A will hear the sound first because sound travels the faster in solids than in liquids and gases.

Question 8.
If 20 waves are produced per second, what is the frequency in hertz?
Answer:
Number of waves per second is known as frequency.
∴ Frequency (v) = 20 Hz.

Question 9.
What is echo?
Answer:
The sound waves produced bounce back or gets reflected from the mountains or buildings and come to our ears, this reflected sound is known as Echo. To hear echo, the barrier reflecting the sound should be atleast at a distance of 17 metres.

Question 10.
What is infrasonic? Give an example.
Answer:
Sound having frequency less than 20 Hz is known as infrasonic sound or infrasonic. Waves produced during earthquake are infrasonic.

Sound Long Answer Type Questions

Question 1.
Establish the relationship between speed of sound, its wave length and frequency. If velocity of sound in air is 340 m s^-1, calculate:

1. wavelength when frequency is 256 Hz.
2. frequency when wavelength is 0.85 m.

Answer:
Derivation of formula ν = v λ

1. 340 = 256 λ ⇒ λ = 1.33 m.
2. 340 = v (0.85) ⇒ v = 400 Hz.

Question 2.
A girl is sitting in the middle of a park of dimension 12 m × 12 m. On the left side of it there is a building adjoining the park and on right side of the park, there is a road adjoining the park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound? Explain answer.
Answer:
If the time gap between the original sound and reflected sound received by the listener is around 0.1 s, only then the echo can be heard. The minimum distance travelled by the reflected sound wave for the distinctly listening the echo, distance = velocity of sound × time interval;
= 344 × 0.1 = 34.4 m.
But, in this case the distance travelled by the sound reflected from the building and then reaching the girl will be (6 + 6 = 12 m),  which is much smaller than the required distance. Therefore, no echo can be heard.

Sound Higher Order Thinking Skills (HOTS)

Question 1.
A key of piano is struck gently and then struck again but much harder this time. What will happen in the second case?
Answer:
In second case, the loudness will increase as this will increase the amplitude of vibration of string. Pitch and frequency will also increase in force or tension in the string.

Question 2.
At a hill station, a child could hear the echo of his voice after 0.2 s. But, when he went to the same place in the afternoon, he could not hear echo at all. What could be the reason?
Answer:
In afternoon, the temperature rises, therefore the speed of sound also increased. The reflected sound will take very less time to travel back and no echo is heard.

Sound Value Based Question

Question 1.
It is not advisable to construct houses near airports, inspite of that many new residential apartments are constructed near airports. Rajesh / Sumit files RTI and also complains the municipal office about the same.

1. Why one should not reside near airport?
2. Name other two places where there is noise pollution.
3. What value of Rajesh is reflected in this act?

Answer:

1. The landing and taking off of the air – planes causes lot of noise-pollution which may lead to deafness, high blood pressure and other health problems.
2. The other two places where there is noise-pollution are residing near the heavy traffic routes and railway stations or lines.
3. Rajesh shows participating citizen and moral responsibility values.

MP Board Class 9th Science Solutions

MP Board Class 9th Special English Grammar Type II

1. In the passage given-below one word has been omitted in each line. Write the missing word along with the word that comes before and the word that comes after it in your answer sheet against the correct blank number. Ensure that the word that forms your answer is underlined. (4 marks)

There was hardly political, social e.g. hardly any political
religious, agrarian, labor any other, (a)
problem which come under (b)
his purview and he did not (c)
deal on his own within the
framework of the (d)
principles which he held be
basic and (e)
fundamental. There hardly any
aspect of life (f)
in India which he influence
and fashion (g)
according his own pattern. (h)
Answer:
(a) labour or any
(b) which didn’t come
(c) and that he
(d) own or within
(e) held to be
(f) There was hardly
(g) he didn’t influence
(h) according to his.

2. In the following paragraph, one word has been omitted in each line. Write the missing word along with the word that comes before and the word that comes after it in your answer sheet against the correct blank number. Ensure that the word that forms your answer is underlined.

The holy man spoke Rama for e.g. spoke to Rama
a while. He then Rama’s mother (a)
that should send him to pray at the (b)
Kali Temple the village. He said (c)
that if prayed sincerely Kali (d)
would appear him and bless him. (e)
This he would make Rama’s fortune. (f)
Answer:
(a) then told Rama’s
(b) that she should
(c) Temple in the village
(d) if he prayed
(e) appear before him
(f) This way he.

3. In the following paragraph, one word has been omitted from each line. Write the missing word along with the word that comes before and the word that comes after it in your answer sheet against the correct blank number. Ensure that the word that forms your answer is underlined.

The holy man to Rama e.g., man spoke to
for a while. He to Rama’s mother (a)
that her son prayed sincerely (b)
to Kali village goddess ‘ (c)
he be blessed by her. (d)
This would his fortune and (e)
she would be free from worries. (f)
Answer:
(a) He said to
(b) that if her
(c) Kali the village
(d) he would be
(e) would change his
(f) from her worries.

4. In the following paragraph, one word has been omitted from each line. Write the missing word along with the word that comes before and the word that comes after it in your answer sheet against the correct blank number. Ensure that the word that forms your answer is underlined. (3 marks)

After the holy man had spoken Rama, e.g. spoken to Rama
he told Rama’s mother that should (a)
be sent to piay the Kali Temple at the (b)
village. He promised good lady that if (c)
this was done, Kali, the goddess appear, (d)
take pity on him and bless. Then all her (e)
worries would over. (f)
Answer:
(a) that he should
(b) pray at the
(c) promised the good
(d) goddess would appear
(e) bless him Then
(f) would be over.

5. In the passage given below one word has been omitted in each line. Write the missing word along with the word that comes before and the word that comes after it in your answer sheet against the correct blank number. Ensure that the word that forms your answer is underlined. (3 Marks)

Man’s increases by hard work. e.g. Man’s practice increases
If he does not it, his practice suffers. (a)
The lawyer is judged the cases that he wins (b)
The writers is examined (c)
by the of the readers and critics; whereas(d)
the typists is tested by his speed (e)
and the accuracy of his work. (f)
Answer:
(a) not do it
(b) judged by the
(c) writer’s capability is
(d) the opinion of
(e) typist’skill is
(f) àccuracy of his.

6. In the passage given below, one word has been omitted in each line. Write the missing word along with the word that comes before and the word that comes after it in your answer sheet against the correct blank number. Ensure that the word that forms your answer is underlined. (1/2 x 6 = 3 Marks)

There was a rich old man who in a palace, e.g. who lived in
There also lived nearby poor man in a hut. (a)
He on crumbs of food given by. others, (b)
But he cheerful and never complained. (c)
Once it happened that the poor man (d)
had to eat for a long while. So he (e)
went to the rich old man help. (f)
Answer:
(a) nearby a poor
(b) He survived on
(c) he remained cheerful
(d) it so happened
(e) had nothing to
(f) man for help

7. In the passage gien below, oie worcilias been omitted in each line. Write the missing word along with the word that comes before and the word that comes after it in your answer sheet against the correct blank number. Ensure that the word that forms your answer is underlined. (5 Marks)

The weather, in first-half e.g. in the first
of the English summer always known (a)
to be quite cold. But, what we aré at the (b)
moment here, is, I think rather exceptional (c)
summer—it is just too coki anyone’s liking. (d)
For, a cricketer, you have icy, chilly (e)
wiñds blowing your face, it certainly (f)
is far from a pleasant experience you are (g)
batting, bowling fielding. And, with (h)
the weather being what is right now, the (i)
visiting Indian team is finding it rather
tough to cope the conditions. (j)
Answer:
(a) summer is always
(b) are discussing at
(c) rather an exceptional
(d) cold for anyone’s’
(e) cricketer, if you
(f) blowing to your
(g) experience when you
(h) bowling or fielding
(i) what it is
(j) cope with the conditions.

8. In the passage given below, one word has been omitted in each line. Write the missing word along with the word that comes before and the word that comes after it in your answersheet against the correct blank number. Ensure that the word that forms your answer is underlined. (5 marks)

Within each the dialect areas, there is considerable e.g. each of the
variaticn in speech according education (a)
and social standing. Ther is
important polarity (b)
of uneducated educated speech in (e)
which former can be identified with (d)
the regional dialect most completely
latter moves (e)
away from dialect usage a form of (f)
English cuts across dialectical boundaries, so (g)
do features of uneducated use; a (h)
prominent example the double negative (j)
as in I don’t no cake. (J)
Answer:
(a) according to education
(b) is an important
(c) uneducated and educated
(d) which the former
(e) completely as latter
(f) usage to a form
(g) English that cuts
(h) do the features
(i) example is the
(j) don’t want no.

MP Board Class 9th English Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.3

Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
Let us draw different pairs of circles as shown below:

We gave,
In figure           Maximum number of common points
(i)             –                     nil
(ii)            –                    one
(iii)           –                    two
Thus, two circles can have at the most two points in common.

Question 2.
Suppose you are given a circle. Give a construction to find its center.
Solution:
Steps of construction:

1. Mark any three points A, B and C on the circle.
   
2. Join AB and BC.
3. Draw the perpendicular bisector of AS. Let it be PQ.
4. Draw the perpendicular bisector of BC. Let it be RS.
5. PQ and RS intersect at point O.
6. O is the center of given circle.

Question 3.
If two circles intersect at two points, prove that their centers lie on the perpendicular bisector of the common chord.
Solution:
Given:
Circles C(O, r) and C(D, r₁) intersect at A and B.
To prove:
OD is perpendicular bisector of AS.

Construction:
Join OA, AD, OB and BD.
Proof:
In ∆AOD and ∆BOD,
OA = OB
DA = DB (Radii of the circle)
OD = OD (Common)
∆AOD ≅ ∆BOD (By SXS)
and ∠1 = ∠2 (By CPCT)
In ∆OAC and ∆OBC,
OA = OB (Radii of the circle)
OC =OC (Common)
∠1 = ∠2 (Proved)
∆AOC = ∆OBC (By SAS)
So ∠3 = ∠4 (By CPCT)
and AC = BC (By CPCT)
∠3 + ∠4 = 180.°
2∠3 = 180°
∠3 = \(\frac{180^{\circ}}{2}\) =90° (UA’s)
OD is perpendicular bisector of AB.

MP Board Class 9th Maths Solutions

MP Board Class 9th Special English Grammar Type VI

1. Look at the following newspaper items. Use the information given in the headlines to complete the paragraph. Write there answers in your answer sheet against the correct blank 07 numbers. Do not copy the whole sentence. (13 marks)

(a) Three more MLAs suspended in the Assembly.
Today on the third day three more MLAs of the ruling party …………………….. for their unruly behavior.
(b) Leander Paes gets Bronze in Olympics.
A bronze …………………….. by Leander Paes in the Olympics for the first time. It is a big achievement.
(c) U.P. elections to begin in September.
General Assembly elections …………………….. in September as per the notification of the Election Commission of India.
Answer:
(a) were suspended from the assembly
(b) medal was won
(c) in U.P. will commence.

2. Look at the newspaper headlines given and then use the information in them to complete the paragraph. Write the answer in your answer sheets against the correct blank number. Do not copy the whole sentence. (5 marks)

(a) 70 die in Peru plane crash
Lima : A Boeing 757 belonging to the …………………….. off the Pacific coast north of Lima on Wednesday after the pilot reported a mechanical failure.
(b) No bull fights in Goa
Panaji : The Mumbai High Court has …………………….. a popular and traditional pastime in Goa.
(c) Man denies hoax charge
Chennai : The man who claimed he could produce petrol from herbs, on Wednesday …………………….. used dishonest means to produce herbal fuel.
(d) Shruti, Rishu set new records.
New Delhi : Shruti Gupta and Rishu Sharma …………………….. their respective events in the Delhi State Aquatic Meet here on Wednesday.
(e) Gooch pulls out
London: Former England captain Graham Gooch …………………….. English team for the forthcoming tour of Australia.
Answer:
(a) Peru Airlines crashed killing 70 passengers
(b) banned bull fight,
(c) denied hoax charges that he
(d) set new records in
(e) has pulled out of the.

MP Board Class 9th English Solutions

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.2

Question 1.
Construct a triangle ABC in which BC = 7 cm, B = ∠75° and AB + AC = 13 cm.
Solution:
BC = 7 cm
∠B = 15°
AB + BC = 13 cm.

Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° AB – AC = 3.5 cm.
Solution:
BC = 8 cm
∠B = 45°
AB – AC = 3.5 cm.

Question 3.
Construct a triangle PQR in which QR = 6 cm. ∠Q = 60° and PR – PQ = 2 cm.
Solution:
QR = 6 cm
∠Q =60°
PR – PQ = 2 cm

Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Solution:
XY + YZ + ZX = 11 cm
∠Y = 30°
∠Z =90°

1. Draw a line segment BC =11 cm.
2. At B construct an angle of 30° and at C, draw angle of 90°.
3. Bisect these angles. Let the bisectors of these angles intersect atX.
4. Draw perpendicular bisectors AC of BX to intersect BC at Y and DZ of XC to intersect BC at Z.
5. Join XY and XZ. XYZ is the required D.

Question 5.
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Solution:
Steps of construction:

1. Draw \(\overline { BC } \) = 12 cm.
2. Construct ∠CBY = 90°.
3. From \(\overline { BY } \), cut off BX = 18 cm.
4. Join CX.
5. Draw PQ, the perpendicular bisector of CX, such that PQ meets BX at A.
6. JoinAC.

Thus, ABC is the required triangle.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.4

Question 1.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution:
Given
Let O and O₁ be the centre of bigger and smaller circle respectively
OA = OB = 5 cm
O₁A – O₁B = 3 cm
OO₁ = 4 cm

To find: AB.
Construction:
Join OA, OB, O₁A and O₁B join AB also.
In ∆OAO₁ and ∆OBO₁
OA = OB (Radii of a circle)
O₁A = O₁B (Radii of a circle)
OO₁ = OO₁ (Common)
so ∆OAO₁ = ∆OBO₁ (By SSS)
and so ∠1 = ∠2 (By CPCT)
In ∆OCA and ∆OCB,
OA = OB (Radii of a circle)
∠1 = ∠2 (Proved)
OC = OC (Common)
∆OCA = ∆OCB (By SAS)
so AC = BC (By CPCT)
and ∠ACO = ∠BCO (By CPCT)
∠ACO + ∠BCO = 180° (LPA’s)
⇒ ∠ACO + ∠BCO = 180°
2∠ACO = 180°
∠ACO = 90°
ar (OAO₁) = \(\frac{1}{2}\) x OO₁ x AC
= \(\frac{1}{2}\) x 4 x AC = 2ACcm² …..(i)
In ∆QAO₁, a = 5 cm, bc = 4 cm, c = 3 cm
12
s = \(\frac{5+4+3}{2}\) = \(\frac{12}{2}\) = 6 cm
s – a = 6 – 5 = 1 cm
s – b = 6 – 4 = 2 cm
s – c = b – 3 = 3 cm
ar(OAO₁) = \(\sqrt{s(s-a)(s -b)(s- c)}\).
= \(\sqrt{6x1x2x3}\)
= 6 cm² …..(ii)
From (i) and (ii), we get
2AC =6
AC = 3 cm
Now, AB = 2AC (∴ AC = BC)
= 2 x 3 = 6 cm.

Method II. By Construction:
Geometrically, AB is the diameter of the circle of radius 3 cm as it passes through centre O₁
AB = 2 x 3 = 6 cm.

Question 2.
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segment of the other chord.
Solution:
Given
C (O, r) is a circle in which AB and CD are two equal chords which intersect at P.
To prove:
CP = BP and AP = DP.
Construction:
Draw OE and OF perpendiculars on AB and CD respectively. Join OP.
Proof:
In ∆OPF and ∆OPE,
OP = OP (Common)
OE = OF (∴ AB = CD)
∠F = ∠E (Each 90°)
∆OPF = ∆OPE (By RHS)
and so PE = PF …..(1) (By CPCT)
AB = CD (Given)

\(\frac{1}{2}\) AB = \(\frac{1}{2}\) CD
BE = CF
and AE = DF …..(2)
Adding (1) and (2), we get,
PE + AE = PF + DF
∴ AP = DP
Subtracting (1) and (2) we get,
BE – PE = CF – PF
∴ BP = CP

Question 3.
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Solution:
Given
AB and CD are two equal chords of a circle which intersect at E.
To prove:
∠1 = ∠2
Construction:
Draw OL ⊥ AB and OM ⊥ CD. Join OE.

Proof:
In ∆OLE and ∆OME,
OE = OE
OL = OM (∴ AB = CD)
∠L = ∠M (Each 90°)
∆OLE = ∆OME (By RHS)
and so∠1 = ∠2 (By CPCT)

Question 4.
If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD. (see Fig. adjacent)

Solution:
Given:
C (O, r) and C (O, r) are two concentric circles.
To prove: AB = CD

Construction: Draw OP ⊥ AD.
Proof:
In circle I, AD is the chord and OP ⊥ AD.
AP = DP …(1)
In circle II, BC is the Chord and OP L BC.
∴ BP = CP …(2)
Subtracting (1) and (2), we get
AP – BP = DP – CP
AB = CD

Question 5.
Three girls Reshma. Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Solution:
Given:
OR = OM = 5 m and SR = SM = 6 m.
To find: MR.

Constrution:
Join OR, OM and OS. Draw ON ⊥ SR. In AORS and AOMS,
OS = OS (Common)
RS = MS (Given)
OR = OM (Given)
∆ORS = ∆OMS (By SSS)
and ∠1 = ∠2 (By CPCT)
SP = SP (Common)
SR = SM (Given)
∠1 = ∠2 (Proved)
∆SPR = ∆SPM (By SAS)
and so PR = PM (By CPCT)
and ∠3 = ∠4 (By CPCT)
∠3 + ∠4 = 180° (LPA’s)
2∠3 = 180°
∠3 = \(\frac{180^{\circ}}{2}\) = 90°
ar (∆OSR) = \(\frac{1}{2}\) x OS x PR …(i)
= \(\frac{1}{2}\) x 5 x PR
ON ⊥ SR
RN = \(\frac{1}{2}\) SR
(Perpendicular drawn from the centre of a circle to a chord bisects the chord)
= \(\frac{6}{2}\) = 3m
In ∆ONR ON² = \(\sqrt{O R^{2}-N R^{2}}\) (Using Pythagoras Theorem)
= \(\sqrt{5^{2}-3^{2}}\) = \(\sqrt{4^{2}}\) = 4m
ar (∆OSR) = \(\frac{1}{2}\) x SR x ON
= \(\frac{1}{2}\) x 6 x \(\frac{1}{2}\) x 4 = 12m² …..(ii)
From (i) and (ii), we get
PR = \(\frac{2×12}{2}\) = 4.8m
MR = 2 PR
= 2 x 4.8
= 9.6 m

Question 6.
A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
Solution:
Given: OS = OA = 20 m and AS = SD = AD
To find: AS, SD and AD.
Construction:
Draw AE ⊥ SD. Join OS.
Let AS = SD = AD = 2x (say)

In equilateral ∠ASD, AE ⊥ SD
⇒ E is the mid-point of SD
SE = \(\frac{2x}{2}\) = x
In ∆AES, AS² = AE² + SE²
(2x)² = AE² + x²
4x² – x² = AE²
AE = \(\sqrt{3x^{2}}\) = \(\sqrt{3}\)
OE = AE – AO
= (\(\sqrt{3}\) – 20)m
In ∆OES, OS² = OE² + SE²
(20)² = [(\(\sqrt{3}\)x) 20]² + x²
400 = (\(\sqrt{3}\)x)² – 2 x \(\sqrt{3}\)x × 20 + (20)² + x²
= 3x² + 400 – 40\(\sqrt{3}\)x + x²
400 – 400 = 4x² – 40\(\sqrt{3}\)x
0 = 4x² – 40\(\sqrt{3}\)x
40\(\sqrt{3}\)x = 4x²
40\(\frac { 40\sqrt { 3 } }{ 4 } \) = x
x = 10\(\sqrt{3}\)m .
2x = 2 x 10\(\sqrt{3}\) = 20\(\sqrt{3}\)m
AS = SD = AD = 20\(\sqrt{3}\)m.

MP Board Class 9th Maths Solutions