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MP Board Class 9th Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4

Question 1.
Give the geometric representations of y = 3 as an equation –

1. In one variable
2. in two variables.

Solution:
1. Linear equation in one variable
y = 3

2. Linear equation in two variables is
0x + y – 3 = 6

Question 2.
Give the geometric representation of 2x + 9 = 0 as an equation –

1. In one variable
2. In two variables.

Solution:
1. Linear equation in one variable
2x + 9 = 0
2x = – 9
x = – 4.5

2. Linear equation in two variables
2x + 0y + 9 = 0

Equations of a Line Parallel to the x – axis and y – axis:
This is a special case when the given point lies on the axes, either x – axis or y – axis. If the point lies on x – axis, then y – coordinate will be 0 and if the point lies on y-axis, then the x-coordinate will be 0.

Example 1.
Draw the graph of the equation represented by a straight line which is parallel to the x – axis and at a distance 3 units below it. (NCERT Exemplar)
Solution:
The equation of a line which is parallel to the x-axis and at a distance of 3 units below. It is given by
y = – 3
The solutions of the equation are:

The graph is shown below.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 1.
Draw the graph of each of the following linear equations in two variables:

1. x + y = 4
2. x – y = 2
3. y = 3x
4. 3 = 2x + y

solution:
1. x + y = 4
Take x = 1
1 + y = 4
∴ y = 3

Take x = 2,
2 + y = 4
∴ y = 2

Take x = 0,
0 + y = 4
∴ y = 4

The solutions are:

The graph is shown below.

2. x – y = 2
Take x = 1,
1 – y = 2
y = 2 – 1
∴ y = – 1

Take x = 2,
2 – y = 2
– y = 2 – 2
∴ y = 0

Take x = 0,
0 – y = 2
∴ y = – 2

3. y – 3x
Take x = 0,
y = 3 x 0 = 0

Take x = 1,
y = 3 x 1 = 3

Take x = 2,
y = 3 x 2 = 6

4. 3 = 2x + y
Take x = 1,
3 = 2 x 1 + y
3 = 2 + y
3 – 2 = y
∴ y = 1

Take x = 0
3 = 2 x 0 + y
∴ y = 3 – 0 = 3

Take x = -1,
3 = 2 x – 1 + y
3 = – 2 + y
∴ y = 3 + 2 = 5

Question 2.
Give the equations of two lines passing through (2, 14). How many more such lines are there and why?
Solution:
Two lines passing through point (2, 14)

1. x + y = 16
2. 2x + y – 18

Infinitely many lines can be drawn through (2, 14).

Question 3.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
Putting the value of x = 3 and y = 4 in 3y = ax + 7, we get
3 x 4 = a x 3 + 7
12 = 3a + 7
3a = 12 – 7
a = \(\frac{5}{3}\)

Question 4.
The taxi fare in a city is as follows: For the first kilometer, the fare is ₹ 8 and for the subsequent distance it is ₹ 5 per km. Taking the distance covered as x km and total fare as ₹ y, write a linear equation for this information and draw its graph.
Solution:
Distance covered = x km
Total fare = ₹ y
Fare of 1st km = ₹ 8
Fare for subsequent kms = ₹ 5 per km

According to question, y = 8 + 5 (x – 1) = 8 + 5x – 5
y = 3 + 5x
Take x = 0,
y = 3 + 5 x 0
∴ y = 3

Take x = 1,
y = 3 + 5 x 1
y = 3 + 5
∴ y = 8

Take x = 2
y = 3 + 5 x 2
∴ y = 13

Question 5.
From the choices given below, choose the equation whose graphs are given in Fig. (a) and Fig. (b)
For Fig. (a)

1. y = x
2. x + y = 0
3. y = 2x
4. 2 + 3y = 7x

For Fig. (b)

1. y = x + 2
2. y = x – 2
3. y = – x + 2
4. x + 2y = 6

Solution:
1. Since (-1, 1), (0, 0) and (1, -1) satisfies the equation x + y = 0.
The equation of the graph is x + y = 0

2. Since (-1,3), (0, 2) and (2, 0) satisfies the equation y = x + 2.
The equation of the graph is y = – x + 2.

Question 6.
If the work done by a body on application of a constant force is directly proportional to the distance traveled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance traveled by the body is:

1. 2 units
2. 0 units.

Solution:
Let y be the work done and x be the distance covered. y ∝ x where k is the constant force.
y = kx, (Given)
k = 5
∴ y = 5x

Take x = 0
y = 5 x 0 = 0

Take x = 1
y = 5 x 1 = 5

Take = 2
y= 5 x 2 = 10

When the distance travelled is –

1. x = 2 units y = 10 units
2. x = 0 units y = 0 units.

Question 7.
Yamini and Fatima, two students of class IX of a school, to gether contributed ₹ 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as ₹ x and ₹ y). Draw the graph of the same.
Answer:
Let the contribution of Yamini be ₹ x and that of Fatima be ₹ y.
According to question
x + y = 100

Take x = 30,
30 + y = 100
y = 100 – 30
y = 70

Take = 40,
40 + y = 100
y = 100 – 40
y = 60

Take = 50,
50 + y = 100
y = 100 – 50
y = 50

Question 8.
In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahren-heit to Celsius.
F = (\(\frac{9}{5}\)) C + 32

1. Draw the graph of the linear equation above using Celsius for x – axis and Fahrenheit for y – axis.
2. If the temperature is 30°C, what is the temperature in Fahrenheit?
3. If the temperature is 95°F, what is the temperature in Celsius?
4. If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
5. Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

1. F = \(\frac{9}{5}\) + 32
Take C = – 5,
F = \(\frac{9}{5}\) x (- 5) + 32
= – 9 + 32 = 23
C = – 10,
F = \(\frac{9}{5}\) x – 10 + 32 = 14
C = – 15,
F = \(\frac{9}{5}\) x – 15 + 32 = 5

2. From the graph, when temperature is 30°C, temperature in °F is 86°.

3. When F = 95°, C = 35°.
\(\frac{63×5}{9}\) = 35°

4. C = 0°, F = 32°
F = 0°, C = – 17.7°

5. F = – 40°, C = – 40°

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2

Question 1.
Which one of the following options is true, and why? y = 3x + 5 has

(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions.

Solution:
The true option is (iii) y = 3x + 5 has infinitely many solutions. Reason. For every value of x, there is a corresponding value of y and vice-versa.

Question 2.
Write four solutions for each of the following equations:

1. 2x + y = 7
2. πx + y = 9
3. x = 4y

Solution:
1. 2x + y = 7
Take x = 0
2 x 0 + y = 7
∴ y = 7
First solution is (0, 7)

Take x = 1,
2 x 1 + y = 7
2 + y = 7
y = 7 – 2
∴ y = 5
Second solution is (1, 5)

Take x = 2,
2 x 2 + y = 7
2 + y = 7
y = 7 – 4
∴ y = 3
Third solution is (2, 3)

Take x = 3,
2 x 3 + y = 7
6 + y = 7
y = 1 – 6
∴ y = 1
Fourth solution is (3, 1)

2. πx + y = 9
Take y = 0,
πx + 0 = 9
πx = 9
∴ x = \(\frac{9}{π}\)
First solution is (\(\frac{9}{π}\), 0)

Take y = 1,
πx + 1 = 9
πx =9 – 1
∴ x = \(\frac{8}{π}\)
Second solution is (\(\frac{8}{π}\), 1)

Take y = 2,
πx + 2 = 9
πx = 9 – 2
∴ x = \(\frac{7}{π}\)
Third solution is (\(\frac{7}{π}\), 2)

Take y = 3,
πx + 3 = 9
πx = 9 – 3
∴ x = \(\frac{6}{π}\)
Fourth solution is (\(\frac{6}{π}\), 3)

3. x = 4y
Take x = 0,
0 = 4y
\(\frac{0}{4}\) = y
∴ y = o
First solution is (0, 0)

Take x = 4,
4 = 4y
y = \(\frac{4}{4}\) = 1
Second solution is (4, 1)

Take x = 8,
8 = 4y
y = \(\frac{8}{4}\) = 2
Third solution is (8, 2)

Take x = 12,
12 = 4y
y = \(\frac{12}{4}\) = 3
Fourth solution is (12, 3)

Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not:

1. (0, 2)
2. (2, 0)
3. (4, 0)
4. (\(\sqrt{2}\), 4\(\sqrt{2}\) )
5. (1, 1)

Solution:
x – 2y = 4
1. Putting x = 0 and y = 2, we get
0 – 2 x 2 = 4
4 = 4
∴ (0, 2) is a solution

2. Putting x = 2 and y = 0, we get
2 – 2 x 0 = 4
2 ≠ 4
∴ (2, 0) is not the solution.

3. Putting x = 4 and y = 0, we get
4 – 2 x 0 = 4
4 = 4
∴ (4, 0) is a solution.

4. Putting x = \(\sqrt{2}\) and y = 4\(\sqrt{2}\), we get
\(\sqrt{2}\) – 2 x 4\(\sqrt{2}\) = 4
\(\sqrt{2}\) – 8\(\sqrt{2}\) = 4
– 7\(\sqrt{2}\) ≠ 4
∴ (\(\sqrt{2}\), 4\(\sqrt{2}\)) is not the solution.

5. Putting x = 1 and y = 1, we get
1 – 2 x 1 = 4
1 – 2 = 4
– 1 ≠ 4
∴ (1, 1) is not the solution.

Question 4.
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
Putting x =2 and y – 1 in 2x + 3y – k, we get
2 x 2 + 3 x 1 = k
4 + 3 = k
∴ k = 1

Graph Of A Linear Equation In Two Variables:
The graph of a linear equation in two variables is a line. To draw a line we need atleast two points. Points are the solutions of the given equation. So to find the graph of a linear equation we will first find out three solutions and then plot these points on a suitable scale to a graph to get a line.

Steps for plotting the graph of Linear Equation in Two variables:

1. Write the linear equation.
2. Express y in terms of x.
3. Choose three value of x and calculate the corresponding values of y from the given equation.
4. Tabulate the values of x and y.
5. Plot the value of x on x – axis and value of y on y – axis to a suitable scale, on a graph paper to get three points.
6. Join the three points by a straight line and extend it in both the directions.
7. The line obtained is the graph of the given equation.

Note:
To draw a graph of a linear equation in two variables, atleast, two solutions are required. In this chapter three solutions are taken to draw the graph for better result. Students can draw the graph by taking two solutions also.

MP Board Class 9th Maths Solutions

MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms

Respiration in Organisms Intex Questions

Question 1.
Bhoojho wants to know if cockroaches, snails, fish, earthworms, ants and mosquitoes also have lungs?
Answer:
Yes.

Question 2.
Boojho has seen in television programmes that whales and dolphins often come up to the water surface? They even release a fountain of water sometimes while moving upwards? Why do they do so?
Answer:
Whales and dolphins take in air during inhalation. They exhale out the air on the surface. The water vapour condenses and we see the condensed water vapour as the fountain.

Question 3.
Paheli wants to know whether roots, which are under ground also take in oxygen? If so, how?
Answer:
Yes. Roots take up air from the air spaces preseht between the soil particles.

Activities

Activity – 1
If you try you can count your rate of breathing. Breathe in and out normally. Find out how many times you breathe in and breathe out in a minute? Did you inhale the same number of times as you exhaled? Now count your breathing rate (number of breaths/minute) after brisk walk and after running. Record your breathing rate as soon as you finish and also after complete rest. Tabulate your findings and compare your breathing rates under different conditions with those of your classmates.

Table:
Changes in breathing rate under different conditions:

Activity – 2
Figure shows the various activities carried out by a person during a normal day. Can you say in which activity, the rate of breathing will be the slowest and in which it will be the fastest? Assign numbers to the pictures in the order of increasing rate of breathing according to your experience.

Activity – 3
Take a deep breath. Measure the size of the chest with a measuring tape and record your observations in Table. Measure the size of the chest again when expanded and indicate which classmate shows the maximum expansion of the chest.
Answer:
Effect of breathing on the chest size of some classmates:

Respiration in Organisms Text Book Exercise

Question 1.
Why does an athlete breathe faster and deeper than usual after finishing the race?
Answer:
During the race, the athlete has to run very fast. The demand for energy at that time increases, which increase the demand for more supply of oxygen. Thus, athlete has to breathe faster and deep to inhale more oxygen.

Question 2.
List the similarities and differences between aerobic and anaerobic respiration?
Answer:
Similarities:
Both aerobic and anaerobic respiration produce energy and give out carbon dioxide.

Differences:
Aerobic respiration require oxygen while anaerobic respiration does not require oxygen. In aerobic respiration large amount of energy is released while in anaerobic respiration small amount of energy is released.

Question 3.
Why do we often sneeze when we inhale a lot of dust laden air?
Answer:
We sneeze to get rid of the unwanted particles like dust from air body. It allows only clean and dust free air to enter our body.

Question 4.
Take three test – tubes. Fill \(\frac{1}{2}\)th of each with water. Label them A, B and C. Keep a snail in test – tube A, a water plant in test – tube B and in C, keep snail and plant both. Which test – tube would have the highest concentration of CO₂?
Answer:
Test – tube A.

Question 5.
Tick the correct answer:

Question (a)
In cockroaches, air enters the body through?
(a) lungs
(b) gills
(c) spiracles
(d) skin.
Answer:
(c) spiracles

Question (b)
During heavy exercise, we get cramps in the legs due to the accumulation of?
(a) carbon dioxide
(b) lactic acid
(c) alcohol
(d) water.
Answer:
(b) lactic acid

Question (c)
Normal range of breathing rate per minute in an average adult person at rest is?
(a) 9 – 12
(b) 15 – 18
(c) 21 – 24
(d) 30 – 33
Answer:
(b) 15 – 18

Question (d)
During exhalation, the ribs?
(a) move outwards
(b) move downwards
(c) move upwards
(d) do not move
Answer:
(b) move downwards

Question 6.
Match the items in Column I with those in Column II:

Answer:

(a) – (iii)
(b) – (iv)
(c) – (i)
(d) – (v)
(e) – (ii)
(f) – (vi)

Question 7.
Mark if the statement is true and if it is false:

1. During heavy exercise the breathing rate of a person slows down.
2. Plants carry out photosynthesis only during the day and respiration only at night.
3. Frogs breathe through their skins as well as their lungs.
4. The fishes have lungs for respiration.
5. The size of the chest cavity increases during inhalation.

Answer:

1. False
2. False
3. True
4. False
5. True

Question 8.
Given below is a square of letters in which are hidden different words related to respiration in organisms. These words may be present in any direction – upwards, downwards, or along the diagonals. Find the words for your respiratory system. Clues about those words are given below the square?

1. The air tubes of insects
2. Skeletal structures surrounding chest cavity
3. Muscular floor of chest cavity
4. Tiny pores on the surface of leaf
5. Small openings on the sides of the body of an insect
6. The respiratory organs of human beings
7. The openings through which we inhale
8. An anaerobic organism
9. An organism with tracheal system

Answer:

1. Trachea
2. Rib
3. Diaphragm
4. Stomats
5. Spiracles
6. Lung
7. Strils
8. Yeast
9. Ant.

Question 10.
The mountaineers carry oxygen with them because:
(a) At an altitude of more than 5 km there is no air.
(b) The amount of air available to a person is less than that available on the ground.
(c) The temperature of air is higher than that on the ground.
(d) The pressure of air is higher than that on the ground.
Answer:
(b) The amount of air available to a person is less than that available on the ground.

Extended Learning – Activities and Projects

Question 1.
Observe fish in an aquarium. You will find flap like structures on both sides of their heads. These are flaps which cover the gills. These flaps open and close alternately. On the basis of these observations, explain the process of respiration in the fish?
Answer:
Do with the help of your subject teacher.

Question 2.
Visit a local doctor. Learn about the harmful effects of smoking. You can also collect material on this topic from other sources. You can seek help of your teacher or parents. Find out the percentage of people of your area who smoke. If you have a smoker in your family, confront him with the material that you have collected?
Answer:
Do yourself.

Question 3.
Visit a doctor. Find out about artificial respiration? Ask the doctor:
(a) When does a person need artificial respiration?
(b) Does the person need to be kept on artificial respiration temporarily or permanently?
(c) From where can the person get supply of oxygen for artificial respiration?
Answer:
Do yourself.

Question 4.
Measure the breathing rate of the members of your family and some of your friends? Investigate:
(a) If the breathing rate of children is different from that of adults?
(b) If the breathing rate of males is different from that of females?
If there is a difference in any of these cases, try to find the reason?
Answer:
Do with the help of your parents.

Respiration in Organisms Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (a)
The life processes that provide energy are?
(a) respiration
(b) nutrition
(c) both respiration and nutrition
(d) none of these.
Answer:
(c) both respiration and nutrition

Question (b)
In …………………… respiration, there is an exchange of gases between the cells and the blood?
(a) aerobic
(b) anaerobic
(c) external
(d) internal.
Answer:
(d) internal.

Question (c)
In the cell, the food (glucose) is broken down into carbon dioxide and water using?
(a) hydrogen
(b) nitrogen
(c) oxygen
(d) none of these.
Answer:
(c) oxygen

Question (d)
Which of the following is not a feature of respiration?
(a) involvement of enzymes
(b) occur outside the cells
(c) release of energy
(d) is a chemical process.
Answer:
(d) is a chemical process.

Question (e)
During heavy exercise, the breathing rate in an adult can increase upto?
(a) 25 times per minute
(b) 30 times per minute
(c) 35 times per minute
(d) none of these.
Answer:
(a) 25 times per minute

Question (f)
The percentage of oxygen and carbon dioxide in inhaled air is –
(a) 21%, 0.04%
(b) 20%, 0.10%
(c) 21%, 1.0%
(d) 20%, 1.0%.
Answer:
(a) 21%, 0.04%

Question (g)
The percentage of oxygen and carbondioxide in exhalted air is –
(a) 16.4%, 4.4%
(b) 21%, 1.0%
(c) 16.4%, 3.4%
(d) 16.4%, 0.04%.
Answer:
(a) 16.4%, 4.4%

Question 1.
Fill in the blanks:

1. All living organisms require ………………………… to perform various life process.
2. The liver and ……………………… are found near the stomach.
3. The exhaled air has a higher percentage of carbon dioxide as compared to the ……………………………… air.
4. Breathing is a process at organ levels, whereas respiration is a ……………………….. process.
5. When breakdown of glucose occurs with the use of oxygen, it is called …………………….. respiration.
6. The taking in of air rich in oxygen into the body is called …………………………
7. The number of times a person breathes in a minute is termed as the ………………………….
8. Lungs are present in the ……………………….. cavity.
9. During inhalation, ribs move up and outwards and diaphragm moves ………………………..
10. Insects have a network of air tubes called ……………………….. for gas exchange.
11. Like all other living cells of the plants, the root cells also need oxygen to ………………………. energy.

Answer:

1. energy
2. pancreas
3. inhaled
4. cellular
5. aerobic
6. inhalation
7. breathing rate
8. chest
9. down
10. tracheae
11. generate.

Question 3.
Which of the following statements are true (T) or false (F):

1. Respiration is a type of combustion at ordinary temperature.
2. Breathing is a process that takes place at the cellular level.
3. Oxygen is released during the process of respiration.
4. During respiration the plants – take CO₂ and release Or
5. Respiration involves on exchange of gases.
6. Cellular respiration takes place in the cells of all organisms.
7. Anaerobic respiration do not takes places in the muscle cells to fulfill the demand of energy.
8. The giving out of air rich in carbon dioxide is known as exhalation.
9. A breathe means one inhalation plus one exhalation.
10. On an average, an adult human being at rest breathes in and out 15 to 18 times in a minute.
11. A cockroach has small openings on the sides of its body.
12. Gills are not supplied with blood vessels for exchange of gases.
13. The end product of anaerobic respiration are carbon dioxide and water.
14. In earthworm, the exchange of gases occurs through the moist skin.

Answer:

1. True
2. False
3. False
4. True
5. True
6. True
7. False
8. True
9. True
10. True
11. True
12. False
13. True
14. True

Respiration in Organisms Very Short Answer Type Questions

Question 1.
Where does cellular respiration takes place?
Answer:
Cells of organisms.

Question 2.
Write the equation for breakdown of food in anaerobic respiration?
Answer:

Question 3.
Write the equation for breakdown of food in aerobic respiration?
Answer:

Question 4.
What are anaerobes?
Answer:
There are some organisms such as yeast that can service in the absence of air. They are called anaerobes.

Question 5.
What is yeast?
Answer:
Yeast is single – celled organisms.

Question 6.
Write the uses of yeast?
Answer:
Yeast respire anaerobically and during this process yield alcohol. So, they are used to make beer and wine.

Question 7.
Which chemical reaction takes place in internal respiration?
Answer:
Glucose + Oxygen → Carbon dioxide + Water + Energy.

Question 8.
Give an example of an oxygen respiration?
Answer:
In human beings.

Question 9.
Name the parts of digestive system of humans?
Answer:
The parts of digestive systems are mouth, oesophagus, stomach, intestine and anus.

Question 10.
Name two processes of respiration?
Answer:
Inhalation and exhalation are the two processes of respiration.

Question 11.
Name the parts of respiratory system of human?
Answer:

1. Nostrils
2. Trachea
3. Lungs with alveali, and
4. Diaphragm.

Question 12.
What waste materials are produced during respiration?
Answer:
Carbon dioxide is produced as the waste material during respiration.

Question 13.
Define respiration?
Answer:
The process of breaking down of food by using oxygen, to form carbon dioxide and release energy required for various life activities, is called respiration.

Question 14.
Name the fuels used for the production of energy during respiration?
Answer:
Glucose is oxidized to give out energy.

Question 15.
Which organs of plants participate in respiration?
Answer:
There is no special organ in plants for breathing.

Question 16.
Write the names of organs in human respiratory system in sequence?
Answer:
Nostrils → Nasal cavity → Pharynx → Tracheae → Lungs.

Question 17.
What is importance of hairs present in the noise?
Answer:
These small hairs present in nose act as filters. These prevent dust particles and harmful germs to enter into respiratory track.

Question 18.
Name the organs of the body from which blood freshly enriched with oxygen goes into the heart?
Answer:
The lungs helps the blood to get freshly enrichment of oxygen.

Question 19.
What happens during breathing?
Answer:
During breathing, oxygen enriched air is inhaled which reaches lungs. Here, oxygen centers blood and unwanted water vapour and carbon dioxide are released out during breathing.

Question 20.
What is breathing?
Answer:
The process of taking oxygen and leaving of carbon dioxide during respiration is called breathing.

Question 21.
Write the name of gases which are involved in breathing?
Answer:
Carbon dioxide and oxygen.

Question 22.
How will you prove that we exhale CO₂ gas during respiration?
Answer:
Pass the exhaled air given out by us into lime water. It will turn milky in colour. We know that CO₂ gas turn lime water milky This confirms that we exhale CO₂ gas in respiration.

Question 23.
What do we exhale?
Answer:
We exhale air rich in carbon dioxide.

Question 24.
Do we exhale only carbon dioxide or a mixture of gases along with it?
Answer:
We exhale a mixture of gases along with carbon dioxide.

Question 25.
How do ribs and diaphragm move during inhalation?
Answer:
During inhalation, ribs move up and outwards and diaphragm moves down.

Question 26.
Which is the respiratory organ for earthworm?
Answer:
Skin.

Question 27.
Can we survive in water?
Answer:
No.

Question 28.
How do fish breathe under water?
Answer:
Gills in fish help them to use oxygen dissolved in water. Gills are projections of the skin. Gills are well supplied with blood vessels for exchange of gases.

Question 29.
What is the function of gills?
Answer:
The fishes and other aquatic animals respire through gills or similar structure.

Question 30.
Can you guess what would happen if a potted plant is over watered?
Answer:
The roots will not get air to respire, so the roots will die and hence the whole plant will also die.

Respiration in Organisms Short Answer Type Questions

Question 1.
What is respiration?
Answer:
All the living organisms perform a number of vital activities. The energy is obtained by the oxidation of food or respiration. When the living organisms are completely at the stage of rest, even then they require some minimum amount of energy for the maintainance of cells and tissues. Thus, respiration is one of the most important process for living organisms.

Question 2.
Define respiration with the help of chemical equation?
Answer:
The process in which the oxidation of absorbed food is takes place by the CO₂ which is inhaled by breathing and energy is released out is called respiration. Chemical equation of nutrition is as follows:
C₆H₁₂O₂ + 6O₂ → 6CO₂ + 6H₂O + 673 kcal (Energy)

Question 3.
Why does our body need a transporting system?
Answer:
Our body needs a transporting system to:

1. Transport oxygen to body cells from lungs.
2. Transport food to body cells from liver.
3. Transport waste material from body cells to excretory organs.
4. Maintains body temperature constant.

Question 4.
Write difference between oxy – respiration and anoxy – respiration?
Answer:
The differences between the oxy – respiration and anoxy – respiration:
oxy – respiration:

1. It takes place in presence of O₂
2. The end products are CO₂ and H₂O.
3. The energy released is more.

Anoxy – respiration:

1. It takes place without oxygen.
2. The end products are ethyl alcohol and CO₂.
3. The energy released

Question 5.
Describe the various types of respiratory organs found in animals?
Answer:
In animals, there are definite respiratory organs for exchange of gases.

1. In earthworm and leech exchange of gases takes place through moist, thin and vascular skin.
2. In insects the trachea are the repiratory organs.
3. In fishes the gills are the respiratory organs.
4. Higher animals like mammal and birds including man have lungs for respiration.

Question 6.
Describe the importance of respiration in plants?
Answer:

1. It takes place in the presence of oxygen.
2. It is completed in cytoplasm and mitochondria of cell.
3. It involves the complete oxidation of glucose into CO_2, and
4. H₂O + C₂H₂O₂ + 6O₂ → 6CO₂ + energy + 6H₂O.
5. It occurs in all the living cell of the organisms.
6. It is day night process.
7. Energy is released in this process.

Question 7.
Name the organs associated with the following functions:

1. digestion
2. absorption of minerals
3. respiration, and
4. excretion of carbon – dioxide in man.

Answer:

1. Digestion. Mouth, stomach, oesophagus, pharynx, small intestine, large intestine.

2. Absorption of minerals.
In Animals : Small intestine
In plants : Root hairs.

3. Respiration.
In Animals : Nose, trachaea, larynx, lung.
In plants. Stomata and lenticells.

4. Excretion of carbon – dioxide.
In Animals : Kidneys, ureters, urinary bladder, urethra.

Question 8.
State the difference between respiration and breathing?
Answer:
Differences between Respiration and Breathing:
Respiration:

1. It takes place inside the cells.
2. The exchange of gases takes place between blood and the tissues of the body.
3. In this process nutrients are oxidised to liberate energy.

Breathing:

1. It takes place at the surface of the respiratory organs.
2. The exchange of gases takes place between the blood and the external environment.
3. The nutrients are not oxidised to liberate energy.

Question 9.
Name the major components of urine?
Answer:
The kidney, ureter, bladder, and urethra are the organs used in the removal of urine from the body. Renal artery carry urea and uric acid along with the large amount of water with the blood into the kindneys, when this blood enters into glomerulus, the solid wastes filter here while the water is diffused out from the network of blood capillaries into the uriniferous tubules. The ques mixture is called as urine.

Question 10.
If a person drinks very little water per day? The volume of urine decreases? In what ways does it affect the body?
Answer:
If a person drinks very little water per day. The volume of urine decreases. Large amount of water will dissolve large quantity of urea in it and large amount of urine will pass out from the body. If someone drinks lesser amounts of water, the concentration of urea in the cells and its larger quantity is very harmful for the body.

Question 11.
What is saliva? What are the functions of saliva?
Answer:
Saliva is a digestive secretion produced by three layered salivary glands, the paroted submaxillary and sunblingual present in our mouth cavity. This soften and lubricants food for easier swallowing and converts starch into reducing sugars. The salivary amylase. enzyme of saliva acts of starch in a neutral medium.

Question 12.
How does the food digested in the stomach?
Answer:
After some time from mouth the food reach inside the stomach. The gastric glands of stomach secrete the gastric juices. The gastric juice containes three enzymes. These are pepsin, renin and HCI. The HCI makes the medium acidic, and inhibites the bacterial growth and prevents the food. The renin curdiles the milk protein to be hydrololysed by pepsin. The pepsin reacts with proties and changes into peptides.

Respiration in Organisms Long Answer Type Questions

Question 1.
Define the respiratory organs in animals?
Answer:
The process of respiratory system in animals possess following organs:

1. Nasal cavity
2. Larynx
3. Trachea
4. Bronchi
5. Lungs.

In animals some organs like gills and lungs are developed for the purpose of exchange of gases. The amount of CO₂ produced after respiration cannot be utilised by animals as in plants. This is also true for the production of oxygen which is required for respiration. Mitochondria are the site of respiration in both plants and animals.

The process of break – down of glucose in the presence of oxygen and some enzymes into CO₂. And water, which is accompanied with release of energy is very complex. Materials like proteins and fats are also consumed during respiration to produce energy.

Question 2.
Write difference between photosynthesis and respiration?
Answer:
The difference between respiration and photosynthesis are:
Photosynthesis:

1. It takes place only in green plants.
2. It requires energy.
3. It requires CO₂ and H₂O.
4. It releases oxygen and make food.
5. It is a building up process.
6. It takes place in the chloroplast of the plant cell.

Respiration:

1. It takes place in all plants and animals.
2. It releases energy.
3. It releases CO₂ and H₂O.
4. It requires oxygen and oxidise the food.
5. It is a breaking down process.
6. It takes place in mitochondira of a cell.

Question 3.
Draw the labelled diagram to show respiratory system in man?
Answer:
The process of respiration is aimed to release energy

Question 4.
What is the difference in the amount of carbon choxide in the inhaled and the exhaled air? How will you test the presence of CO₂ in the exhaled air?
Answer:
Excess carbon dioxide is present in exhaled air as compared inhaled air.

Test to indicate the presence of CO₂ in the exhaled air:
Take two test tubes. Fill each of them half with freshly prepared lime water. Fix stoppers with two holes in both the test tubes. Insert glass tubes in both the stoppers. The lime water through which exhaled air is passed turns milky. This shows that more CO₂ is present in exhaled air.

Question 5.
Draw diagrams to show movements of rib and diaphragm during breathing?
Answer:

Question 6.
How do the following organism breathe? Amoeba, fish, frog, grasshopper, earthworm?
Answer:
Amoeba:
Amoeba breathes by diffusion of gases in between body surface and the water.

Fish:
Fish breathe with gills. The gills are special organs. They help fish to extract dissolved oxygen from the water.

Frog:
Frog can breathe through skin and lungs. In water if breathes through skin whereas in air through lungs.

Grasshopper:
The grasshopper and other insects have holes and air tubes those help them to breathe.

Earthworm:
It breathes through its moist body surface.

Question 7.
Describe the breathing in cockroach with diagram?
Answer:
A cockroach has small openings on the sides of its body. Other insects also have similar openings. These openings are called spiracles, bisects have a network of air tubes called tracheae for gas exchange. Oxygen rich air rushes through spiracles into the tracheal tubes, diffuses into the body tissue, and reaches every cell of the body. Similarly, carbon dioxide from the cells goes into the tracheal tubes and moves out through spiracles. These air tubes or tracheae are found only in insects and not in any other group of animals.

MP Board Class 7th Science Solutions

MP Board Class 9th Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.1

Question 1.
The cost of notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
Solution:
Let cost of pen be ₹ x and cost of a notebook be ₹ y
y = 2x
y – 2x = 0.

Question 2.
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

1. 2x + 3y = 9.3\(\overline { 5 } \)
2. x – \(\frac{y}{5}\) – 10 = 0
3. -2x + 3y = 6
4. x = 3y
5. 2x = – 5y
6. 3x + 2 = 0
7. y – 2 = 0
8. 5 = 2x

Solution:
1. 2x + 3y = 9.3\(\overline { 5 } \)
2x + 3y – 9.3\(\overline { 5 } \) = 0
a = 2, b = 3, c = – 9.3\(\overline { 5 } \)

2. x – \(\frac{y}{5}\) – 10 = 0
a = 1, b = – \(\frac{1}{5}\), c = – 10

3. -2x + 3y = 6
– 2x + 3y – 6 = 0
a = – 2, b = 3, c = – 6

4. x = 3y
1. x – 3y + 0 = 0
a – 1, b = – 3, c = 0

5. 2x = – 5y
2x + 5y + 0 = 0
a = 2, b = 5, c = 0

6. 3x + 2 = 0
3x + 0y + 2 = 0
a = 3, b = 0, c = 2

7. y – 2 = 0
0x + y – 2 = 0
a = 0, b = 1, c = – 2

8. 5 = 2x
– 2x + 0y + 5 = 0
a = – 2, b = 0, c = 5

Solution of a Linear Equation:
Consider a Linear equation x + 2y = 6
Let x = 2 and y = 2.
Then L.H.S. of the equation = x + 2y = 2 + 2 x 2 = 6
and R.H.S. of the equation = 6 (given)
i.e., LHS. = R.H.S. for x = 2 and y = 2.
Therefore, x = 2 and y = 2 i.e., (2, 2) is the solution of the given equation x + 2y = 6.
Any pair of values of x and y which satisfies the given equation is called a solution of the equation. A linear equation in two variables has infinitely many solutions.

Note:
To find the solution of an equation, assure a value of one of the variable and calculate the value of second variable from the given equation.

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 3 Atoms and Molecules

Atoms and Molecules Intext Questions

Atoms and Molecules Intext Questions Page No. 32 – 33

Question 1.
In a reaction 5.3g of sodium carbonate reacted with 6g of ethanoic acid. The products were 2.2g of carbon dioxide. 0. 9g water and 8.2g of sodium ethanoate. Show that these observations are in agreement with the law of co serration of mass.
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Answer:
The reaction is,
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
5.3g + 6g → 8.2g + 2.2g + 0.9g
Now,
Total mass of reactants = (5.3 + 6)g = 11.3g
And, total mass of products = (8.2 + 2.2 + 0.9)g = 11.39g
So, Mass of reactants = Mass of product
It shows the law of conservation of mass.

Question 2.
Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3g of hydrogen gas?
Answer:
Ratio of hydrogen and oxygen in water = 1 : 8
So, oxygen is 8 times that of hydrogen by mass.
Let, xgrams of oxygen will react with 3g of hydrogen.
Then,
1 : 8 = 3 : x
x = 8 × 3
x = 24g
∴ 24g of oxygen gas required.

Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer:
Postulate of Dalton’s theory based on law of conservation of mass is “Atoms are indivisible particles which can neither be created nor be destroyed in a chemical reaction.”

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:
Postulate is the relative number and kinds of atoms remain constant in a given compound.

Atoms and Molecules Intext Questions Page No. 35

Question 1.
Define the atomic mass unit.
Answer:
Atomic mass unit is the mass unit equal to \(\frac { 1 }{ 12 }\)th mass of one carbon-12 atom.

Question 2.
Why is it not possible to see an atom with naked eyes?
Answer:
The size of an atom is very very small that we can see with naked eyes. The size of an atoms lies in nano meters (nm).

Atoms and Molecules Intext Questions Page No. 39

Question 1.
Write down the formulae of:
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide.
Answer:
(i) Sodium Oxide

• Symbol → NaO
• Charge → +1-2
• Formula → Na₂O

(ii) Aluminium Chloride

• Symbol → AlCl
• Charge → +3-1
• Formula → AlCl₃

(iii) Sodium Sulphate

• Symbol → NaS
• Charge → +1-2
• Formula → Na₂S

(iv) Magnesium Hydroxide

• Symbol → MgOH
• Charge → +2-1
• Formula → 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

Answer:

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?
Answer:
It is the representation of composition of a compounds in the form of symbols of elements present in it.

Question 4.
How many atoms are present in a:

1. H₂S molecule and
2. PO₄^3- ion?

Answer:

1. H₂S Molecule -2 atoms of H + 1 atom of S = Total 3 atoms.
2. PO₄^3- 1 atom of phosphorus + 4 atoms of oxygen total 5 atoms.

Atoms and Molecules Intext Questions Page No. 40

Question 1.
Calculate the molecular masses of:

1. H₂
2. O₂
3. Cl₂
4. CO₂
5. CH₄
6. C₂H₆
7. C₂H₄
8. NH₃
9. CH₃OH.

Answer:

1. H₂ = (2 × 1)u = 2u
2. O₂ = (2 × 16)u = 32u
3. Cl₂ = (2 × 35.5)u = 71u
4. CO₂ = (1 × 12 + 2 × 16)u = (12 + 32)u = 44u
5. CH₄ = (1 × 12 + 4 × 1)u = (12 + 4)u = 16u
6. C₂H₆ = (2 × 12 + 6 × 1)u = (24 + 6)u = 30u
7. C₂H₄= (2 × 12 + 4 × 1)u = (24 + 4)u = 28u
8. NH₃ = (1 × 14 + 3 × 1)u = (14 + 3)u = 17u
9. CH₃OH = (1 × 12 + 3 × 1 + 1 × 16 + 1 × 1)u = (12 + 3 + 16 + 1)u = 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.
Answer:
Formula unit mass of

1. ZnO = (1 × 65 + 1 × 16)u = (65 + 16)u = 81u
2. Na₂O = (2 × 23 + 1 × 16)u = (46 + 16)u = 62u
3. K₂CO₃ = (2× 39 + 1 × 12 + 3 × 6)u = (78 + 12 + 48)u = 138u

Atoms and Molecules Intext Questions Page No. 42

Question 1.
If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?
Answer:
1 mole of carbon atoms = 6.022 × 10²³ atoms
Also, 1 mole of carbon atoms = 12g
6.022 × 10²³ atoms of carbon weigh = 12g
1 atom of carbon weigh = 1.99 × 10²³g.

Question 2.
Which has more number of atoms, 100 grams of sodium or 100 grams of ion (given atomic mass of Na = 23u, Fe = 56u)?
Answer:
1 mole of sodium = 23g of Na
Atoms = 6.022 × 10²³ atoms
23g of Na = 6.022 × 10²³ atoms
100g of Na = \(\frac { 100 }{ 23 }\) × 6.022 × 10²³ = 2.617 × 10²⁴ atoms 8 23
Now, 1 mole of iron atoms = 56g of Fe = 6.022 × 10²³ atoms
56g of Fe = 6.022 × 10²³ atoms
100g of Fe = \(\frac { 100 }{ 56 }\) × 6.022 × 10²³ = 1.075 × 10²⁴ atoms
So, 100 g of Na contains more atoms.

Atoms and Molecules NCERT Textbook Exercises

Question 1.
A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.
Answer:
Percentage composition of boron

Percentage composition of oxygen

Question 2.
When 3.0g of carbon is burnt in 8.00g oxygen, 11.00g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00g of oxygen? Which law of chemical combination will govern your answer?
Answer:
We know that 3g of carbon is burnt in 8g of oxygen to form 11g of carbon dioxide.
When 3g of carbon is burnt in 50g of oxygen, then 11g of carbon dioxide will be formed.
And oxygen remain unreacted, (50 – 8)g = 42g.
Law of constant proportion governs here.

Question 3.
What are polyatomic ions? Give examples.
Answer:
The ions which contain more than one type of atoms (same kind or different kinds) as a single unit are called polyatomic ions.
Examples:

• Sulphate ion (SO₄^-2)
• Nitrate ion (NO₃^-2)
• Carbonate ion (CO₃^-2).

Question 4.
Write the chemical formulae for the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Answer:
(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
Answer:
(a) Quick lime (calcium oxide): Elements: Calcium and oxygen.
(b) Hydrogen bromide: Elements: Hydrogen and bromide.
(c) Baking powder (sodium hydrogen carbonate) Elements: Sodium, hydrogen, carbon, oxygen.
(d) Potassium sulphate: Elements: 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.
Answer:
(a) Ethyne, C₂H₂ = (2 × 12 + 2 × 1)g = (24 + 2)g = 26g
(b) Sulphur molecules, S₈ = (8 × 32)g = 256g
(c) Phosphorus molecule, P₄ = (4 × 31)g = 124g
(d) Hydrochloric acid, HCl = (1 × 1 + 1 × 35.5)g = (1 + 35.5)g = 36.5g
(e) Nitric acid, HNO₃ = (1 × 1 + 1 × 14 + 3 × 16)g = (1 + 14 + 48)g = 63g

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)?
Answer:
(a) 1 Mole of nitrogen atoms = 14g.

(b) Mass of 1 mole of aluminium atoms = 27g.
So, mass of 4 moles of aluminium atoms = (4 × 24)g = 108g.

(c) Mass of 1 mole of sodium sulphide
= (2 × 23 + 32 × 1 + 3 × 16)g = (46 + 32 + 48)g = 126g
∴ Mass of 10 moles of Na₂SO₃ = (126 × 10)g = 1260g.

Question 8.
Convert into mole:
(a) 12g of oxygen gas
(b) 20g of water
(c) 22g of carbon dioxide.
Answer:
(a) Mass of oxygen gas = 12g
Now, 1 mole of oxygen gas = 32g moles mass
So, 32g molar mass of oxygen = 1 moles
∴ 1g of oxygen gas = \(\frac { 1 }{ 32 }\) moles
and 12 g of oxygen gas = \(\frac { 1 }{ 32 }\) moles × 12 moles = 0.375 moles.

(b) 1 Mole of water = 18g molar mass
So, 18g moles mass of water = 1 mole
∴ 1g of water = \(\frac { 1 }{ 18 }\) moles
and 20 g of water = \(\frac { 1 }{ 18 }\) moles × 20 moles = 1.11 moles.

(c) 1 Mole of carbon dioxide = 44g molar mass
So, 44g molar mass of = 1 mole carbon dioxide
∴ 22g of carbon dioxide = \(\frac { 22 }{ 44 }\) moles = 0.5 mole.

Question 9.
What is the mass of
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer:
(a) 1 mole of oxygen atoms = 16g
∴ 0.2 moles of oxygen atoms = (16 × 0.2)g = 3.2g

(b) 1 mole of water molecules = 18g
∴ 0.5 mole of water molecule = (18 × 0.5)g = 9g.

Question 10.
Calculate the number of molecules of sulphur (S₈) present in 16g of solid sulphur.
Answer:
256g of sulphur = 1 mole of sulphur molecules
So, 16g of sulphur = \(\frac { 1 }{ 16 }\) × 16 moles of sulphur molecules
= 0.0625 mole of sulphur molecule
Also, 1 mole of sulphur molecules = 6.023 × 10²³ molecules
So, 0.0625 moles of sulphur molecules = 6.025 × 10²³ × 0.0625 molecules
= 3.76 × 10²² molecules.

Question 11.
Calculate the number of aluminium ions present in 0.051g 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 = 27u).
Answer:
Molar mass of aluminium oxide (Al₂O₃) = (2 × 27)g + (3 × 16)g
= (54 + 48)g = 102g
∴ 10²g of aluminium = 6.022 × 10²³ oxide contains aluminium ions
So, 0.051g Al₂O₃ contains aluminium ions = 6.022 × 10²⁰ aluminium ions.

Atoms and Molecules Additional Questions

Atoms and Molecules Multiple Choice Questions

Question 1.
Atomic theory of matter was proposed by.
(a) Newton
(b) John Dalton
(c) Rutherford
(d) Lavoisier.
Answer:
(b) John Dalton

Question 2.
Law of Conservation of mass was given by.
(a) Lavoisier
(b) Dalton
(c) Kennedy
(d) Faraday.
Answer:
(a) Lavoisier

Question 3.
The term mole was first introduced by.
(a) Grahm
(b) Dalton
(c) Ostwald
(d) Boyle.
Answer:
(c) Ostwald

Question 4.
Atomic radius of an atom is measured in.
(a) Micrometre
(b) Millimetre
(c) Nanometre
(d) Centimetre.
Answer:
(c) Nanometre

Question 5.
Law of constant proportions was proposed by.
(a) Dalton
(b) Bezelius
(c) Proust
(d) Lavoisier.
Answer:
(c) Proust

Question 6.
Latin name of an atom is argentum. The English name of this element is.
(a) Argon
(b) Gold
(c) Silver
(d) Mercury.
Answer:
(c) Silver

Question 7.
Phosphorus molecule is.
(a) Diatomic
(b) Triatomic
(c) Tetra – atomic
(d) Mono – atomic.
Answer:
(c) Tetra – atomic

Question 8.
The atom chosen for reference for measuring atomic masses is.
(a) C-14
(b) C-12
(c) H-2
(d) O-12.
Answer:
(b) C-12

Question 9.
IUPAC is.
(a) Indian Union of Pacific and Applied Chemistry.
(b) International Union of Permanent and Applied Chemicals.
(c) International Union of Pure and Applied Chemistry.
(d) Indian Union of Pure and Applied Chemistry.
Answer:
(c) International Union of Pure and Applied Chemistry.

Question 10.
Atomic mass is measured in.
(a) Grams
(b) Atomic mass radius
(c) Centigrams
(d) Atomic mass unit.
Answer:
(d) Atomic mass unit.

Question 11.
1 A.M.U. is equal to.
(a) 1.65 × 10^-23g
(b) 1.63 × 10^-25g
(c) 1.66 × 10^-24g
(d) 1.66 × 10^-22g.
Answer:
(c) 1.66 × 10^-24g

Question 12.
The naming of elements from first or first and second letter was introduced by.
(a) Berzellius
(b) Dalton
(c) Proust
(d) Lavoisier.
Answer:
(a) Berzellius

Question 13.
The combining capacity of an element is called.
(a) Atomicity
(b) Valency
(c) Reactivity
(d) None of these.
Answer:
(b) Valency

Question 14.
The cation of element has.
(a) More electrons than normal atom
(b) Equal electrons than normal atom
(c) Less electrons than normal atom
(d) Equal proton than normal atom.
Answer:
(c) Less electrons than normal atom

Question 15.
The formula of a compound is A₅B₄. Then valency of A and B will be.
(a) 5 and 4
(b) 5 and 9
(c) 4 and 9
(d) 4 and 5.
Answer:
(d) 4 and 5.

Question 16.
1 Mole has.
(a) 6.012 × 10²³ particles
(b) 6.022 × 10²³ particles
(c) 6.022 × 10²² particles
(d) 6.022 × 10²⁵ particles
Answer:
(b) 6.022 × 10²³ particles

Question 17.
Which of the following has the maximum number of atoms?
(a) 18g of CH₄
(b) 18g of H₄O
(c) 18g of CO₂
(d) 18g of O_2.
Answer:
(a) 18g of CH₄

Question 18.
Atomic mass of C₆H₁₂O₆ is.
(a) 24
(b) 80
(c) 100
(d) 12
Answer:
(a) 24

Question 19.
1 atomic mass unit is equal to.
a) \(\frac { 1 }{ 14 }\) mass of a C – 12 atom
(b) \(\frac { 1 }{ 16 }\)mass of a C – 12 atom
(c) \(\frac { 1 }{ 18 }\) mass of a C – 12 atom
(d) \(\frac { 1 }{ 12 }\)mass of a C – 12 atom.
Answer:
(d) \(\frac { 1 }{ 22 }\)mass of a C – 12 atom.

Question 20.
Atomicity of Sulphate (SO₄^2-) ion.
(a) Mono – atomic
(b) Tri – atomic
(c) Poly – atomic
(d) Tetra – atomic
Answer:
(c) Poly – atomic

Question 21.
Formula of number of moles of a substance is.

Answer:

Atoms and Molecules Very Short Answer Type Questions

Question 1.
Name the scientist who gave atomic theory of matter.
Answer:
John Dalton.

Question 2.
Name the scientist who gave the law of conservation of mass.
Answer:
Antoine Lavoisier

Question 3.
Name the scientist who gave the law of constant proportions.
Answer:
Joseph Proust.

Question 4.
How many metres are in 1nm?
Answer:
1nm = 10^-9m.

Question 5.
Give two examples of polyatomic molecules of elements.
Answer:
P₄ (Phosphorus) and S₈ (Sulphur).

Question 6.
Write full form of IUPAC.
Answer:
International Union of Pure and Applied Chemistry.

Question 7.
An element A and B has valency of 3 and 4. Write its chemical formula.
Answer:
A₄B_3.

Question 8.
Name the scientist who introduced the term ‘mole’.
Answer:
Wilhem Ostwald.

Question 9.
How many particles exist in 1 mole of atom?
Answer:
1 mole = 6.022 × 10²³ particles.

Question 10.
Write the formula of aluminium sulphate.
Answer:
Al₂(SO₄)₃.

Question 11.
Name the charged particles formed by gaining of electrons.
Answer:
Anion.

Question 12.
Name the charged particles formed by loosening of electrons.
Answer:
Cation.

Question 13.
What is numerical value of Avogadro number or Avogadro’s constant?
Answer:
6.022 × 10²³.

Question 14.
Write the Latin names of iron, gold and copper.
Answer:

1. Iron – Ferrum
2. Gold – Aurum
3. Copper – Cuprum.

Question 15.
What is 1 amu?
Answer:
1 amu = \(\frac { 1 }{ 2 }\)th mass of a carbon-12 atom.

Question 16.
Calculate the number of atoms of oxygen present in its 3.5 moles.
Answer:
1 mole of oxygen atoms = 6.022 × 10²³
∴ 3.5 moles of oxygen atoms = 3.5 × 6.022 × 10²³ atoms
= 2.10 × 10²⁴ atoms.

Atoms and Molecules Short Answer Type Questions

Question 1.
Define:
(a) Law of conservation of mass.
(b) Law of constant proportions.
(c) Atomic mass.
(d) Molecules of element.
(e) Molecules of compound.
(f) Atomicity.
(g) Ion.
(h) Chemical formula.
(i) Valency.
(j) Molecular mass.
(k) Formula unit mass.
(l) Mole.
Answer:
(a) Law of conservation of mass: Matter is neither created nor destroyed during a chemical reaction i.e., total mass of reactants is equal to the total mass of products in a chemical reaction.

(b) Law of constant proportions: In a chemical compound, the elements are always present in a definite proportion by mass.
e.g.:

• In CO₂ the ratio of mass of carbon to the mass of oxygen is always 7 : 16.

(c) Atomic mass: The Atomic mass of an element is defined as the relative mass of its atom as compared with the mass of a carbon 12 atom taken as 12 units.

(d) Molecules of element: It contain two or more similar kinds of atoms chemically combined together.
e.g.:

• O₂, N₂, P₄.

(e) Molecules of compound: It contains two or more different kinds of atoms chemically combined together.
e.g.:

• SO₄, CO₂, H₂O.

(f) Atomicity: It is the total number of atoms present in a molecule.
e.g.:

• Atomicity of H₂ is 2
• Atomicity of P₄ is 4 and
• Atomicity of CO₂ is 3.

(g) Ion: It is the positively or negatively charged atom or group of atoms. It is formed by either loosening or gaining of electrons. Positively charged ion is called cation and negatively charged ion is called anion.
e.g.:

• Cation – Na⁺, Mg²⁺, NH₄⁺
• Anion – Cl^–, Br^–,OH^–, SO₄^2-

(h) Chemical formula: Chemical formula of a compound is representation in the form of symbols of elements present in it.
e.g.:

• Sodium chloride (NaCl)
• Calcium oxide (CaO).

(i) Valency: It is the combining power (or capacity) atom or group of atoms. It tells number of electrons lost or gained by the atom during the Chemical reaction.
e.g.:

(j) Molecular mass: It is the sum of atomic masses of all the atoms in a molecule of the substance.
e.g.:
Molecular mass of CO₂ is = Atomic mass of C + 2 × atomic mass of O
= 12 + 2 × 16
= 12 + 32 = 44u
It is expressed in atomic mass unit (u).

(k) Formula unit mass: It is the sum of atomic masses of all atoms in a formula unit of the compound containing constituents as ions. It is also expressed in atomic mass unit (u).
e.g.:
Formula unit mass of NaCl = Atomic mass of Na + atomic mass of Cl.
= (1 × 23 + 1 × 35.5)u
= (23 + 35.5)u = 58.5u.

(l) Mole: One mole of any substance like atoms or molecules or ions is that quantity which has mass equal to its atomic or molecular mass in grams and also contains 6.022 x 10²³ particles of that substance.
e.g.:
1 mole of H₂ molecule = (1 × 2)g = 2g and also,
1 Mole of H₂ molecule = 6.022 × 10²³ molecules of hydrogen.

Question 2.
Write the postulates of Dalton’s atomic theory.
Answer:
Postulates:

1. All matter is made up of very tiny particles called atoms.
2. Atoms are indivisible.
3. Atoms can neither be created nor be destroyed in a chemical reaction.
4. Atoms of a given element are identical in mass and chemical properties.
5. Atoms combine in the ratio of small whole numbers to form compounds.
6. The number of atoms and kind of atoms is fixed in a given compound.

Question 3.
What were the drawbacks of Dalton’s atomic theory?
Answer:
Drawbacks:

1. According to theory, atoms were indivisible but they can be divided in ions in electrons, protons and neutrons under special conditions.
2. According to theory, atoms of an element have masses but it is found that atoms of some elements can have some different masses.
3. According to theory, atoms of different elements have different masses but it is found that atoms of different elements can have same masses.

Question 4.
Calcium carbonate decomposes on heating to form calcium oxide and carbon dioxide. When 20g calcium carbonate is decomposed completely then 11.2g of calcium oxide is formed. Calculate the mass of carbon dioxide formed. Which law of chemical combination will be applied there?
Answer:
Law of conservation of mass is applied here.
Now,

Let, x grams of CO₂ is formed
Now, according to law of conservation of mass.
Total mass of reactants = Total mass of product
⇒ 20 = 11.2 + x
⇒ x = (20 – 11.2) = 8.89
So, Mass of CO₂ formed is 8.8g.

Question 5.
In an experiment 9.8g of copper oxide was obtained from 7.84g of copper. In another experiment 9.1g of copper oxide was obtained on reduction 7.28g of copper. Show with the help of calculations that these figures verify the law of constant proportions.
Answer:
Case I:
Mass of copper oxide = 9.8g
Mass of copper = 7.84g
Mass = (9.8 – 7.84)g = 1.96g
So, Ratio of copper to oxygen = \(\frac { 7.84 }{ 1.96 }\) = \(\frac { 4 }{ 1 }\) = 4 : 1

Case II:
Mass of copper oxide = 9.1g
Mass of copper = 7.28g
∴ Mass of oxygen = (9.1 – 7.28)g = 1.82g
So, Ratio of copper to oxygen = \(\frac { 7.28 }{ 1.82 }\) = \(\frac { 4 }{ 1 }\) = 4 : 1
Since, ratio of copper and oxygen in the two samples is same.
The law of constant proportions is verified.

Question 6.
Write the atomicity of following compounds.
(a) H₂O
(b) CaCO₃
(c) H₂SO₄
(d) C₆H₁₂O₆
(e) S_8.
Answer:

Question 7.
Write the chemical formula of:
(a) Hydrogen oxide
(b) Calcium carbonate
(c) Magnesium chloride
(d) Aluminium sulphate
Answer:

Question 8.
Calculate the molecular ma
(a) CH₃OH
(b) C₆H₁₂O₆
(c) NH₃
(d) H₂SO₄
Atomic masses:
C = 12u
H = 1u
O = 16u
N = 14u
S = 32u
Answer:
(a) Molecular mass of
CH₃OH = (1 × 12 + 3 × 1 + 1 × 16 + 1 × 1)u
= (12 + 3 + 16 + 1 )u = 32u.

(b) Molecular mass of
C₆H₁₂O₆ = (6 × 12 + 12 × 1 + 6 × 16)u
= (72 + 12 + 96)u = 180u.

(c) Molecular mass of
NH₃ = (1 × 14 + 3 × 1)u
= (14 + 3)u = 17u.

(d) Molecular mass of
H₂SO₄ = (1 × 2 + 1 × 32 + 4 × 16)u
= (2 + 32 + 64)u
= 98u.

Question 9.
Calculate the number of molecules of:
(a) SO₂ in its 3 moles.
(b) O₂ in its 2.5 moles.
Answer:
(a) 1 mole of SO₂ molecule
= 6.022 × 10²³ molecules of SO₄
= 3 moles of SO₄ molecule
= 3 × 6.022 × 10²³ molecules
= 18.066 × 10²³ molecules.

(b) 1 mole of O₂ molecule
= 6.022 × 10²³ molecules of O₂
= 2.5 moles of O₂ molecule
= 2.5 × 6.022 × 10²³ molecules of O₂
= 15.055 × 10²³ molecules
= 1.5055 × 10²⁴ molecules.

Question 10.
Calculate the number of moles in:
(a) 12.044 × 10²³ molecules of CO₂.
(b) 24.012 × 10²⁴ molecules of H₂
Answer:
(a) 1 mole of CO₂ = 6.022 × 10²³ particles of CO₂.
⇒ 6.022 × 10²³ particles = 1 mole of CO_2.
So, 1 particle of CO₂ molecule
= 1 / 6.022 × 10²³ mole and,
12.044 × 10²³ particles of CO₂ molecules
= 1 / 6.022 × 10²³ × 12.044 × 10²³
= 12.044 × 10²³ × 1 / 6.022 × 10²³ moles = 2 moles.

(b) 1 mole of H₂ contains = 6.022 × 10²³ particles of H₂ molecule
or 6.022 × 10²³ particles of H₂ = 1 mole of H₂
1 particle of H₂ molecule = 1 / 6.022 × 10²³ mole of H₂.
∴ 24.012 × 10²³ particles of H₂ = 1 / 6.022 × 10²³ × 24.012 × 10²³ moles of H₂
= 3.98 moles of H2.

Atoms and Molecules Long Answer Type Questions

Question 1.
Calculate the ratio of following elements by mass in the given compound:
(a) Hydrogen and oxygen in water (H₂O).
(b) Carbon and oxygen in carbon dioxide (CO₂).
(c) Carbon and hydrogen in ethene (C₂H₄).
Answer:
(a) In H₂O molecule,
Mass of H = 2 × 1g = 2g
Mass of O = 1 × 16g = 16g
So, Ratio of hydrogen and oxygen by mass = \(\frac { 2 }{ 16 }\) =\(\frac { 1 }{ 8 }\) = 3 : 8

(b) In CO₂ molecule,
Mass of C= 1 × 12g = 12g
Mass of O = 2 × 16g = 32g
.;. Ratio of carbon and oxygen by mass = \(\frac { 12 }{ 32 }\) = 3 : 8

(c) In ethene (C₂H₄) molecule,
Mass of C = 2 × 12g = 24g
Mass of H = (4 × 1)g = 4g
Ratio of carbon and hydrogen by mass = \(\frac { 24 }{ 4 }\) = 6 : 1

Question 2.
Calculate the formula unit mass of the following Ionic Compounds:
(a) Sodium Chloride (NaCl)
(b) Calcium Oxide (CaO)
(c) Copper Sulphate (CuSO₄)
(d) Calcium Nitrate [Ca(NO₃)₂]
[Atomic masses: Na = 23u, Cl = 35.5u, Ca = 40u, O = 16u, Cu = 63.5u, S = 32u, N = 14u]
Answer:
(a) Formula unit mass of NaCl Molecule = (1 × 23 + 1 × 35.5)u
= (23 + 35.5)u = 58.5u.

(b) Formula unit mass of CaO molecule = (1 × 40 + 1 × 16)u
= (40 + 16)u = 56u.

(c) Formula unit mass of CuSO₄ = (1 × 63.5 + 1 × 32 + 4 × 16)u
= (63.5 + 32 + 64)u = 159.5u.

(d) Formula unit mass of [Ca(NO₃)₂]
= 1 × 40 + 2 [1 × 14 4 + 3 × 16]
= 40 + 2 [14 4 + 48]u
= (40 + 124)u
= 164u.
MOLE CONCEPT (Graus to uoles):

Question 3.
Find the number of moles in:
(a) 20g of H₂O
(b) 140g of CO₂
(c) 200g of CaCO₃.
Answer:
(a) 1 mole of H₂O = Molar mass of H₂O
= (2 × 1 + 16)g
So, 1 Mole = 18g of H₂O
or 18g of H₂O = 1 Mole
1g of H₂O = \(\frac { 1 }{ 18 }\) Mole
and, 20g of H₂O = \(\frac { 1 }{ 18 }\) × 20 = 1.11 Mole

(b) 1 mole of CO₂ = Molar mass of CO₂ = (1 × 12 + 2 × 16)g
= (12 + 32)g = 44g
So, 1 mole = 44g of CO₂
or 44g of CO₂ = 1 mole
1 g of CO₂ = \(\frac { 1 }{ 44 }\) mole
and, 140 g of CO₂ = \(\frac { 1 }{ 44 }\) × 140 moles
= 3.18 moles.

(c) 1 mole of CaCO₃ = Molar mass of CaCO₃
= (1 × 40 + 1 × 12 + 3 × 16)g
= (40 + 12 + 48)g = 100g
So, 1 mole = 100g of CaCO₃
or 100g of CaCO₃ = 1 mole
1g of CaCO₃ = \(\frac { 1 }{ 100 }\) mole
and, 200 g of CaCO₃ = \(\frac { 1 }{ 100 }\) × 200 moles = 2 moles.
Moles to Grams

Question 4.
Calculate the mass in grams of:
(а) 3 moles of H₂O
(b) 2 moles of H₂SO₄
(c) 1.5 moles of carbon atoms (C atom).
Answer:

(a) 1 mole of H₂O = Molar mass of H₂O
= (2 × 1 + 16)g of H₂O
= 18g of H₂O
∴ 3 mole of H₂O = (3 × 18)g of H₂O
H₂O = 54g of H₂O

(b)1 mole of H₂SO₄ = Molar mass of H₂SO₄
= (2 × 1 + 1 × 32 + 4 × 16)g = 98g
2 mole of H₂SO₄ = (2 × 98)g = 196g

(c) 1 mole of C atom = Molar mass of C atom = 12g
∴ 1.5 moles of C -atom = (1.5 × 12)g = 18g.
Moles to Number of Particles.

Question 5.
Calculate the number of particles in:
(a) 115g of H₂O.
(b) 60g of CO₂.
Answer:
(a) We know,
1 mole of H₂O = molar mass of H₂O
= (2 × 1 + 16)g = 18g
It means, 18g of H₂O = 1 mole of CO₂
1g of H₂O = \(\frac { 1 }{ 18 }\) mole = 6.38 moles
Now, 1 mole of H₂O = 6.022 × 10²³ particles
6.38 moles of H₂O = 6.38 x 6.022 × 10²³ particles
= 3.84 × 10²⁴ particles.

(b) We know,
1 mole of CO₂ = Molar mass of CO₂
= (12 + 2 × 16)g = (12 + 32)g = 44g
It means, 44g of CO₂ = 1 mole of CO₂
1g of CO₂= \(\frac { 1 }{ 44 }\) mole of CO₂
and, 60g of CO₂ = \(\frac { 1 }{ 44 }\) × 60 moles of CO₂
= 1.3636 moles
Number of Particles to Mass.

Question 6.
Calculate the mass of:
(a) O₂ in its 36.48 × 10²⁵ particles.
(b) NH₃ in its 4.012 × 10²⁴ particles.
Answer:
(a) We know,
1 mole of O₂ = 6.023 × 10²³ particles
or 6.023 × 10²³ particles = 1 mole of O₂

Now, We also know,
1 mole of O₂ contains = molar mass of O₂
So, 60.57 moles of O₂ contains = 60.57 × 32g = 1938.49g.

(b) We know,
1 mole of ammonia (NH₃) = 6.022 X 1023 particles
or 6.022 × 1023 particles of ammonia (NH₃) = 1 mole

Now, We also know that
1 mole NH₃ contains = Molar mass of NH₃
= (14 + 3 × 1)g = (14 + 3)g
= 17g
So, 6.66 moles of NH₃ contains = (6.66 × 17)g = 113.25g

Question 7.
(a) Define atomicity and poly atomic ions.
(b) Find out the atomicity of following:
CO₂, NH₃, S₈, CaCO₃, H₂SO₄, Ca(OH)₂, K₂SO₄, Al₂(SO₄)₃, NaCl.
(c) Write 2 divalent and 2 trivalent polyatomic ions.
Answer:
(a) Atomicity: It is total number of atoms present in a molecule.
e,g.,

• Atomicity of H2 is 2.

Polyatomic ions: Ions which are formed from group of atoms are called polyatomic ions.
e,g.,

• Carbonate (CO₃^2-), Sulphate (SO₂^2-).

(b)

(c) Divalent polyatomic ions:

• Carbonate ion (CO₃s^2-)
• Sulphate ion (SO₄^2-).

Trivalent polyatomic ions:

• Phosphate ion (PO₄^3-)
• Phosphite ion (PO₃³).

Atoms and Molecules Higher Order Thinking Skills (HOTS)

Question 1.
A thermos P contains 0.5 mole of oxygen gas. Another thermos Q contain 0.4 mole of ozone gas. Which of the two thermos contain greater number of oxygen atom?
Answer:
1 molecule of oxygen (O₂) = 2 atoms of oxygen
1 molecule of ozone (O₃) = 3 atoms of oxygen
In thermos P:
1 mole of oxygen gas = 6.022 × 10²³ molecules
0.5 mole of oxygen gas = 6.022 × 10²³ × 0.5 molecules
= 6.022 × 10²³ × 0.5 × 2 atoms
= 6.022 × 10²³ atoms

In thermos Q:
1 mole of ozone gas = 6.022 × 10²³ molecules
0.4 mole of oxygen gas = 6 .022 × 10²³ × 0.4 molecules
= 6.022 × 10²³ × 0.4 x 3 atoms
= 7.32 × 10²³ atoms.

Question 2.
On analysing an impure sample of sodium chloride, the percentage of chlorine was found to be 45.5. What is the percentage of pure sodium chloride in the sample?
Answer:
Molecular mass of pure NaCl
= Atomic mass of Na + Atomic mass of Cl
= 23 + 35.5 = 58.5u
Precentage of chlorine in pure NaCl
Now, if chlorine is 60.6 parts
NaCl =100 parts
If chlorine is 45.5 parts,
Thus, percentage of pure NaCl = 75%.

Question 3.
Write the chemical formulae of the following:

1. Ammonium phosphate
2. Iron sulphate
3. Calcium nitrate
4. Magnesium nitride
5. Ammonium sulphate
6. Aluminium chloride
7. Copper Nitrate
8. Aluminium sulphate
9. Sodium carbonate
10. Barium chloride
11. Calcium nitrate
12. Potassium chloride
13. Hydrogen sulphide
14. Magnesium hydroxide
15. Zinc sulphate.

Answer:

1. Ammonium phosphate – (NH₄)₃PO₄
2. Iron sulphate – -Fe₂(SO₄)₃
3. Calcium nitrate – Ca(NO₃)₂
4. Magnesium nitrate – Mg(NO₃)₂
5. Ammonium sulphate – (NH₄)₂SO₄
6. Aluminium chloride – AlCl₃
7. Copper nitrate – CU(NO₃)₂
8. Aluminium sulphate – Al₂(SO₄)₃
9. Sodium carbonate – Na₂CO₃
10. Barium chloride – BaCl₂
11. Calcium nitrate – Ca(NO₃)₂
12. Potassium chloride – KCl
13. Hydrogen sulphide – H₂S
14. Magnesium hydroxide – Mg(OH)₂
15. Zinc sulphate – ZnSO₄

Atoms and Molecules Value Based Question 

Question 1.
Jolly buys gold ornaments and she is told that the ornaments has 90% gold and the rest is copper. She has been given a bill which amounts 100% charges of gold. Jolly refused to pay
the bill for 100% gold but settles the bill for 90% gold?
(a) How many atoms of gold are present in 1 gram of gold?
(b) Find out the ratio of gold and copper in the ornaments.
(c) What value of Jolly is seen in the above discussion?
Answer:
(a) 1 gram of gold will contain \(\frac { 90 }{ 100 }\) = 0.9 g of gold.

∴ 0.046 mol of gold will contain = 0.046 × 6.022 × 10²³ = 2.77 × 10²¹ atoms

(b) Ratio of gold : Copper 90 : 10

(c) Value of responsible behaviour and self – awareness is seen.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 3 Coordinate Geometry Ex 3.2

Question 1.
Write the answer of each of the following questions:

1. What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?
2. What is the name of each part of the plane formed by these two lines?
3. Write the name of the point where these two lines intersect.

Solution:

1. The x – axis and the y – axis.
2. Quadrants.
3. The origin.

Question 2.
See figure and write the following:

1. The coordinates of B.
2. The coordinates of C.
3. The point identified by the coordinates (-3, -5)
4. The point identified by the coordinates (2, -4).
5. The abscissa of the point D.
6. The ordinate of the point H.
7. The coordinates of the point L.
8. The coordinates of the point M.

Solution:

1. B → (- 5, 2 )
2. C → 4 (5, – 5 )
3. E
4. G
5. 6
6. – 3
7. L → (0, 5 )
8. M → (-3, 0 )

Question 3.
In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1, 0), (1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane.
Solution:

1. The point (-2, 4) lies in the II quadrant.
2. The point (3, -1) lies in the IV quadrant.
3. The point (-1, 0) lies on the negative x-axis.
4. The point (1, 2) lies in the I quadrant.
5. The point (-3, -5) lies in the III quadrant.

Question 4.
Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.

Solution:

Plotting of points on Graph Paper:
Steps:

1. Draw a horizontal and vertical line mutually perpendicular to each other.
2. Mark the point of inter section of two lines as ‘O’. This represent origin and horizontal line XOX’ vertical line as YOY’.
3. The line XOX’ will show X – axis and vertical line YOY’ will Y – axis.
4. Choose a suitable scale and marks the points on X – axis and Y – axis.
5. Obtain the coordinates of the given point and mark point.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 3 Coordinate Geometry Ex 3.1

Question 1.
How will you describe the position of a table lamp on your study table to another person?
Solution:
Consider the lamp as a point and table as a plane. Choose any two perpendicular edges of the table. Measure the distance of the lamp from the longer edge, suppose it is 25 cm. Again, measure the distance of the lamp from the shorter edge, and suppose it is 30 cm. You can write the position of the lamp as (30, 25) or (25, 30), depending on the order you fix.

Question 2.
(Street Plan):
A city has two main roads which cross each other at the center of the city. These two roads are along the North – South direction and East – West direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1 cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single line.

There are many cross – streets in your model. A particular cross-street is made by two streets, one running in the North – South direction and another in the East – West direction. Each cross – street is referred to in the following, manner: If the 2nd street running in the North – South direction and 5th in the East – West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find:

1. How many cross-streets can be referred to as (4, 3)?
2. How many cross-streets can be referred to as (3, 4)?

Solution:
Both the cross-streets are marked in the figure given. They are uniquely found because of the two reference lines we have used for locating them.

Cartesian Coordinate Axes:
These are two mutually perpendicular reference lines shown in Fig. which locate a point is a plane.

Convention of Signs:
OX is the positive direction and OX’ is the negative direction of x – axis while OY is positive and OY’ is negative.
In other words,

On the X – axis:
Values to the right are positive and values to the left are negative.

On the Y – axis:
Values above are positive and those below are negative.

Cartesian System:
The system used for describing the position of a point in a plane is called the Cartesian system.

Origin:
The point at which the two coordinate axes meet is called the origin. when a point is in

I. Quadrant: Y – value + and Y – value +
II. Quadrant: Y – value – and Y – value +
III. Quadrant: Y – value – and Y – value –
IV. Quadrant: Y – value + and Y – value –

The X – coordinate of a point is perpendicular distance from Y – axis, is known as abscissa. The Y – coordinate of a point is known as ordinate is perpendicular distance from Y – axis.

MP Board Class 9th Maths Solutions

MP Board Class 7th Science Solutions Chapter 1 Nutrition in Plants

Nutrition in Plants Intext Questions

Question 1.
Boojho wants to know how plants prepare their own food?
Answer:
Plants are the only that can prepare food for themselves by using water, carbon dioxide and minerals.

Question 2.
Paheli wants to know why our body cannot make food from carbon dioxde, water and minerals like plants do?
Answer:
Our body do not have chlorophyll.

Question 3.
Boojho wants to know how water and minerals absorbed by roots reach the leaves?
Answer:
Water and minerals are transported to the leaves by the vessels which run like pipes throughout the roots, stems and leaves of the plant. They form a continuous path or passage for the nutrients to reach the leaves.

Question 4.
Paheli wants to know what is so special about the leaves that they can synthesize food but other parts of the plant cannot?
Answer:
Because the leaves have a green pigment called chlorophyll.

Question 5.
Boojho has observed some plants with deep red, violet or brown leaves. He wants to know whether these leaves also carry out photosynthesis?
Answer:
No.

Question 6.
Paheli wants to know whether mosquitoes, bed bugs, lice and leeches that suck our blood are also parasites?
Answer:
Lice are parasites. Mosquitoes are not parasites because they suck blood to incubit their eggs and not for nutrition.

Question 7.
Boojho is confused. If the pitcher plant is green and carries out photosynthesis, then why does it feed on insects?
Answer:
Because these plants do not get enough nutrition from the soil as required.

Question 8.
Boojho wants to know how these organisms acquire nutrients. They do not have mouths like animals do. They are not like green plants as they lack chlorophyll and cannot make food by photosynthesis?
Answer:
These organisms acquire food from dead organisms.

Question 9.
Paheli is keen to know whether her beautiful shoes, which she wore on special occasions, were spoiled by fungi during the rainy season. She wants to know how fungi appear suddenly during the rainy season?
Answer:
The fungal spores are generally present in the air. When they land on wet and warm things they germinate and grow During rainy season, there are more chances of things getting wet So, fungi spoil more things in rainy season.

Question 10.
Boojho says once his grandfather told him that his wheat fields were spoiled by a fungus. He wants to know if fungi cause diseases also?
Answer:
Yes, fungi causes diseases in plants, animals and humans. However, some fungi are also used is medicines.

Nutrition in Plants Text book Exercises

Question 1.
Why do organisms need to take food?
Answer:
Food is needed by all living organisms for the following purposes:

1. Get energy to do work.
2. Build up body.
3. Improve resistance power against diseases and protects us from infections.
4. Replacement and repairing damaged part in the body.
5. Maintain the functions of the body.

Question 2.
Distinguish between a parasite and a saprotroph?
Answer:
Distinguish Parasite and Saprotroph:
Parasite:

1. They derives nutrients from the body of some other living organisms.
2. They use the heterotrophic mode of nutrition.
3. They mostly live on or in the host.
4. Examples: Tapeworm, Round warm, Cuscutta, Puccinia, etc.

Saprotroph:

1. They derives nutrients from dead and decaying organisms.
2. They use saprotrophic mode of nutrition.
3. They live on dead and decaying stuff.
4. Examples: Mushrooms, Bacteria, Yeast, etc.

Question 3.
How would you test the presence of starch in leaves?
Answer:
Starch Test:
Take the green leaf to be tested. Boil it in water for 5 minutes (approximately). Keep it in the 60% angle amyle alcohol at 60°C till it becomes colourless. Now take the colourless leaf out from alcohol and wash it with cold water. Also pour few drops of dilute iodine solution on the leaf. The leaf becomes very blue with the solution which proves the presence of starch is the leaf.

Question 4.
Give a brief description of the process of synthesis of food in green plants?
Answer:
The leaves have a green pigment called chlorophyll. It helps leaves to capture the energy of the sunlight. This energy is used to synthesize (prepare) food from carbon dioxide and water. Since the synthesis of food occurs in the presence of sunlight, it is called photosynthesis (Photo: light: synthesis: to combine).

So we find that chlorophyll, sunlight, carbon dioxide and water are necessary to carry out the process of photosynthesis. It is a unique process on the earth. The solar energy is captured by the leaves and stored in the plant in the form of food. Thus, sun is the ultimate source of energy for all living organisms.

During photosynthesis, chlorophyll containing cells of leaves, in the presence of sunlight, use carbon dioxide and water to synthesise carbohydrates. The process can be represented as an equation.

During the process oxygen is released. The carbohydrates ultimately get converted into starch. The presence of starch in leaves indicates the occurrence of photosynthesis. The starch is also a carbohydrate.

Question 5.
Show with the help of a sketch that the plants are the ultimate source of food?
Answer:

Question 6.
Fill in the blanks:

1. Green plants are called …………….. since they synthesise their own food.
2. The food synthesised by the plants is stored as ……………..
3. In photosynthesis solar energy is captured by the pigment called ……………..
4. During photosynthesis plants take in …………….. and release

Answer:

1. Autotrophs
2. Starch
3. Chlorophyll
4. Carbon dioxide, oxygen.

Question 7.
Name the following:

1. A parasitic plants with yellow, slender and tubular stem.
2. A plant that has both autotrophic and heterotrophic mode of nutrition.
3. The pores through which leaves exchange gases.

Answer:

1. Cuscuta (Amarbel)
2. Pitcher plant
3. Stomata.

Question 8.
Tick the correct answer:
(a) Amarbel is an example of –
(i) Autotroph
(ii) Parasite
(iii) Saprotroph
(iv) Host.
Answer:
(ii) Parasite

(b) The plant which trape and feeds on insects is –
(i) Cuscuta
(ii) China rose
(iii) Pitcher plant
(iv) Rose.
Answer:
(iii) Pitcher plant.

Question 9.
Match the items given in Column I with those in Column II:

Answer:

(i) (d)
(ii) (a)
(iii) (e)
(iv) (b)
(v) (c)

Question 10.
Mark “T” if the statement is true and “F” if it is false:

1. Carbon dioxide is released during photosynthesis. (T/F)
2. Plants which synthesise their food themselves are called saprotrophs. (T/F)
3. The product of photosynthesis is not a protein (T/F)
4. Solar energy is converted into chemical energy during photosynthesis. (T/F)

Answer:

1. False (F)
2. False (F)
3. True (T)
4. True (T)

Question 11.
Choose the correct option from the following:
Which part of the plant gets carbon dioxide from the air for photosynthesis –
(i) Root hair
(ii) Stomata
(iii) Leaf veins
(iv) Sepals.
Answer:
(ii) Stomata.

Question 12.
Chose the correct option from the following:
Plants take carbon dioxide form the atmosphere mainly through their –
(i) Roots
(ii) Stem
(iii) Flowers
(iv) Leaves.
Answer:
(iv) Leaves.

Nutrition in Plants Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (i)
The green coloured pigment present in plants is ………….
(a) Xanthophyll
(b) Haemoglobin
(c) Chlorophyll
(d) None of these.
(c) Chlorophyll

Question (ii)
The life processes that provide energy are ……………
(a) Respiration
(b) Nutrition
(c) Both respiration and nutrition
(d) None of these.
Answer:
(c) Both respiration and nutrition

Question (iii)
Which of these are autotrophs …………..
(a) Green plants
(b) All plants
(c) All animals
(d) None of these.
Answer:
(a) Green plants

Question (iv)
………….. changes solar energy into chemical energy.
(a) Oxygen
(b) Carbon – di – oxide
(c) Water
(d) Chlorophyll.
Answer:
(d) Chlorophyll.

Question (v)
………….. is saprophyte.
(a) Fungus
(b) Cuscuta
(c) Money plant
(d) Mosquito.
Answer:
(a) Fungus

Question 2.
Fill in the blanks:

1. Resin, gum, latex are ……………. substances of plants.
2.  …………….. is the mode of taking food by an organism and its utilisation by the body.
3. The plant on which it climbs is called a ……………
4. Plants which use saprotrophic mode of nutrition are called …………..
5. Oxygen is produced during ……………..

Answer:

1. Excretory
2. Nutrition
3. Host
4. Saprotrophs
5. Photosynthesis.

Question 3.
Which of the fallowing statements are true (T) or false (F):

1. All organisms take food and utilise it to get energy for the growth and maintenance of their bodies.
2. Chlorophyll and sunlight are not the essential requirements for photosynthesis.
3. Solar energy is stored by the leaves with the help of chlorophyll.
4. Oxygen is not produced during photosynthesis.
5. Only a few plants adopt other modes of nutrition like parasitic and saprotraphic.

Answer:

1. True (T)
2. False (F)
3. True (T)
4. False (F)
5. True (T).

Question 4.
Match the items is Column A with Column B:

Answer:

(i) (b)
(ii) (d)
(iii) (a)
(iv) (b).

Question 5.
Name the following :

1. A plants food factory.
2. Living on another organisms and derive food from them.
3. A chlorophyll containing partner, which is an alga, and a fungus live together.

Answer:

1. Leaf
2. Parasitic
3. Lichens

Nutrition in Plants Very Short Answer Type Questions

Question 1.
How many types of nutritions are there?
Answer:
There are two types of nutritions. These are holophytic nutrition and holozoic nutrition.

Question 2.
How do symbiotic live?
Answer:
Symbiotic live with host and parasite which depend on host benefit.

Question 3.
What are autotrophs? Give examples.
Answer:
The organism who can prepare their own food utilising sunlight, carbon dioxde and water, are known as autotrophs.
Examples: All green plants are autotrophs.

Question 4.
Define photosynthesis?
Answer:
The process by which the green plants prepare their food using carbon dioxide and water in the presence of chlorophyll and light is called photosynthesis.

Question 5.
What is the function of chlorophyll?
Answer:
Chlorophyll helps leaves to capture the energy of the Sun.

Question 6.
What are algae?
Answer:
The slimy green patches in ponds or in other stagnant water bodies are called algae.

Question 7.
What is the purpose of starch test?
Answer:
To confirm the presence of starch in the green plants.

Question 8.
Name two herbivorous animals?
Answer:
Deer, cow.

Question 9.
Name two omnivorous animals?
Answer:
Dog, cat.

Question 10.
Name two carnivorous animals?
Answer:
Tiger, lion.

Question 11.
Name two leguminous plants?
Answer:
Gram, pea.

Question 12.
Name two insectivorous plants?
Answer:
Sundew, Aldrovenda.

Question 13.
Which bacteria can convert nitrogen into soluble term?
Answer:
Rhizobium.

Question 14.
Define autotrophs?
Answer:
Green plants synthesise their food themselves by the process of photosynthesis. They are autotrophs.

Nutrition in Plants Short Answer Type Questions

Question 1.
How do the exchange of gases occur in plants?
Answer:
In case of plants the lower surface of leaves have small pores called stomata. These are provided by two guard cells which control the opening or closing of the stomata. When the concentration of O₂ gas increases during photosynthesis the guard cells open and O₂ gas is given out and if concentration of CO₂ gas increases during respiration, the guard cells cause CO₂ gas to go out of the cells. This is how the exchange of gases occur in plants.

Question 2.
How many types of heterotrophs are there? Give examples.
Answer:
Organisms which depend upon plants and other organisms for their food are called heterotrophs. Heterotrophs can further be classified as:

1. Herbivorous Animals:
The organisms or animals who eat plants and plant products.
Examples: cow, horse, goat, etc.

2. Carnivorous Animals:
The animals who eat flesh of other animals are called carnivorous.
Examples: lion, tiger, wolf, etc.

3. Omnivorous Animals:
The animals who eat both plant and animals are called ominivorous animals.
Examples: man, cat, dog, crow, etc.

Question 3.
W’hat is the difference between ‘heterotrophs and autotrophs’?
Answer:
The differences between two are:

Heterotroph:

1. These are organism which can not make their own food.
2. They do not have chloroplast.

Autotroph:

1. They can make their own food.
2. They have chloroplast.

Question 4.
Give two examples of insectivorous plants?
Answer:
The plants which have the special system to trap the insects and kill them are called insectivorous plants. The pitcher plant and venus faly trap plants. In the pitcher plant, the leaf is modified in form of a pitcher. When any insect visits this pitcher it is trapped and killed in it.

Question 5.
What is the difference between holophytic and holozoic type of nutrition?
Answer:
Those living beings such as plants who prepare their own food are called autotrophic and this type of nutrition is called as holophytic type of nutrition.

Those living beings who cannot prepare their own food but depend on food prepared by some other living beings are called heterotopic. This type of nutrition is called holozoic nutrition.

Question 6.
How is Sun the ultimate source of energy for all the living beings?
Answer:
Green plants prepare food utilising sunlight. All other organisms depend on green plants directly or indirectly for their nutrition. Thus, Sun is the ultimate source of energy.

Question 7.
How is holophytic nutrition different from holozoic nutrition?
Answer:
1. Holophytic nutrition is found in plants and lower forms of animals they consume liquid food as they lack digestive systems.

2. Holozoic nutrition is found in man and other higher forms of animals. There is a well developed digestive system in all of them. Hence, they can consume solid food.

Nutrition in Plants Long Answer Type Questions

Question 1.
Draw a diagram of stomata?
Answer:

Question 2.
Give an experiment to demonstrate that light is necessary for photosynthesis?
Answer:
Take a broad leaved potted plant and keep it under dark for 24 – 48 hours. The plant is kept in dark to make the plant free from starch. After this fix a leaf still attached to the plant with a paper clip having paper black as shown in figure. Now keep the plant in light for few hours and test the leaf for starch. To test the leaf for starch, pluck the lgaf and kill its cells in boiling water. Remove the chlorophyll by boiling in alcohol. Wash the boiled leaf in water and treat with iodine solution.

It is believed that the portion of the leaf, exposed to sunlight turned blue in colour while the covered portion did not undergo any change. You know that starch give blue colour with iodine solution. This was because the covered portion did not receive any sunlight. This shows that sunlight is necessary for photosynthesis.

Question 3.
Describe the process of nutrition in hydra?
Answer:
In hydra, the tentacles help in ingesting the food (taking the food inside). The cells inside the body cavity wall secrete certain chemicals and enzymes to digest the food. The digested food is absorbed by the cells of the wall in the body cavity by diffusion. On eating food the hydra grows and reproduces by forming buds.

Question 4.
Define cell with their structure
Answer:
The bodies of living organisms are made of tiny units called cells. Cells can be seen only under the microscope. Some organisms are made of only one cell. The cell is enclosed by a thin outer boundary, called the cell membrane. Most cells have a distinct, centrally located spherical structure called the nucleus (Fig.). The nucleus is surrounded by a jelly – like substance called cytoplasm.

MP Board Class 7th Science Solutions

MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.5

Question 1.
Use suitable identities to find the following products:

1. (x + 4) (x + 10)
2. (x + 8) (x – 10)
3. (3x + 4) (3x – 5)
4. (y² + \(\frac{3}{2}\)) (y² – \(\frac{3}{2}\))
5. (3 – 2x) (3 + 2x)

Solution:
1. (x + 4)(x + 10)
= (x)² + (4 + 10)x + 4 x 10
= x² + 14x + 40

2. (x + 8) (x – 10)
= (x)² + (8 – 10) x + 8 x – 10
= x² – 2x – 80

3. (3x + 4) (3x – 5)
= (3x)² + (4 – 5) (3x) + 4 x (-5)
= 9x² – 3x – 20

4. (y² + \(\frac{3}{2}\)) (y² – \(\frac{3}{2}\))
= (y²)² + (\(\frac{3}{2}\) – \(\frac{3}{2}\))y + \(\frac{3}{2}\) x \(\frac{-3}{2}\)
= y⁴ – \(\frac{9}{4}\)

5. (3 – 2x) (3 + 2x)
= (3)² – (2x)²
= 9 – 4x²

Question 2.
Evaluate the following products without multiplying directly:

1. 103 x 107
2. 95 x 96
3. 104 x 96

Solution:
1. 103 x 107
= (100 + 3) (100 + 7)
= (100)² + (3 + 7) x 100 + 3 x 7
[Using identity (x + a) (x + b) = x² + (a + b) x + ab]
= 10000 + 1000 + 21
= 11021

2. 95 x 96
= (100 – 5) (100 – 4)
[Using identity (a – b)² = a² + b² – 2ab]
= (100)² + (- 5 – 4) x 100 + (- 5) x (- 4)
= 10000 – 900 + 20
= 9120

3. 104 x 96
= (100 + 4) (100 – 4)
= (100)² – (4)²
[Using identity (a + b) (a – b) – a² – b²]
= 10000 – 16
= 9984

Question 3.
Factorize the following using appropriate identities:

1. 9x² + 6xy + y²
2. 4y² – 4y + 1
3. x² – \(\frac{y^{2}}{100}\)

Solution:
1. 9x² + 6xy + y²
= 9x² + 6xy + y²
= (3x)² + 2 x 3x × y + (y)²
= (3x + y)²
= (3x + y) (3x + y)

2. 4y² – 4y + 1
= (2y)² – 2 x 2y x 1 + (1)²
= (2y – 1)²
= (2y – 1) (2y – 1)

3. x² – \(\frac{y^{2}}{100}\)
= (x)² – (\(\frac{y^{2}}{10}\))
= (x + \(\frac{y}{10}\))(x – \(\frac{y}{10}\))

Question 4.
Expand each of the following using suitable identities:

1. (x + 2y + 4z)²
2. (2x – y + z)²
3. (- 2x + 3y + 2z)²
4. (3a – 7b – c)²
5. (- 2 + 5y – 3z)²
6. (\(\frac{1}{4}\)a – \(\frac{1}{2}\)b + 1)

Solution:
1. (x + 2y + 4z)²
= (x)² + (2y)² + (4z)² + 2x(2y) + 2(2y)(4z) + 2 (4z)(x)
= x² + 4y² + 16z² + 4xy + 16yz + 8zx

2. (2x – y + z)v
= (2x)² + (-y)² + (z)² + 2(2x)(- y) + 2(- y)(z) + 2(z)(2x)
= 4x² + y² + z² – 4xy – 2yz + 4zx

3. (- 2x + 3y + 2z)²
= (-2x)² + (3y)² + (2z)² + 2(- 2x) (3y) + 2(3y)(2z) + 2 (2z) (- 2x)
= 4x² + 9y² + 4z² – 12xy + 12zy – 8zx

4. (3a – 7b – c)²
= (3a)² + (- 7b)² + (- c)² + 2(3a)(- 7b) + 2 (- 7b)(- c) + 2(- c)(3a)
= 9a² + 49b² + c² – 42ab + 14bc – 6ca

5. (-2x + 5y – 3z)2
= (-2x)² + (5y)² + (- 3z)² + 2(-2x)(5y) + 2(5y) (- 3z) + 2 (-3z) (- 2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

6. (\(\frac{1}{4}\)a – \(\frac{1}{2}\)b + 1)

Question 5.
Factorize:

1. 4x² + 9y² + 16z² + 12xy – 24yz – 16 xz
2. 2x² +y² + 8z² – 2√2xy + 4√2yz – 8xz

Solution:
1. 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
= (2x)² + (3y)² + (- 4z)² + 2(2x)(3y) + 2 (3y)(- 4z) + 2 (- 4z) (2x)
= (2x + 3y – 4z)²
= (2x + 3y – 4z) (2x + 3y – 4z)

2. 2x² + y² + 8z² – 2√2xy + 4√2 yz – 8xz
= (- √2 x)² + (y)² + (2√2z)² + 2x (-√2x) (y) + 2(y) (2√2z) + 2 (2√2z ) (-√2x ) = (-√2 x + y + 2√2z)²
= (-√2x + y + 2√2z) (-√2 x + y + 2√2 z)

Question 6.
Write the following cubes in expanded form:

1. (2x + 1)³
2. (2a – 3b)³
3. (\(\frac{3}{2}\)x + 1)³
4. (x – \(\frac{2}{3}\)y)³

Solution:
1. (2x + 1)³ = (2x)³ + (1)³ + 3(2x)(1)(2x + 1)
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 1 + 12x³ + 6x

2. (2a – 3b)³ = (2a)³ – (3b)³ – 3(2a)(3b)(2a – 3b)
= 8a³ – 27b³ – 18ab(2a – 3b)
= 8a³ – 27b³ – 36a²b + 54ab²

3. (\(\frac{3}{2}\)x + 1)³

4. (x – \(\frac{2}{3}\)y)³

Question 7.
Evaluate the following using suitable identities:

1. (99)³
2. (102)³
3. (998)³

Solution:
1. (99)³ = (100 – 1)³
= (100)³ – (1)³ – 3(100)(1)(100 – 1)
= 1000000 – 1 – 300 (100 – 1)
= 1000000 – 1 – 300 x 99
= 999999 – 29700 = 970299

2. (102)³ = (100 + 2)³
= (100)³ + (2)³ + 3 (100)(2)(100 + 2)
= 1000000 + 8 + 600 (100 + 2)
= 1000008 + 61200
= 1061208

3. (998)³ = (1000 – 2)³
= (1000)³ – (2)³ – 3(1000)(2)(1000 – 2)
= 1000000000 – 8 – 6000 x 998
= 994011992

Question 8.
Factorize each of the following:

1. 8a³ + b³ + 12a²b + 6ab²
2. 8a³ – b³ – 12a²b + 6ab²
3. 27 – 125a³ – 135a + 225a²
4. 64a³ – 27b³ – 144a²b + 108ab²
5. 27p³ – \(\frac{1}{216}\) – \(\frac{9}{2}\)p² + \(\frac{1}{4}\)p

Solution:
1. 8a³ + b³ + 12a²b + 6ab²
= (2a)³ + (b)³ + 3(2a)(b)(2a + b)
= (2a + b)³
= (2a + b)(2a + b)(2a + b)

2. 8a³ – b³ – 12a²b + 6ab²
= (2a)³ – (b)³ – 3(2a)(b)(2a – b)
= (2a – b)³
= (2a – b) (2a – b) (2a – b)

3. 27 – 125a³ – 135a + 225a²
= (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)
= (3 – 5a)³
= (3 – 5a)(3 – 5a)(3 – 5a)

4. 64a³ – 27b³ – 144a²b + 108ab²
= (4a)³ – (3b)³ – 3(4a)(3b)(4a – 3b)
= (4a – 3b)³
= (4a – 3b)(4a – 3b)(4a – 3b)

5. 27p³ – \(\frac{1}{216}\) – \(\frac{9}{2}\)p² + \(\frac{1}{4}\)p

Question 9.
Verify:

1. x³ + y³ = (x + y) (x² – xy + y²)
2. x³ – y = (x – y) (x² + xy + y²)

Solution:
1. x³ + y³ = (x + y) (x² – xy + y²)
RHS = (x + y) (x² – xy + y²)
= x(x² – xy + y²) + y(x² – xy + y²)
= x² – x²y + xy² + x²y – xy² + y³
= x³ + y² = LHS

2. x³ – y³ = (x – y)(x² + xy + y²)
RHS = (x – y) (x² + xy + y²)
= x(x² + xy + y²) – y(x² + xy + y²)
= x³ + x²y + xy² – x²y – xy² – y³
= x³ – y³ = LHS .

Question 10.
Factorize each of the following:

1. 27y³ + 125z³
2. 64m³ – 343n²

Solution:
1. 21y³ + 125z³ = (3y)³ + (5z)³
= (3y + 5z) [(3y)² – (3y)(5z) + (5z)²]
= (3y + 5z) (9y² – 15yz + 25z²)

2. 64m³ – 343n³ = (4m)³ – (7n)³
= (4m – 7n) [(4m)² + (4m)(7n) + (7n)²]
= (4m – In) (16m² + 28mn + 49n²)

Question 11.
Factorize: 27x³ + y³ + z³ – 9xyz
Solution:
27x³ + y³ + z³ – 9xyz
= (3x)³ + (y)³ + (z)³ – 3 x 3x × y x z
= (3x + y + z) {(3x)² + (y)² + (z)² – 3xy – yz – 3xz}
= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3xz)

Question 12.
Verify that x³ + y³ + z³ – 3xyz = \(\frac { 1 }{ 2 } \) (x + y + z)[(x-y)2 + (y – z)2 + (z – x)2]
Solution:
x³ + y³ + z³ – 3xyz
= \(\frac { 1 }{ 2 } \) (x + y + z) [(x – y)² + (y – z)² (z – x)²]
RHS = \(\frac { 1 }{ 2 } \) (x + y + z) [(x – y)² + (y – z)² + (z – x)²]
= \(\frac { 1 }{ 2 } \) (x + y + 2) [(x² + 2xy + y²) + (y² – 2yz + z²) + (z² – 2zx + x²)]
= \(\frac { 1 }{ 2 } \) (x + y + z) [2x² + 2y² + 2z² – 2xy – 2yz – 2zx]
= \(\frac { 1 }{ 2 } \) (x + y + z) x 2(x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz = LHS

Question 13.
If x + y + z = 0, show that x³ + y³ + z³ = 3xyz
Solution:
We know x³ + y³ + z³ – 3xyz
= (x + y + z) (x² + y² + z² – xy – yz – xz)
= x³ + y³ + z³ – 3xyz
= 0 x (x² + y² + z² – xy – yz – xz) (∴ x + y + z = 0)
x³ + y³ + z³ – 3xyz = 0
x³ + y³ + z³ = 3xyz

Question 14.
Without actually calculating the cubes, find the value of each of the following:

1. (-12)³ + (7)³ + (5)³
2. (28)³ + (-15)³ ÷ (-13)³

Solution:
1. (-12)³ + (7)³ + (5)³
a + b + c = – 12 + 7 + 5
= – 12 + 12 = 0
(- 12)³ + (7)³ + (5)³
= 3 (- 12) (7) (5) = – 1260

2. (28)³ + (- 15)³ + (- 13)³
a + b + c = 28 + (- 15) + (- 13)
= 28 – 28 = 0
∴ (28)³ + (-15)³ + (- 13)³
= 3(28) (- 15) (- 13) = 16380

Question 15.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

1. Area: 25a² – 35a + 12 …..(i)
2. Area: 35y² + 13y – 12 …..(ii)

Solution:
1. A = 25a² – 35a + 12
= 25a² – 20a – 15a + 12
= 5a (5a – 4) – 3 (5a – 4)
= (5a – 4) (5a – 3)
Length = (5a – 4) and breadth = 5a – 3

2. A = 35y² + 13y – 12
= 35y² + 28y – 15y – 12
= 7y (5y + 4) – 3 (5y + 4)
= (7y – 3) (5y + 4)
Length = (7y – 3) and breadth = (5y + 4)

Note: In the above questions out of the two factors, any one factor can be taken as length and the other factor will be breadth.

Question 16.
What are the possible expressions for the dimensions of the cuboids, whose volumes are given below:

1. Volume: 3x² – 12x …(i)
2. Volume: 12ky² + 8ky – 20k …(ii)

Solution:
1. V = 3x² – 12x
= 3x (x – 4) = 3 × x × (x – 4)
l = 3
b = x
h = (x – 4)

2. V = 12ky² + 8ky – 20k
= 4k (3y² + 2y – 5)
= 4k (3y² + 5y – 3y – 5)
= 4k [y (3y + 5) – 1 (3y + 5)]
= 4k (3y + 5) (y – 1)
l = 4 k
b = (3y + 5)
h = (y – 1)

MP Board Class 9th Maths Solutions