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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

MP Board Class 9th Science Solutions Chapter 2 Is Matter Around Us Pure

Is Matter Around Us Pure Intext Questions

Is Matter Around Us Pure Intext Questions Page No. 15

Question 1.
What is meant by a pure substance?
Answer:
It is the substance which is made up of only one kind of matter. It cannot be separated into other kinds of matter by any physical process.
Examples:

• Water
• Sugar.

Question 2.
List the points of differences between homogeneous and heterogeneous mixtures.
Answer:

Is Matter Around Us Pure Intext Questions Page No. 18

Question 1.
Differentiate between homogeneous and heterogeneous mixtures with examples.
Answer:
Homogeneous mixtures:

1. These mixtures have uniform composition throughout their mass in which substances are completely mixed with one another.
2. They show no visible boundaries of separation between the constituents.
3. They do not show tyndall effect.
4. Examples: Sugar solution, salt solution.

Heterogeneous mixtures:

1. These mixtures have non – uniform composition throughout their mass and its substances remain separated and do not completely mix with one another.
2. They show visible boundaries of separation between the constituents.
3. They show tyndall effect.
4. Examples: Sugar – soil mixture, water – oil mixture.

Question 2.
How are sol, solution and suspension different from each other?
Answer:
Sol:

1. They appear to be homogeneous but actually they are heterogeneous.
2. The size of solute particles is more than true solution (1 nm) but less than suspension (100 nm).
3. Particles cannot be separated by Alteration.
4. Particles are not visible to naked eyes.
5. They scatter a beam of light.
6. Examples: Smoke, blood and ink.

Solution:

1. They are homogeneous mixtures.
2. The size of solute particles is less than 1 nm in diameter.
3. Particles cannot be separated by filteration.
4. Particles are not visible to naked eyes.
5. They do not scatter a beam of light.
6. Examples: Sugar in water and salt in water.

Suspension:

1. They are heterogeneous mixtures.
2. The size of solute particles is more than 100 nm in diameter.
3. Particles can be separated by filteration.
4. Particles are visible to naked eyes.
5. They scatter a beam of light.
6. Examples: Dust in air, mud in water.

Question 3.
To make a saturated solution, 36 g of sodium chloride is dissolved in 100g of water at 293K. Find its concentration at this temperature.
Answer:
We know,

Now, mass of solute sodium chloride = 36g
Mass of Solvent (water) = 100g
So, Mass of Solution = mass of solute + mass of solvent
⇒ (36 + 100)g = 136g
Then,
Concentration of solution = 3600 / 136 × 100
⇒ 26.47% by mass.

Is Matter Around Us Pure Intext Questions Page No. 24

Question 1.
How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25°C) which are miscible with each other?
Answer:
The difference in the boiling points of kerosene and water is 25°C. So, they can be separated by the process of simple distillation.
Method:

1. Take the mixture in a distillation flask and fit a thermometer in it.
2. Make the arrangement as shown.
3. Heat slowly and observe.
4. Petrol has lower boiling point, so it vaporises first and get condensed in the condenser and finally get collected in the receiver.
5. The kerosene has higher boiling point, so it will be left behind.
   

Question 2.
Name the technique to separate:

1. Butter from curd.
2. Salt from sea – water.
3. Camphor from salt.

Answer:

1. Centrifugation
2. Evaporation
3. Sublimation.

Question 3.
What type of mixtures are separated by the technique of crystallisation?
Answer:
Crystallisation is used to separate impurities present in the solvent from mixtures to obtain pure substance as a crystal.
Example:

• Obtaining salt from sea – water
• purification of copper sulphate.

Is Matter Around Us Pure Intext Questions Page No. 24

Question 1.
Classify the following as chemical or physical changes:
(i) Cutting of trees.
(ii) Melting of butter in a pan.
(iii) Rusting of almirah.
(iv) Boiling of water to form steam.
(v) Passing of electric current through water and the water breaking down into hydrogen and oxygen gases.
(vi) Dissolving common salt in water.
(vii) Making a fruit salad with raw fruits and
(viii) Burning of paper and wood.
Answer:
Chemical changes: (iii), (v), (viii)
Physical changes: (i), (ii), (iv), (vi), (vii)

Question 2.
Try segregating the things around you as pure substances or mixtures.
Answer:

1. Pure substances: Salt, sugar, water, silver, diamond, alcohol.
2. Mixtures: Milk, air, plastic, cold – drink, bronze.

Is Matter Around Us Pure NCERT Textbook Exercises

Question 1.
Which separation techniques will you apply for the separation of the following?

1. Sodium chloride from its solution in water.
2. Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
3. Small pieces of metal in the engine oil of a car.
4. Different pigments from an extract of flower petals.
5. Butter from curd.
6. Oil from water.
7. Tea leaves from tea.
8. Iron pins from sand.
9. Wheat grains from husk.
10. Fine mud particles suspended in water.

Answer:

1. Evaporation
2. Sublimation
3. Alteration
4. Chromatography
5. Centrifugation
6. Using separating funnel
7. Filteration
8. Magnetic separation
9. Winnowing
10. Decantation and filtration.

Question 2.
Write the steps you would use for making tea.
Use the words: solution, solvent, solute, dissolve, soluble, insoluble, filterate and residue.
Answer:
Steps:

1. Take 2 – 3 cups of water in a container as a solvent and boil it for few minutes.
2. Add 3 – 4 teaspoons of sugar, 2 teaspoons of tea-leaves as a solute making it a solution and boil it.
3. Add 1 cup of milk as a solute which will make a dark brown colour mixture. Leave it to boil.
4. Sugar will get dissolved in solution but tea-leaves remain insoluble. So, it will not get dissolved in the mixture.
5. Boil the mixture for few more minutes.
6. Filter the solution through tea-strainer and collect the filterate in a cup.
7. The tea – leaves being insoluble will be left as residue.

Question 3.
Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table as grams of substances dissolved in 100 grams of water to form a saturated solution).

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313K?
(b) Pragya makes a saturated solution of potassium chloride in water at 353K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
(c) Find the solubility of each salt at 293K. Which salt has the highest solubility at this temperature?
(d) What is the effect of change of temperature on the solubility of a salt?
Answer:
From the table,
(a) Mass of KnO₃ to make saturated solution, at 313 K in 100g water at 313K = 62g
So, In 50 g of water mass of KnO₃ required = (62 / 2)g = 31g.

(b) The solubility of potassium chloride (KCl) decreases on cooling and crystal formation will start.

(c) Solubility of each salt at 293 K is:

• Potassium nitrate : 32g / 100g of water
• Sodium chloride : 36g / 100g of water
• Potassium chloride : 35g / 100g of water
• Ammonium chloride : 37g / 100g of water

So, Ammonium chloride has highest solubility at 293K.

(d) On increasing the temperature, the solubility increases and on decreasing the temperature, the solubility decreases.

Question 4.
Explain the following giving examples:
(a) Saturated solution
(b) Pure substance
(c) Colloid
(d) Suspension
Answer:
(a) Saturated solution: The solution in which no more solute can be dissolved at a particular temperature is called saturated solution.
Example:

• In aqueous sugar solution, if no more sugar can be dissolved at that temperature, then it will be called as saturated solution.

(b) Pure substance: It is the substance which contains only one kind of particle and its composition is same throughout.
Examples:

• Water
• sugar.

(c) Colloid: It is the heterogeneous mixture in which the size of particles lie between true solutions and suspensions. The particles can not be seen through naked eyes but they can scatter a beam of light and possess Tyndall effect.
Examples:

• Smoke
• milk
• blood.

(d) Suspension: It is a heterogeneous mixture in which solute particles remain undissolved and get suspended within the medium. The particles scatter a beam of light and are visible through naked eyes.
Examples:

• Mud in water
• sand in water.

Question 5.
Classify each of the following as a homogeneous or heterogeneous mixtures soda water, wood, air, soil, vinegar, filtered tea.
Answer:

1. Homogeneous mixture: Soda water, air, vinegar, filtered tea.
2. Heterogeneous mixture: Wood, soil.

Question 6.
How would you confirm that a colourless liquid given to you is pure water?
Answer:
Boil the water at 100°C. If it boils at 100°C and left no residue behind, then it is pure water.

Question 7.
Which of the following materials fall in the category of a “pure substance”?
(a) Ice
(b) Milk
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
(g) Brick
(h) Wood
(i) Air.
Answer:
Pure substances are
(a) Ice
(c) Iron
(e) Calcium oxide
(f) Mercury.

Question 8.
Identify the solutions among the following mixtures.
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water.
Answer:
Solutions are:
(b) Sea water
(e) Soda water
(c) Air.

Question 9.
Which of the following will show “Tyndall effect”?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution.
Answer:
(b) Milk
(d) Starch solution.

Question 10.
Classify the following into elements, compounds and mixtures:
(a) Sodium
(b) Soil
(c) Sugar solution
(d) Silver
(e) Calcium carbonate
(f) Tin
(g) Silicon
(h) Coal
(i) Air
(j) Soap
(k) Methane
(l) Carbon dioxide
(m) Blood.
Answer:
Elements:
(a) Sodium
(d) Silver
(f) Tin
(g) Silicon.

Compounds:
(e) Calcium carbonate
(k) Methane
(l) Carbon dioxide
(j) Soap.

Mixtures:
(c) Sugar solution
(b) Soil
(h) Coal
(i) Air
(m) Blood.

Question 11.
Which of the following are chemical changes?
(a) Growth of a plant.
(b) Rusting of iron.
(c) Mixing of iron filings and sand.
(d) Cooking of food.
(e) Digestion of food.
(f) Freezing of water.
(g) Burning of a candle.
Answer:
(a) Growth of a plant.
(b) Rusting of iron.
(d) Cooking of food.
(e) Digestion of food.
(g) Burning of a candle.

Is Matter Around Us Pure Additional Questions

Is Matter Around Us Pure Multiple Choice Questions

Question 1.
Which of the following is a pure substance?
(a) Paint
(b) Sodium chloride
(c) Soil
(d) Milk.
Answer:
(b) Sodium chloride

Question 2.
Which of the following is not a mixture?
(a) Air
(b) Milk
(c) Kerosene
(d) Petroleum.
Answer:
(c) Kerosene

Question 3.
A mixture is made up of __________ .
(a) Two or more compounds.
(b) Two or more elements.
(c) Two or more elements or compounds.
(d) None of these.
Answer:
(c) Two or more elements or compounds.

Question 4.
Husk in water is an example of __________ .
(a) Colloid
(b) Solution
(c) Suspension
(d) Saturated solution.
Answer:
(c) Suspension

Question 5.
Which of the following is a metalloid?
(a) Iron
(b) Boron
(c) Zinc
(d) Carbon.
Answer:
(b) Boron

Question 6.
A colloidal solution is also called:
(a) True solution
(b) Sol
(c) Solvent
(d) Real solution.
Answer:
(b) Sol

Question 7.
When a ray of light passes through a solution and particles are not visible, then it is called:
(a) True solution
(b) Suspension
(c) Colloid
(d) Heterogeneous solution.
Answer:
(a) True solution

Question 8.
Which of the following is a true solution?
(a) Muddy water
(b) Salt solution
(c) Blood
(d) Curd.
Answer:
(b) Salt solution

Question 9.
The zig – zag motion of colloidal particles is known as __________ .
(a) Brownian movement
(b) Tyndall effect
(c) Rectilinear movement
(d) Electrolysis.
Answer:
(a) Brownian movement

Question 10.
A separating funnel is used for separating __________ .
(a) Solids
(b) Miscible liquids
(c) Immiscible liquids
(d) None of these.
Answer:
(c) Immiscible liquids

Question 11.
Particles of mixture of ammonium chloride and sand can be separated by __________ .
(a) Sublimation
(b) Evaporation
(c) Centrifugation
(d) Decantation.
Answer:
(a) Sublimation

Question 12.
A solution that can be separated by filtration is called __________ .
(a) Colloid
(b) Suspension
(c) True solution
(d) None of these.
Answer:
(b) Suspension

Question 13.
The separation technique which involves the difference in their densities is __________ .
(a) Centrifugation
(b) Sublimation
(c) Distillation
(d) Filteration.
Answer:
(a) Centrifugation

Question 14.
Milk of Magnesia is a:
(a) True solution
(b) Colloid
(c) Suspension
(d) Homogeneous mixture.
Answer:
(c) Suspension

Question 15.
Concentration of solution is __________ .
(a) mass of solute / mass of solvent × 100
(b) mass of solvent / mass of solute × 100
(c) mass of solute / mass of solution × 100
(d) mass of solution / mass of solute × 100.
Answer:
(c) mass of solute / mass of solution × 100

Question 16.
Which one of the following is not a chemical change?
(a) Freezing of water
(b) Ripening of fruit
(c) Formation of curd
(d) Corrosion of metals.
Answer:
(a) Freezing of water

Question 17.
Tincture of iodine is the solution of __________ .
(a) Iodine in water
(b) Iodine in acid
(c) Iodine in alcohol
(d) Iodine in bromine.
Answer:
(c) Iodine in alcohol

Question 18.
An alloy is a __________ .
(a) Heterogeneous mixture
(b) Homogeneous mixture
(c) Colloid
(d) Both (a) and (b).
Answer:
(b) Homogeneous mixture

Question 19.
Tyndall effect in forest is seen due to the presence of:
(a) Air in forest
(b) Dust in water
(c) Water droplets in air
(d) All of the above.
Answer:
(c) Water droplets in air

Question 20.
The compound FeS is not attracted by magnet because __________ .
(a) It is a compound
(b) It is black
(c) Iron has lost its properties
(d) It is a liquid.
Answer:
(c) Iron has lost its properties

Is Matter Around Us Pure Very Short Answer Type Questions

Question 1.
Define pure substance.
Answer:
The substance which is made up of only one kind of particle.

Question 2.
Give four examples of pure substance.
Answer:
Iron, aluminium, water, sodium chloride.

Question 3.
Define metalloids.
Answer:
Elements which show properties of metals as well as non – metals.

Question 4.
Name the scientist who first used the term element.
Answer:
Robert Boyle (1661).

Question 5.
Name two elements which are liquid.
Answer:
Mercury and bromine.

Question 6.
Give two examples of homogeneous mixtures.
Answer:
Salt solution, copper sulphate solution, brass alloy.

Question 7.
Give examples of heterogeneous solution.
Answer:
Sugar and sand, mud in water, iron filings in powder.

Question 8.
Give examples of colloidal solution.
Answer:
Blood, ink, milk.

Question 9.
Name the types of colloids.
Answer:
Sol, solid sol, aerosol, emulsion, foam, gel, solid foam.

Question 10.
Name the solutions which show Tyndall effect.
Answer:
Muddy water, milk of magnesia, starch solution, milk and ink.

Question 11.
Name the process used in separation of cream from milk.
Answer:
Centrifugation.

Question 12.
Name the technique used in separation of oil from water.
Answer:
Separation using separating funnel.

Question 13.
What is emulsion?
Answer:
If both dispersed phase and dispersing medium is liquid, it is called emulsion. Example: Milk.

Question 14.
Calculate the concentration of solution which contains 3 g of salt dissolved in 60 g of water.
Answer:
Concentration = Mass of solute / mass of solution × 100
⇒ (3 / 60 + 3) × 100%
⇒ 300 / 63%
⇒  4.76%.

Question 15.
What happens to solubility of solid in liquid when the temperature increases?
Answer:
Increases.

Question 16.
What are the other names of solute and solvent?
Answer:
Dispersed phase and dispersed medium.

Question 17.
Give two examples of aerosol.
Answer:
Fog, hairspray.

Question 18.
Which one has smallest particles – colloid, suspension or solution?
Answer:
Solution.

Question 19.
Name the substances that can be separated by sublimation.
Answer:
Iodine, ammonium chloride and camphor.

Question 20.
Which process is used to separate ink from coloured component (dye)?
Answer:
Evaporation.

Is Matter Around Us Pure Short Answer Type Questions

Question 1.
Define:
(a) Pure substance
(b) Mixture
(c) Metalloid
(d) Solution
(e) Solute
(f) Solvent
(g) Concentration of solution
(h) Saturated solution
(i) Unsaturated solution
(j) Emulsion
(k) Distillation
(l) Tyndall effect
(m) Crystallisation.
Answer:
(a) Pure substance: A pure substance is made up of only one kind of particle. It includes all elements or compounds.
Examples:
Iron, gold, silver, water, sugar, hydrochloric acid.

(b) Mixture: It is the impure substance which is made up of two or more different types of pure substances like elements or compounds in any ratio.
Examples:
Milk, rocks air, salt solution and soil.

(c) Metalloid: These are the elements which show the properties of metals and non – metals. They are also called semi – conductors.
Examples:
Silicon (Si), germanium (Ge) and boron (B).

(d) Solution: It is the homogeneous mixture of two or more substances.
Examples:
Air, alloys, salt solution.

(e) Solute: It is the component of solution which is dissolved in the medium.
Example:
salt in salt solution.

(f) Solvent: It is the component of solution which dissolves the solute.
Example:
water in salt solution.

(g) Concentration of solution: It is the quantity of solute present in the given quantity of solution.

(h) Saturated solution: It is the solution in which no more solute can be dissolved at that particular temperature.

(i) Unsaturated solution: It is the solution is which more amount of solute can be dissolved at the temperature to make it saturated.

(j) Emulsion: It is a colloid in which a liquid is dispersed in another liquid but they are immiscible.
Examples:
Milk, beauty cream.

(k) Distillation: It is the technique which is used to separate two miscible liquids which are having different boiling points.
Examples:
Separating pure water from sea – water, separation of mixture of acetone and water.

(l) Tyndall effect: It is the effect caused due to scattering of light by colloidal particles.
Example:
Sunlight when passes through forest get scattered due to tiny water droplets present in the air.

(m) Crystallisation: It is the process of obtaining pure solid substances in the form of crystals from its saturated solution by cooling it down slowly.

Question 2.
Give two examples of each:

1. Element
2. mixture
3. homogeneous mixture
4. heterogeneous mixture
5. solution
6. suspension
7. colloid.

Answer:

1. Element: Iron, gold.
2. Mixture: Soil with powder, sand in water.
3. Homogeneous mixture: Sugar solution, salt solution.
4. Heterogeneous mixture: Soil in water, milk.
5. Solution: Sugar solution, salt solution.
6. Suspension: Milk of magnesia, chalk with water.
7. Colloid: Ink, milk.

Question 3.
Write all types of colloidal solution and give examples.
Answer:
Sol : Ink, paints
Aerosol : Fog, spray paint
Emulsion : Butter
Foam : Fire – extinguisher foam, shaving foam
Gel : Hair gel
Solid sol : Colourful gemstones.

Question 4.
Write down the properties of a true solution.
Answer:

1. It is a homogeneous mixture.
2. The particles are not visible to naked eyes even not with a microscope.
3. Particles passes through filter paper.
4. True solution does not show Tyndall effect.

Question 5.
Write down the properties of a suspension.
Answer:

1. It is a heterogeneous mixture.
2. The particles are visible to naked eyes.
3. Particles do not pass through a filter paper.
4. It shows Tyndall effect.

Question 6.
Define physical and chemical change and give two examples of each.
Answer:
(i) Physical change:  It is a change in which no new substance is formed and particles do not change their characteristics.
Examples:

• Formation of ice from water
• cutting of paper.

(ii) Chemical change: It is a change in which new substance is formed and particle changes their characteristics.
Examples:

• Ripening of fruits
• burning of wood.

Question 7.
Define chromatography. Give examples where it is used.
Answer:
Chromatography: It is the method of separation of solid solutes which are dissolved in the same solvent. A special paper called “Chromatography paper” is used in this technique.
Examples:

• Separation of dyes present in the ink
• Separation of pigments from natural colours.

Question 8.
Name the separation technique used in the separation of:
(a) Iron pins from soil.
(b) Butter from curd.
(c) Copper sulphate from impure sample.
(d) Pigments from extract of flower.
(e) Different gases from air.
(f) Mixture of salt – water.
(g) Mixture of salt and ammonium chloride.
(h) Mixture of immiscible liquid – petrol and water.
Answer:
(a) Magnetic separation.
(b) Centrifugation.
(c) Crytallisation.
(d) Chromatography.
(e) Fractional distillation.
(f) Distillation.
(g) Sublimation.
(h) Separating funnel.

Question 9.
What is Brownian movement?
Answer:
The particles of colloids move randomly in a zig – zag manner. This movement is called Brownian movement. It happens due to collision of particles of dispersed phase and dispersion medium.

Question 10.
Write the dispersed phase and dispersion medium of the following colloidal substances: Mist, milk, paint, sponge, coloured gemstone, ink, smoke, face cream.
Answer:

Question 11.
21.6g of sodium chloride dissolves in 60g of water at 20°C. Calculate its solubility.
Answer:
Now,
60g of water dissolves = 21.6 g of sodium chloride
So, 100g of water dissolves = (21.6 / 60) × 100g of sodium chloride
⇒ 36g of sodium chloride
Thus, solubility of sodium chloride is 36g at 20°C.

Question 12.
As solution contains 60 ml of alcohol mixed with 200 ml of water. Calculate the concentration of this solution.
Answer:
Concentration of solution = (volume of solute / volume of solution) × 100
= (60 / 200) × 100%
= 30% (by volume)

Question 13.
What are the advantages and disadvantages of crystallisation method over evaporation method in separation of mixture?
Answer:
1. Advantages:

• Proper crystals of substance are obtained.
• Soluble impurities can be removed easily.

2. Disadvantages:

• Some substance get remain with the solvent.
• Solvent needs to be purified again for further use.

Is Matter Around Us Pure Long Answer Type Questions

Question 1.
Explain the Centrifugation process? Mention its applications also.
Answer:
Centrifugation process: It is the process of separation of suspended particles of a substance from a liquid by churning the liquid at high speed. In this process, denser particles are forced to settle at the bottom and lighter particles stay at the top. The following figure shows the process of centrifugation:
Applications:

• In washing machine, for washing clothes.
• In diagnostic labs, to test blood.
• In dairies, to separate butter from milk.

Question 2.
How can we separate the mixture of two immiscible liquids – kerosene oil and water? Explain the method with a labelled diagram.
Answer:
Mixture of two immiscible liquids – kerosene oil and water, is separated by separating funnel method. This method is based on the principle of difference in their densities.

Method:

1. Take the mixture of kerosene oil and water in a separating funnel as shown in the figure.
2. Leave the medium undisturbed for sometime till separate layers of oil and water are formed.
3. Open the stop – cock of the funnel and collect the water (denser liquid) in the collection beaker.
4. Close the cock when the kerosene oil reaches the stop – cock.
5. Collect the oil in a separate beaker.

Question 3.
How can mixture of salt and ammonium chloride be separated? Name the technique and the principle involved. Mention one application of this technique.
Answer:
Mixture of salt and ammonium chloride can be separated by the process of sublimation (as shown in figure). The principle of sublime nature is involved in this.

Method:

1. The mixture is kept in a china dish placed on a stands shown in figure.
2. An inverted funnel is kept over the mixture with cotton plug to close it.
3. Place the burner below it and heat the mixture.
4. On heating, ammonium chloride directly changes from solid to gaseous state and gets collected at the stem of the funnel.
5. The common salt is left behind.

Application:

• Separation of naphthalene balls’ powder and Common salt.

Question 4.
Elaborate the technique to separate dyes in black ink using chromatography ?
Answer:
Chromatography is the method of separation of solid solutes which are dissolved in the same solvent.
Method of separation of dyes in black ink:

1. Take a thin and long strip filter paper or chromatography paper.
2. Make a line on it at 3 cm from the lower end.
3. Put a small drop of black ink on it at the centre of line.
4. Dry the paper.
5. Dip the paper from lower edge into a tall, glass jar containing same water in a way that the ink drop remains just above the water and leave it undisturbed.
6. The water start rising in the paper by capillary action and the ink drop starts splitting into the constituents.
7. The constituent which is more soluble in water will rise more than less soluble constituent
   

Chromatography technique is used in separation of:

• Drugs from blood.
• Pigments from natural colours.
• Sugar from urine.

Question 5.
What is crystallisation? How can we obtain pure copper sulphate from impure copper sulphate solution?
Answer: Crystallisation is the process of obtaining pure solid substance in the form of crystals from its saturated solution by cooling down slowly.
Method to obtain pure copper sulphate from its impure solution:

1. Take some quantity of impure copper sulphate in water in a china dish.
2. Filter the solution to remove insoluble impurities.
3. Take a beaker Riled half with water and place the china dish over it.
4. Heat the beaker to heat the copper sulphate solution through water bath.
5. The copper sulphate solution gets evaporated till the solution becomes saturated.
6. Cool down the saturated solution slowly to obtain the crystals of pure copper sulphate.
7. Separate the copper sulphate crystals from solution by Alteration and dry them.
   

Question 6.
What is distillation technique? How can the mixture of acetone and water be separated?
Answer:
Distillation: It is the technique used to separate two miscible liquids by conversion of liquid into vapours by heating and then condensing the vapours into the pure liquid.

Types:
(a) Simple distillation: This technique is used to separate only two liquids where difference of boiling points is more than (20 – 25°C).
(b) Fractional distillation: This technique is used to separate more than tw o liquids and also where difference of boiling points of liquids is less than (20 – 25°C).
Procedure to separate the mixture of acetone and water:

1. Take the mixture of acetone and water in a distillation flask.
2. Fix a thermometer to the flask.
3. Heat the mixture slowly and observe the temperature of thermometer.
4. Acetone vaporises first and get condensed in the condenser and finally get collected in the beaker.
5. Water is left behind in the flask.
   

Question 7.
What is Magnetic separation method? Write the procedure to separate iron filings from sulphur powder. Write some applications.
Answer:
Magnetic separation method is used to separate magnetic component of a mixture using strong magnets like horse-shoe magnet or electro magnets.
Separation of iron filing from sulphur powder:

1. Take the mixture of iron filings and sulphur powder in a china-dish.
2. Take a horse shoe magnet and move it over the mixture.
3. The iron filings get attached to the magnet leaving the sulphur powder in the china dish.
   

Applications:

• Separation of iron materials from waste.
• Removing iron materials from metal scrap.
• Removing of iron foreign particles from eyes using electro-magnets by the doctors.

Is Matter Around Us Pure Higher Order Thinking Skills (HOTS)

Question 1.
With the help of diagram explain the water purification system in water works.
Answer:

Procedure:

1. Firstly, the water is collected in the reservoir.
2. Then it is passed to sedimentation tank to allow solid impurities to get settle down.
3. Then, it is passed to loading tank where the suspended impurities are loaded to settle down as sediment.
4. Then, water is filtered into the filtration tank where it is passed through different layers of sand and gravel.
5. Finally, it is passed through chlorinated tank where it is mixed with chlorine to kill bacteria or germs present in it. Now, it is called disinfected water.
6. This disinfected water is pumped to high storage tank from where it is supplied to city through network of pipes.

Question 2.
Name all the separation techniques used to separate different mixtures indicating the type of mixtures as well.
Answer:
A. Separation of solid – solid mixture:

• Solvent method: Sugar and sand.
• Magnetic method: Iron filings from sulphur powder.
• Sublimation: Common salt and ammonium chloride.

B. Separation of solid – liquid mixture:

• Filtration method: Chalk and water, sand and water.
• Centrifugation method: Clay particles from water, cream from milk.
• Crystallisation method: Pure copper sulphate from impure solution.
• Chromatography: Dyes from the black ink, pigments in natural colours, sugar from blood.
• Distillation: Acetone and water, gaseous components of air.

C. Separation of liquid – liquid mixture:

• Separating funnel method: Petrol and water, chloroform and water.

Is Matter Around Us Pure Value Based Questions

Question 1.
Kritika was asked to bring the jar of common salt from kitchen. But accidentally jar got slipped into the tub of water. She got tensed of scolding by mother. She went to his brother mentioning the problem. Her brother advised her to place the tub in open sunlight for long hours for two – three days. She did that and observed that common salt is left behind.
(a) What is the technique used above to separate the common salt from water?
(b) Name the type of mixtures that can be separated from this technique.
(c) Mention the values of Kritika’s brother mentioned here.
Answer:
(a) Evaporation technique is used here to separate the common salt from water.

(b) Mixtures in which solid substance is dissolved in water can be separated by this technique.
Example:
Coloured component (dye) from ink.

(c) Krikita’s brother showed his intelligent and helpful behaviour.

Question 2.
Vaibhav and Hritik’s teacher asked them to separate the mixture of acetone and water contained in the jar. Vaibhav brought a beaker and Hritik brought a separating funnel to separate the mixture. After few minutes, Hritik come with water and acetone contained in separate containers but Vaibhav come with the same mixture. Teacher congratulate Hritik for separating the mixture and advised Vaibhav to use the same technique followed by Hritik.
(a) Which technique is followed by Hritik for separation?
(b) Name the principle involved in this technique.
(c) Mention the value of Hritik seen here.
Answer:
(a) Separating funnel method is followed by Hritik for separation.
(b) Principle of difference of densities is applied in this technique.
(c) Values of obedience and wiseness is seen by Hritik.

Question 3.
Three friends Sanjana, Saumya and Shruti were sitting in a room. Sanjana told Shruti and Saumya about the visibility of dust particles in air. But, they were doubting her statement Then Sanjana closed the room and opened a window slightly. She told diem to observe the path of light and other things in that path. Saumya and Shruti were very surprised to see the presence of minute dust particles in air.
(a) Which effect is illustrated by Sanjana?
(b) Where this phenomenon can be seen naturally?
(c) What values did Sanjana show to her friends?
Answer:
(a) Tyndall effect is illustrated by Sanjana.
(b) This phenomenon is seen naturally in dense forest, in winter season through fog.
(c) Sanjana showed the values of wiseness and friendliness.

Question 4.
Pranjal went to her friend’s house and bring a black ink and a special kind of paper with her. Her friend asked her about that paper. Pranjal perform an activity with the paper by applying the dot of ink on it. And, placed the paper in ajar of water in such a way that the dot just touched the upper level of water. Finally, they observed the separation of different coloured constituents of that ink.
(a) Name the technique used by Pranjal.
(b) Describe the technique mentioned and name the paper used.
(c) Write some other applications of this technique.
(d) Write the values shown by Pranjal.
Answer:
(a) Chromatography technique is used here.
(b) Chromatography is the method of separation of solid solutes which are dissolved in the same solvent. Chromatographic paper is used in this technique.
(c) This technique is used in separation of drugs from blood, pigments from natural colours and separating sugar from urine.
(d) Pranjal showed the values of intelligence and friendliness.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 1.
Determine which of the following polynomials has (x + 1) as factor:

1. x³ + x² + x + 1
2. x⁴ + x³ + x² + x + 1
3. x⁴ + 3x² + 3x² + x + 1
4. x³ – x² – (2 + √2 ) x + √2

Solution:
1. p(x) = x³ + x² + x + 1
x + 1 = 0 ∴ x = – 1
p(-1) = (-1)³ + (-1)² + (-1) + 1
= – 1 + 1 – 1 + 1 = 0
(x + 1) is a factor of p(x).

2. p(x) = x⁴ + x³ + x² + x + 1
x + 1 = 0, x = – 1
P(- 1) = (- 1)⁴ + (-1)³ + 3(-1)² + (-1) + 1
= 1 – 1 + 1 – 1 + 1 = 1
(x + 1) is not a factor of p(x).

3. p(x) = x⁴ + 3x³ + 3x² + x + 1
x + 1 = 0,
x = – 1
p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)³ + (-1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
(x + 1) is not a factor of p(x).

4. p(x) = x³ – x² – (2 + √2 ) x + √2
x + 1 = 0, x = – 1
p(-1) = (-1)³ – (-1)² – (2 + √2) x (-1) + √2
= – 1 – 1 + 2 + √2 + √2 = 2√2
(x + 1) is not a factor of p(x).

Question 2.
Use the factor theorem to determine whether g(xr) is a factor of p(x) in each of the following cases:

1. p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1
2. p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
3. p(x) = x² – 4x² + x + 6, g(x) = x – 3

Solution:
1. p(x) = 2x³ + x² – 2x – 1, g(x) = 0
x + 1 = 0 x = – 1
p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1
= – 2 + 1 + 2 – 1 = 0
(x + 1) is a factor of p(x).

2. p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
g(x) = 0
x + 2 = 0, x = – 2
zero of g(x) is – 2.
Now, p(-2)
= (-2)³ + 3 (-2)² + 3(-2) + 1 ,
= – 8 + 12 – 6 + 1 = – 1 ≠ 0
By factor theorem, g(x) is not a factor of p(x).

3. p(x) = x³ – 4x² + x + 6
g(x) = x – 3 = 0, x = 3
p(3) = (3)³ – 4(3)² + (3) + 6
= 27 – 36 + 3 + 6
= 36 – 36 = 0
(x – 3) is a factor of p(x).

Question 3.
Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

1. p(x) = x² + x + k
2. p(x) = 2x² + kx + √2
3. p(x) = kx² – √2x + 1
4. p(x) = kx² – 3x + k.

Solution:

1. p(x) = x² + x + k
If x – 1 is a factor of p(x), then p( 1) = 0
(∴ By factor theorem)
(1)² + (1) + k = 0
1 + 1 + k = 0
2 + k = 0
k = – 2

2. p(x) = 2x² + kx + √2
g(x) = x – 1, x = 1
P(1) = 2(1)² + k(1) + √2 = 0
2 + k + √2 = 0
k = – (2 + √2)

3. p(x) = kx² – √2 x + 1
g(x) = x – 1,
x = 1
p(1) = k(1)² – √2(1) + 1
k – √2 + 1 = 0
k = √2 – 1

4. p(x) = kx² – 3x + k
g(x) = x – 1, x = 1
p(1) = k(1)² – 3(1) + k = 0
k – 3 + k = 0
2k – 3 = 0

Question 4.
Factorise:

1. 12x² – 7x + 1
2. 2x² + 7x + 3
3. 6x² + 5x – 6
4. 3x² – x – 4

Solution:
1. 12x² – 7x + 1
12x² – 7x + 1 = 12x² – 4x – 3x + 1
= 4x(3x – 1) – 1 (3x – 1)
= (3x – 1) (4x – 1)

2. 2x² + 7x + 3
2x² + 7x + 3 = 2x² + 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
= (x + 3) (2x + 1)

3. 6x² + 5x – 6
6x² + 5x – 6 = 6x² + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3) (3x – 2)

4. 3x² – x – 4
3x² – x – 4 = 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4) (x + 1)

Question 5.
Factorize:

1. x³ – 2x² – x + 2
2. x³ – 3x² – 9x – 5
3. x² + 13x² + 32x + 20
4. 2y³ + y² – 2y – 1.

Solution:
1. x³ – 2x² – x + 2
Let p(x) = x³ – 2x² – x + 2
We find that
p(1) = (1)³ – 2(1)² – (1) + 2
= 1 – 2 – 1 + 2 = 0
By factor theorem, (x – 1) is a factor of p(x).
Now,
x³ – 2x² – x + 2 = x²(x – 1) – x(x – 1) – 2(x – 1)
= (x – 1) (x² – x – 2)
= (x – 1) (x² – 2x + x – 2)
= (x – 1) {x(x – 2) + 1(x – 2)}
= (x – 1) (x – 2) (x + 1).

2. p(x) = x³ – 3x² – 9x – 5
x = – 1
p(-1) = (-1)³ – 3(-1)² – 9(- 1) – 5
= – 1 – 3 + 9 – 5 = 0
(x + 1) is a factor of p(x).

Now, p(x) = (x + 1) (x² – 4x – 5)
= (x + 1) (A2 – 5A + A – 5)
= (x + 1) [x(x – 5) + 1 (x – 5)]
= (x + 1) (x – 5) (x + 1)

3. p(x) = x³ + 13x² + 32x + 20
x = – 1,
p(-1) = (-1)³ + 13(-1)² + 32(-1) + 20
= – 1 + 13 – 32 + 20 = 0
(x + 1) is a factor of p(x)

p(x) = (x + 1) (x² + 12x + 20)
= (x + 1) (x² + 10x + 2x + 20)
= (x + 1) [x (x + 10) + 2 (x + 10)]
= (x + 1) (x + 2) (x + 10)

4. p(y) = 2y³ + y² – 2y – 1
y = 1
P(1) = 2(1)³ + (1)² – 2(1) – 1
= 2 + 1 – 2 – 1
= 0
(y – 1) is a factor of p(y)

p(x) = (y – 1) (2y² + 3y + 1)
= (y – 1) (2y² + 2y + 1y + 1)
= (y – 1)[2y(y + 1) + 1 (y + 1)]
= (y – 1) (y + 1) (2y + 1)

Algebraic Identities:

1. (x + y)² = x² + 2xy + y²
2. (x – y)² = x² – 2xy + y²
3. x² – y² = (x + y) (x – y)
4. (x + a) (x + b) = x² + (a + b) x + ab

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.3

Question 1.
Find the remainder when x³ + 3x³ + 3x + 1 is divided by

1. x + 1
2. x – \(\frac{1}{2}\)
3. x
4. x + π
5. 5 + 2x.

Solution:
Let p(x) = x³ + 3x² + 3x + 1

1. x + 1
x + 1 = 0
x = – 1
Remainder
= p(- 1)= (- 1)³ + 3 (- 1)² + 3 (-1) +1
= – 1 + 3 – 3 + 1 = 0.

2. x – \(\frac{1}{2}\) = 0
x =π
When p(x) is divided by ( x – \(\frac{1}{2}\)) the remainder is P(\(\frac{1}{2}\))

3. x = 0
When p(x) is divided by x, the remainder is p(0).
p(0) = (0) + 3(0)² + 3(0) + 1
= 0 + 0 + 0 + 1 = 1

4. x + π = 0
x = – π
When p(x) is divided by (x + π), the remainder is p(- π) p(- K) = (- π)³ + 3 (- π)² + 3 (- π) + 1
= – π³ + 3π² – 3π + 1

5. 5 + 2x = 0 x = \(\frac{-5}{2}\)
When p(x) is divided by (5 + 2x), the remainder is p(\(\frac{-5}{2}\))

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Solution:
Let p(x) = x² – ax² + 6x – 1
x – a = 0
x = a
Remainder = (a)³ – a(a)² + 6(a) – a
= a² – a² + 6a – a
= 5a.

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Solution:
p(x) = 3x³ + 7x
7 + 3X = 0

(7 + 3x) is not a factor of p(x).

Factorization of Polynomials:

Factor Theorem:
Let p{x) be any polynomial of degree equal to or greater than 1 and ‘a’ be any real number such that –

1. If p(a) = 0, then (x – a) is a factor of p(x)
2. If (x – a) is a factor then p(a) = 0

Proof:
Let p(x) be a polynomial of degree ≥ 1 and be a real number.

1. If p(a) = 0, then (x – a) is a factor of p(x)
Suppose q(x) be the quotient when p(x) is divided by (x – a). By remainder theorem, remainder = p(a).
P(x) = (x – a) x q(x) + p(a)
p(x) = (x – a) x q(x) (∴ p(a) = 0, given)
= (x – a) is a factor of p(x).

2. If (x – a) is a factor then p(a) = 0
=> p(x) when divided by (x – a) gives remainder = 0 …(i)
But by remainder theorem
When p(x) is divided by (x – a), remainder = p(a) …(ii)
From (i) and (ii), we get
p(a) = 0 Proved.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.2

Question 1.
Find the value of the polynomial 5x – 4x² + 3 at

1. x = 0
2. x = – 1
3. x = 2.

Solution:
Let f(x) = 5x – 4x² + 3

1. Value of f(x) at x = 0
= f(0) = 5(0) – 4(0)² + 3 = 3

2. Value of f(x) at x = – 1
= f(- 1) = 5(- 1) – 4(- 1)² + 3
= – 5 – 4 + 3 = – 6

3. Value of f(x) at x = 2
= f(2) = 5(2) – 4(2)² + 3
= 10 – 16 + 3 = – 3.

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:

1. p(y) = y² – y + 1
2. p(t) = 2 + t + 2t² – t³
3. p(x) = x³
4. p(x) = (x – 1)(x + 1)

Solution:
1. p(y) = y² – y + 1
P(0) = (0)2 – (0) + 1
= p(1) = (1)² – (1) + 1 = 1,
and p(2)² = (2)² – (2) + 1
= 4 – 2 + 1 = 3.

2. p(t) = 2 + t + 2t² – t³
p(0) = 2 + 0 + 2(0)² – (0)² = 2
p(1) = 2 + 1 + 2(1)² – (1)²
= 2 + 1 + 2 – 1 = 4
and p(2) = 2 + 2 + 2(2)² – (2)³
= 2 + 2 + 8 – 8 = 4

3. p(x) = x³
p(0) = (0)² = 0,
p(1) = (1)² = 1,
and p(2) = (2)³ = 8.

4. p(x) = (x – 1) (x + 1)
p(0) = (0 – 1) (0 + 1) = (- 1) (1) = – 1
P(1) = (1 – 1) (1 + 1) = (0) (2) = 0 and
p(2) = (2 – 1) (2 + 1) = (1) (3) = 3

Question 3.
Verify whether the following are zero of the polynomial, written against them.

1. p(x) = 3x + 1, x = – \(\frac{1}{3}\)
2. p(x) = 5x – π, x = \(\frac{4}{5}\)
3. p(x) = x2 -1, x = 1, – 1
4. p(x) = (x + 1) (x – 2), x – 1, 2
5. p(x) = x², x = 0
6. p(x) = lx + m, x = – \(\frac{m}{l}\)
7. p(x) = 3x² – 1, x = – \(\frac{1}{\sqrt{3}}\), \(\frac{2}{\sqrt{3}}\)
8. p(x) = 2x + 1, x = \(\frac{1}{2}\)

Solution:
1. p(x) = 3x + 1
x = – \(\frac{1}{3}\)
p(- \(\frac{1}{3}\)) = 3(- \(\frac{1}{3}\)) + 1 = 0
= – 3 x \(\frac{1}{3}\) + 1 = – 1 + 1 = 0
As p(- \(\frac{1}{3}\)) = 0, – \(\frac{1}{3}\) is the zero of p(x).

2. p(x) = 5x – π
x = \(\frac{4}{5}\)
p(\(\frac{4}{5}\)) = 5 x \(\frac{4}{5}\) – π = 4 – π
As p(\(\frac{4}{5}\)) ≠ 0,
\(\frac{4}{5}\) is not the zero of p(x).

3 p(x) = x² – 1, x = 1, -1
P(1) = (1)² – 1 = 1 – 1 = 0
p(-1) = (-1)² – 1 = 1 – 1 = 0
As p(1) = 0 and p(-1) = 0
Both 1 and -1 are the zeros of p(x).

4. p(x) = (x + 1)(x – 2)
x = – 1, 2
p(- 1) = (-1 + 1) (- 1 – 2)
= (0) (-3) = 0
p( 2) = (2 + 1) (2 – 2)
= 3 x 0 = 0
As p(- 1) = 0 and p(2) = 0,
Both -1 and 2 are the zeros of p(x).

5. p(x) = x²
x = 0
P(0) = (0)² = 0
As p(0) = 0, 0 is a zero of p(x)

6. p(x) = lx + m, x = – \(\frac{m}{l}\)
p(- \(\frac{m}{l}\)) = l x – \(\frac{m}{l}\) + m
As p(- \(\frac{m}{l}\)) = 0, (-\(\frac{m}{l}\)) is a zero of p(x).

7.

8.

Question 4.
Find the zero of the polynomial in each of the following cases:

1. p(x) = x + 5
2. p(x) = x – 5
3. p(x) = 2x + 5
4. p(x) = 3x – 2
5. p(x) = 3x
6. p(x) = ax, a ≠ 0
7. p(x) = cx + d, c ≠ 0, c, d are real numbers.

Solution:
1. p(x) = x + 5
p(x) = 0
x + 5 = 0
x = – 5
∴ – 5 is a zero of the polynomial p(x).

2. P(x) = x – 5
p(x) = 0
x – 5 = 0 x = 5
∴ 5 is a zero of the polynomial p(x).

3. p(x) = 2x + 5
2x + 5 = 0
x = – \(\frac{5}{2}\)
∴ – \(\frac{5}{2}\) is the zero of p(x).

4. p(x) = 3x – 2
3x – 2 = 0
x = \(\frac{2}{3}\)
∴ \(\frac{2}{3}\) is the zero of p(x)

5. p(x) = 3x
3x = 0
x = \(\frac{0}{3}\)
∴ 0 is the zero of p(x).

6. p(x) = ax where a ≠ 0
ax = 0
x = \(\frac{0}{a}\)
∴ 0 is the zero of p(x).

7. p(x) = cx + d where c ≠ 0, c, d are real numbers
cx + d = 0
x = – \(\frac{d}{c}\)
∴ – \(\frac{d}{c}\) is the zero of p(x).

Remainder Theorem:
If a polynomial p{x) is divided by another polynomial (x – a), the remainder isp(a). Where p(x) is any polynomial of degree greater than or equal to one and also degree p(x) > degree p(a) ‘a’ be any real number.

Dividend = divisor x quotient + remainder

Proof:
Let p(x) be any polynomial with degree greater than or equal to 1, suppose that when p(x) is divided by x – a, the quotient is q(x) and remainder is r(x).
p(x) – (x – a) x q(x) + r(x)
p(x) = (x – a) x q(x) + r(x)…..(i)

Case I:
If r(x) = 0
Equation reduce to
p(x) = (x – a) q(x) ……(ii)
On putting x = a, in (ii),
We get p(a) = (a – a) q(a)
p(a) = 0 = Remainder

Case II:
If r(x) ≠ 0
Equation (i) reduced to
p(x) = (x – a) q(x) + r …..(iii)
On putting x = a in (iii), we get
p(a) =(a – a) x q(x) + r
p(a) = r = Remainder
∴ So the remainder is p(a) when p(x) is divided by (x – a).

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 1 Matter in Our Surroundings

Matter in Our Surroundings Intext Questions

Matter in Our Surroundings Intext Questions Page No. 3

Question 1.
Which of the following are matter? Chair air, love, smell, hate, almonds, thought, cold, cold – drink smell of perfume.
Answer:
Chair, air, almonds and cold – drink.

Question 2.
Give reasons for the following observation: The smell of hot sizzling food reaches you several metres away, but to get the smell from cold food you have to go close.
Answer:
The particles of hot food have more kinetic energy due to higher temperature, so their rate of diffusion is more and they move several meters away as compared to the particles of cold food.

Question 3.
A diver is able to cut through water in a swimming pool. Which property of matter does this observation show?
Answer:
This shows that the particles of matter have space between them and a weak force of attraction between them.

Question 4.
What are the characteristics of the particles of matter?
Answer:
Following are the characteristics of the particles of matter:

1. Particles of matter are very small.
2. Particles of matter have space between them.
3. Particles of matter are continuously moving.
4. Particles of matter attract each other.

Matter in Our Surroundings Intext Questions Page No. 6

Question 1.
The mass per unit volume of a substance is called density, (density = mass / volume). Arrange the following in order of increasing density: air, exhaust from chimneys, honey, water, chalk, cotton and iron.
Answer:
The order of increasing densities:
air < exhaust from chimneys < cotton < water < honey < chalk < iron.

Question 2.
(a) Tabulate the differences in the characteristics of states of matter.
(b) Comment upon the following: Rigidity, Compressibility, Fluidity, Filling a gas container, Shape, Kinetic energy and Density.
Answer:
(a)

(b)
(i) Rigidity: It is the tendency of matter to retain or maintain its shape when an outside force is applied.

• Solids are rigid in matter

(ii) Compressibility: It is the property of matter due to which it can be compressed to lower volume.

• Solids have minimum compressibility, but gases have maximum.

(iii) Fluidity: It is the tendency of a matter or particles of a matter to flow.

• Liquid and gases have fluidity so they are fluids.

(iv) Filling a gas container: Gas particles fill the container completely due to negligible intermolecular force and maximum space between particles.

(v) Shape: It is the property of a matter which indicates its boundaries.

• Solid have fixed shape but liquids and gases have no fixed shape.

(vi) Kinetic energy: The energy of matter or particles of matter due to their motion is called kinetic energy.

• Particles of gases have highest kinetic energy and liquids have less kinetic energy than gases but more than solids.

(vii) Density: It is mass per unit volume.

• Solids have highest density, liquids have low density than solids but greater than gases.

Question 3.
Give reasons:
(a) A gas fills completely the vessel in which it is kept.
(b) A gas exerts pressure on the walls of the container.
(c) A wooden table should be called a solid.
(d) We can easily move our hand in air but to do the same through a solid block of wood, we need a karate expert.
Answer:
(a) The molecules of gas have very less force of attraction and possess high kinetic energy due to which they move in all directions and fill the vessel completely.

(b) The particles of gas move freely and randomly in all directions. So, they collide with each other and also with the walls of the container due to which they exert a Pressure on its walls.

(c) A wooden table has a fixed shape, fixed volume. It is rigid and cannot be compressed. So it should be called a solid.

(d) The particles of Air have very less force of attraction between them, so we can easily move our hand, but particles of wood have strong force of attraction, so we have to apply a greater amount of force to break it and pass through solid wood.

Question 4.
Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why.
Answer:
We know that ice is solid and water is liquid but the structure of ice is cage like due to which the molecules of water are not closely packed and have vacant space between them due to which ice has low density than water and floats over it.

Matter in Our Surroundings Intext Questions Page No. 9

Question 1.
Convert the following temperature to celsius scale:
(a) 300 K
(b) 573 K
Answer:
(a) 300 K = (300 – 273)°C = 27°C.
(b) 573 K = (573 – 273)°C = 300°C.

Question 2.
What is the physical state of water at:
(a) 250°C
(b) 100°C?
Answer:
(a) Gaseous state.
(b) Liquid and gaseous both.

Question 3.
For any substance, why does the temperature remain constant during the change of state?
Answer:
During the change of state of any substance, the temperature remains constant because the heat supplied to the substance is used in overcoming the force of attraction between the particles and change its state. This hidden heat is called latent heat.

Question 4.
Suggest a method to liquefy atmospheric gases.
Answer:
The atmospheric gases can be liquefied by cooling and applying pressure on them in a closed chamber or cylinder.

Matter in Our Surroundings Intext Questions Page No. 10

Question 1.
Why does a desert cooler cool better on a hot dry day?
Answer:
On a hot dry day, the rate of evaporation is high because of low humidity. So, water sprinkling on the pads of cooler gets evaporated from outside which results in making its walls cool and then we receive cool air.

Question 2.
How does the water kept in an earthen pot (matka) become cool during summer?
Answer:
The earthen pot has a lot of pores on its surface. So, water comes out and gets evaporated from these pores and cools the water inside the pot.

Question 3.
Why does our palm feel cold when we put some acetone or petrol or perfume on it?
Answer:
When we put acetone, petrol or perfume on our palm, then these liquids absorb energy from our palm and get evaporated which cause cooling effect on our palm.

Question 4.
Why are we able to sip hot tea or milk faster from a saucer rather than a cup?
Answer:
Saucer has a large surface area than a cup. So the rate of evaporation is more in saucer than cup which causes faster cooling of the hot tea. Hence, we can sip hot tea from a saucer faster than cup.

Question 5.
What type of clothes should we wear in summer?
Answer:
Light coloured cotton clothes should be worn in summer because light colour reflects heat and cotton absorbs sweat quickly and evaporates it easily which makes our body feel cool and dry.

Matter in Our Surroundings NCERT Textbook Exercises

Question 1.
Convert the following temperatures to the Celsius scale.
(a) 293 K
(b) 470 K.
Answer:
(a) 293 K into°C
⇒ 993 – 273 = 20°C

(b) 470 K into°C
⇒ 470 – 273 = 197°C

Question 2.
Convert the following temperatures to the Kelvin scale.
(a) 25°C
(b) 373°C
Answer:
(a) 25°C = 25 + 273 = 298 K
(b) 373°C = 373 + 273 = 646 K.

Question 3.
Give reason for the following observations.
(a) Naphthalene balls disappear with time without leaving any solid.
(b) We can get the smell of perfume sitting several metres away.
Answer:
(a) Naphthalene balls disappear with time without leaving any solid, because naphthalene balls sublime and directly change into vapour state without leaving any solid.

(b) We can get the smell of perfume sitting several metres away because perfume contains volatile solvent i.e., gaseous particles, which have high speed and large space between them and diffuse faster and can reach people sitting several metres away.

Question 4.
Arrange the following substances in increasing order of forces of attraction between the particles – water, sugar, oxygen.
Answer:
Oxygen → water → sugar.

Question 5.
What is the physical state of water at:
(a) 25°C
(b) 0°C
(c) 100°C?
Answer:
(a) 25°C is liquid.
(b) 0°C is solid or liquid.
(c) 100°C is liquid and gas.

Question 6.
Give two reasons to justify:
(a) water at room temperature is a liquid.
(b) an iron almirah is a solid at room temperature.
Answer:
(a) Water at room temperature is a liquid because its freezing point is 0°C and boiling point is 100°C.
(b) An iron almirah is a solid at room temperature because melting point of iron is higher than the room temperature.

Question 7.
Why is ice at 273 K more effective in cooling than water at the same temperature?
Answer:
Ice at 273 K will absorb heat energy or latent heat from the medium to overcome the heat of fusion to become water. Hence, the cooling effect of ice is more than the water at same temperature because water does not absorb this extra heat from the medium.

Question 8.
What produces more severe burns, boiling water or steam?
Answer:
Steam at 100°C will produce more severe burns as extra heat is hidden in it called latent heat. Whereas, the boiling water does not have this hidden heat.

Question 9.
Name A, B, C, D, E and F in the following diagram showing change in its state

Answer:
A → Liquefication / melting / fusion
B → Vaporisation / evaporation
C → Condensation
D → Solidification
E → Sublimation
F → Sublimation

Matter in Our Surroundings Additional Questions

Matter in Our Surroundings Multiple Choice Questions

Question 1.
Which of the following substances is solid?
(a) Plasma
(b) BEC
(c) Wood
(d) Mercury
Answer:
(c) Wood

Question 2.
Fusion is the phenomenon of changing of state of __________ .
(a) Liquid to gas
(b) Solid to liquid
(c) Solid to gas
(d) Gas to plasma.
Answer:
(b) Solid to liquid

Question 3.
Diffusion is not possible in the case of __________ .
(a) Liquid into Solid
(b) Gas into liquid
(c) Gas into Gas
(d) Solid into Solid.
Answer:
(d) Solid into Solid.

Question 4.
Compressibility is highest in __________ .
(a) Liquid
(b) Solid
(c) Gas
(d) Plasma.
Answer:
(c) Gas

Question 5.
We can smell odour of deodrant several metres away due to __________ .
(a) Diffusion
(b) Evaporation
(c) Fusion
(e) None.
Answer:
(a) Diffusion

Question 6.
Which of the following do not exhibit sublimation?
(a) Water
(b) Camphor
(c) Naphthalene
(d) Dry ice.
Answer:
(a) Water

Question 7.
Ice floats on water because of __________ .
(a) Higher density than water
(b) Equal density than water
(c) Lower density than water
(d) None of these.
Answer:
(c) Lower density than water

Question 8.
As the pressure of air increases then boiling point of liquid __________ .
(a) Increases
(b) Decreases
(c) Remains same
(d) None of these.
Answer:
(a) Increases

Question 9.
The atmospheric pressure equals to __________ .
(a) 1.01325 × 10⁸ Pa
(b) 101.3 25 × 10⁴ Pa
(c) 1.01325 × 10⁵ Pa
(d) 1.01325 × 10² Pa.
Answer:
(c) 1.01325 × 10⁵ Pa

Question 10.
Cooking at high altitudes is difficult because __________ .
(a) Boiling point decreases
(b) Boiling point increases
(c) Freezing point reduced
(d) None of the above.
Answer:
(a) Boiling point decreases

Question 11.
Density of water is highest at __________ .
(a) 3°C
(b) 5°C
(c) 4°C
(d) 6°C.
Answer:
(a) 3°C

Question 12.
Dry ice is __________ .
(a) Solid carbon monoxide (CO)
(b) Solid Nitrogen dioxide (NO₂)
(c) Solid carbon dioxide (CO₂)
(d) Solid Ammonia (NH₃).
Answer:
(c) Solid carbon dioxide (CO₂)

Question 13.
The amount of heat required to change the state of 1 kg of substance is called __________ .
(a) Calorific heat
(b) Latent heat
(c) Thermal heat
(d) Conversion heat.
Answer:
(b) Latent heat

Question 14.
Evaporation takes place at __________ .
(a) Boiling point
(b) Melting point
(c) Freezing point
(d) At all temperatures.
Answer:
(d) At all temperatures.

Question 15.
Evaporation generally takes place at __________ .
(a) Below boiling point
(b) Above boiling point
(c) At boiling point
(d) None.
Answer:
(a) Below boiling point

Question 16.
Evaporation causes __________ .
(a) Heating effect
(b) Boiling effect
(c) Cooling effect
(d) Absorbing effect.
Answer:
(c) Cooling effect

Question 17.
Latent heat of vaporisation of water is __________ .
(a) 2.25 × 10⁶ J/kg
(b) 3.34 × 10⁶ J/kg
(c) 22.5 × 10⁶ J/kg
(d) 33.4 × 10⁵ J/kg.
Answer:
(a) 2.25 × 10⁶ J/kg

Question 18.
Which factor does not affect the rate of evaporation?
(a) Humidity
(b) Colour
(c) Wind Speed
(d) Surface area.
Answer:
(b) Colour

Question 19.
The rate of evaporation is minimum in __________ .
(a) Dry day
(b) Humid day
(c) Hot day
(d) Stormy day.
Answer:
(b) Humid day

Question 20.
Which of the following processes consumes heat?
(i) Melting
(ii) Freezing
(iii) Vaparisation
(iv) Condensation.

(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) and (iv).
Answer:
(c) (i) and (iii)

Matter in Our Surroundings Very Short Answer Type Questions

Question 1.
Define Matter.
Answer:
Matter is defined as something which occupies space and has mass.

Question 2.
What is the melting point of ice?
Answer:
0°C or 273 K.

Question 3.
Name the phenomenon by which two substances intermix with each other.
Answer:
Diffusion.

Question 4.
Name the physical State of matter which can be highly compressed.
Answer:
Gaseous state.

Question 5.
Name the tiny particles of which matter is made up of.
Answer:
Atoms.

Question 6.
Name the process of conversion of solid state into gaseous state directly.
Answer:
Sublimation.

Question 7.
Name the Indian scientist who calculated the fifth state of matter.
Answer:
Satyendra Nath Bose.

Question 8.
Name all the states of matter.
Answer:
Solid, liquid, gas, plasma, BEC (Bose Einstein Condensate).

Question 9.
What is Chemical name of dry ice?
Answer:
Solid Carbon Dioxide.

Question 10.
What is the main effect produced by evaporation?
Answer:
Cooling effect.

Matter in Our Surroundings Short Answer Type Questions

Question 1.
Define diffusion? On which features does it depend?
Answer:
It is the mixing of particles of different matters. It depends on the state, temperature and kinetic energy of particles.

Question 2.
Define:
(a) Latent heat of fusion
(b) Latent heat of vaporisation.
Answer:
(a) Latent heat of fusion is the amount of heat energy required to convert 1 kg of a solid into liquid at its melting point.

(b) Latent heat of vaporisation is the amount of heat energy required to convert 1 kg of liquid into gas at its boiling point.

Question 3.
Explain the following terms:
(a) Sublimation
(b) Evaporation.
Answer:
(a) Sublimation: The phenomenon of change of state directly from solid to gas or vapours or vice-versa without changing into liquid state.
Examples:
Dry ice, camphor.

(b) Evaporation: The process of changing matter from liquid state into the vapour state at any temperature below its boiling point.
Examples:
Drying of clothes, volatile liquids like petrol, kerosene.

Question 4.
What are the different factors on which evaporation of liquids depend?
Answer:
The factors on which evaporation depends are:
(a) Temperature
(b) Surface area
(c) Humidity
(d) Wind speed

Question 5.
Write four substances which show sublimation.
Answer:
(a) Camphor
(b) Naphthalene balls
(c) Dry ice
(d) Ammonium chloride

Question 6.
Why the temperature of ice remains constant when it starts melting?
Answer:
When the ice melts, then it absorbs the latent heat of fusion which changes its states. And, that heat is consumed in changing the state, keeping the temperature constant.

Question 7.
How can we liquefy gas? How is it useful to us?
Answer:
Gases can be liquefied by cooling them and by applying pressure into a closed chamber like cylinder. This property of compressibility is useful for us and we use it through different ways e.g., LPG (Liquefied Petroleum Gas), CNG (Compressed Natural Gas) and Liquefied O₂ in welding.

Question 8.
Differentiate between:
(a) Freezing and Melting.
(b) Evaporation and Boiling.
Answer:
(a) Freezing and Melting:

(b) Evaporation and Boiling:

Question 9.
Why solid carbon dioxide is called dry ice?
Answer:
It is because carbon dioxide is pressurised in the form of solid state which looks cool like ice. But when it is heated, it directly get converted into carbon dioxide gas.

Question 10.
Change the following temperatures accordingly.
(a) -173°C
(b) 250 K
(c) 2°C
(d) 0°K
Answer:
(a) We know that,
Temperature on kelvin = Temperature on Celsius + 273
So, -173°C = (-173 + 273) K = 100 K

(b) 250 K = (250 – 273)°C = -23°C

(c) 2°C = (2 + 273) K = 275 K

(d) 0°K = (0 – 273)°C = -273°C.

Question 11.
Give one example of each: Melting, Vaporisation, Condensation, Sublimation, Evaporation.
Answer:
Melting: Ice to water
Vaporisation: Water to steam
Condensation: LPG
Sublimation: Naphthalene balls, solid iodine
Evaporation: Volatile liquids like petrol.

Matter in Our Surroundings Long Answer Type Questions

Question 1.
Give reasons:
(а) The smell of hot tasty cooked food reaches us from far.
(b) The smell of perfume reaches us several metres away.
(c) The fragrance of an incense stick spreads in entire hall quickly.
(d) Steam is more severe than boiling water.
(e) We see water droplets collected on outer surface of steel glass containing cold water.
(f) We feel lot of perspiration on a humid day.
(g) Evaporation causes cooling.
(h) It is advised to use pressure cooker at high altitudes.
(i) We should wear light colour cotton clothes in summer.
(j) A gas fills the container completely.
(k) Clothes take more time to dry in a humid day.
(l) Kerosene oil is kept in a cool place.
Answer:
(a) The particles of hot tasty food have high kinetic energy due to raised temperature. So, they can easily diffuse and move to long distances.

(b) The particles of perfume are in vapour form. So, they possess high kinetic energy and move randomly and finally reach us several metres away.

(c) The particles of an incense stick are in gaseous state. So, they have high kinetic energy and diffuse quickly into atmosphere and cover the room.

(d) Steam has the heat in the form of latent heat as compared to boiling water. So, it is more severe.

(e) The water vapours present in the atmosphere get condensed when it comes in contact with the chilled container and get deposited in the form of liquid droplets.

(f) The rate of evaporation decreases on humid day as there are sufficient water vapours present in the atmosphere. So, our sweat does not get dried and we perspire a lot.

(g) During the evaporation, the particles on the surface of the fluid take energy and heat from the surface get vaporised which create cooling effect.

(h) It is advised to use pressure cooker at high altitudes because the atmospheric pressure is low and water boils quickly. So, to increase the pressure and to cook the food properly, pressure cooker is required.

(i) Light colour cotton clothes are good reflectors of heat as well as good absorbers of sweat. When sweating occurs, the cotton absorbs it quickly and evaporates it faster.

(j) The particles of gas have very less intermolecular force of attraction and have large intermolecular space. So, they possess high kinetic energy and expand quickly resulting in filling the container completely.

(k) The rate of evaporation is minimum on a humid day. So, cloth takes more time to get dried.

(l) Kerosene a volatile substance and it evaporates quickly and also there is a risk of breaking of bottle due to the pressure created by its particles. So, it is stored in a closed container and kept in a cool place.

Question 2.
(a) What is evaporation? On what factors does it depend?
(b) How is it useful to us?
Answer:
(a) Evaporation:
The phenomenon of changing of a matter from liquid state into the vapour state or gaseous state at any temperature below its boiling point is known as evaporation. The rate of evaporation depends on the following factors:

1. Surface area: If the surface area of the liquid increases, then evaporation becomes faster.
2. Temperature: When the temperature of the liquid increases then the kinetic energy of the particles increases which causes faster evaporation.
3. Wind speed: When the speed of wind increases, it covers the vacant space in the atmosphere created by moving wind, the evaporation gets faster. Also, we have seen that clothes dry faster in a windy day than a normal day.
4. Humidity: If the humidity is least then rate of evaporation is faster but when humidity is high, then Atmosphere can’t hold much water vapours which causes slow rate of evaporation.

(b) Uses of evaporation:

1. Drying of clothes.
2. Separation of different mixtures like water and salt.
3. Desert coolers etc.

Question 3.
Give one example of each:
(a) Water vapours present in the atmosphere.
(b) Diffusion of a gas into liquid.
(c) Diffusion of a liquid into liquid.
(d) Diffusion of a gas into gas.
(e) Two sublime substances.
Answer:
(a) Presence of water droplets is seen on the outer surface of steel container containing cold water.
(b) Presence of dissolved oxygen in marine water which is used up by aquatic organisms.
(c) Mixing of ink and water.
(d) Burning of incense stick.
(e) Ammonium chloride, Naphthalene balls.

Question 4.
(a) What is matter? What characteristics does a matter possess?
(b) Explain all the five states of matter.
Answer:
(a) Matter: Matter is anything which occupies space and has mass.
(b) Matter has five states:

1. Solid: It is the state of matter which has a fixed shape, volume, high density, incompressible and cannot flow.
2. Liquid: It is the state of matter which possesses indefinite shape, definite volume, low compressibility and can flow easily.
3. Gas: It is the state of matter which possesses indefinite shape, indefinite volume, high compressibility and can flow easily.
4. Plasma: This state of matter consists of free electrons . and ions in the form of ionised gases.
5. BEC (Bose – Einstein Condensate): This state of matter is achieved by cooling gas of extremely to low density of about one – hundredth thousand the density of a normal air to super low temperature.

Matter in Our Surroundings Higher Order Thinking Skills (HOTS)

Question 1.
Osmosis is a special kind of diffusion, Comment
Answer:
In diffusion, molecules of a substance move from higher concentration to lower concentration. But, during Osmosis, the water or solvent molecules move from their higher concentration to the place of their lower concentration. Therefore, Osmosis is a special kind of diffusion.

Question 2.
Why does honey diffuse in water at a slower rate than ink?
Answer:
Honey diffuse at a slower rate than ink since the density of honey is greater than that of water.

Matter in Our Surroundings Value Based Questions

Question 1.
Gaurav is preparing for his cricket match in summer camp. He took full sleeves black colour silk shirt for the match. But his mother advised him to wear white half sleeve cotton shirt
(a) Why did mother advise him to wear white cotton shirt?
(b) What type of clothes we should wear in summer?
(c) What values of mother are reflected here?
Answer:
(a) Mother advised him to wear white cotton shirt because white is a good reflector of heat and cotton is a good absorber of sweat and helps in quick evaporation, which will make Gaurav feel cool during the match.

(b) Light coloured cotton clothes should be worn in summer.

(c) Mother showed her values of caring and insightful.

Question 2.
Aakansha parked her scooty in a parking of her society. In the evening, she saw her scooty tyre was burst. Then, she immediately went to mechanic to repair the puncture and advised the mechanic not to inflate the tyres fully.
(а) Why did the scooty punctured?
(b) Why she advised mechanic not to inflate the tyres fully?
(c) What values of Aakansha is reflected here?
Answer:
(a) The scooty got punctured due to bursting of tyre because the particles of air expanded and gained kinetic energy due to heat and pressurised the walls of tyre resulting in bursting.

(b) She advised mechanic not to inflate the tyres fully as the summer days were hot which caused the expansion of inflated air and create excessive pressure in the tyres.

(c) Aakansha showed her values of intelligence and awareness.

Question 3.
Ansh had a “Pooja” in his house. His mother told him to bring camphor required for “Pooja” from the market. On the “Pooja” day when he opened up the packet, he got surprised to see it empty. Then, Ansh got alert and brought a new packet.
(a) Why did the packet got empty?
(b) Name the property of matter mentioned here.
(c) Why did Ansh keep new camphor packet in a closed container?
(d) What values of Ansh are mentioned here?
Answer:
(a) Packet got emptied due to the evaporation of camphor due to its sublime nature.
(b) The property of sublimation is shown here.
(c) To prevent sublimation of camphor.
(d) Ansh shows the values of curiosity, wisdom and duty.

Question 4.
Sanjana went to a farm house with her family to spend her summer vacations. But the farm house faced the frequent problem of power cut. So, the refrigerator was not chilling the water bottles. Then, she brought two earthen pots for making drinking water, cool.
(a) How did the earthern pot, cool the water in it?
(b) Name the phenomenon involved in it
(c) Mention the values of Sanjana depicted here.
Answer:
(a) Earthen pots have small pores through which water gets evaporated making the water cool inside the pot.
(b) The process of evaporation is involved in it.
(c) Sanjana showed her intelligent, caring and responsible behaviour.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.1

Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

1. 4x² – 3x + 7
2. y² + √2
3. 3√t + t√2
4. y + \(\frac{2}{y}\)
5. x¹⁰ + y³ + t⁵⁰.

Solution:
1. 4x² – 3x + 7
This expression is a polynomial in one variable .r because in the expression there is only one variable (x) and all the indices of x are whole numbers.

2. y² + √2
This expression is a polynomial in one variable y because in the expression there is only one variable (y) and all the indices of y are whole numbers.

3. 3√t + t√2
This expression is not a polynomial because in the term 3√t, the exponent of t is \(\frac{1}{2}\), which is not a whole number.

4. y + \(\frac{2}{y}\)
This expression is not a polynomial because in the term \(\frac{2}{y}\) the exponent of y is (-1) which is not a whole number.

5. x¹⁰ + y³ + t⁵⁰
This expression is not a polynomial in one variable because in the expression, three variables (x, y and t) occur.

Question 2.
Write the coefficients of x2 in each of the following:

1. 2 + x² + x
2. 2 – x² + x³
3. \(\frac{π}{2}\)x² + x
4. \(\sqrt{2x}\) – 1

Solution:
1. 2 + x² + x
Coefficient of x² = 1

2. 2 – x² + x³
Coefficient of x² = – 1

3. \(\frac{π}{2}\)x² + x
Coefficient of x² = \(\frac{π}{2}\)

4. \(\sqrt{2x}\) – 1
Coefficient of x² = 0.

Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
One example of a binomial of degree 35 is 3x³⁵ – 4.
One example of a monomial of degree 100 is √2y¹⁰⁰

Question 4.
Write the degree of each of the following polynomials:

1. 5x³ + 4x² + 7x
2. 4 – y²
3. 5x³ – √7
4. 3

Solution:
1. 5x³ + 4x² + 7x
Term with the highest power of x = 5x³
Exponent of x in this term = 3
∴ Degree of this polynomial = 3.

2. 4 – y²
Term with the highest power of y = – y²
Exponent of y in this term = 2
∴ Degree of this polynomial = 2.

3. 5t – √7
Term with the highest power of t = 5t
Exponent of t in this term = 1
∴ Degree of this polynomial = 1.

4. 3
It is a non-zero constant. So the degree of this polynomial is zero.

Question 5.
Classify the following as linear, quadratic and cubic polynomials:

1. x² + x
2. x – x³
3. y + y² + 4
4. 1 + x
5. 3t
6. r²
7. 7x³

Solution:

1. Quadratic
2. Cubic
3. Quadratic
4. Linear
5. Linear
6. Quadratic
7. Cubic.

Zero’s of a Polynomial:
1. A zero of a polynomial p(x) is a number c such that p(c) = 0. Here p(x) = 0 is a polynomial equation and c is the root of the polynomial equation.

2. A non-zero constant polynomial has no zero. Every real number is a zero of the zero polynomial.

Some Observations:

1. A zero of a polynomial need not be 0.
2. 0 may be a zero of a polynomial.
3. Every linear polynomial has one and only one zero.
4. A polynomial can have more than one zero.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 1 Number Systems Ex 1.6

Question 1.
find:

1. 64^{\(\frac{1}{2}\)}
2. 32^{\(\frac{1}{5}\)}
3. 125^{\(\frac{1}{3}\)}

Solution:

1. 64^{\(\frac{1}{2}\)} = (8²)^{\(\frac{1}{2}\)}
   
2. 32^{\(\frac{1}{5}\)} = (2⁵)^{\(\frac{1}{5}\)}
   
3. 125^{\(\frac{1}{3}\)} =(5)^{3x\(\frac{1}{2}\)} = 5

Question 2.
Find:

1. 9^{\(\frac{3}{2}\)}
2. 32^{\(\frac{1}{5}\)}
3. 16^{\(\frac{3}{4}\)}
4. 125^{\(\frac{-1}{3}\)}

Solution:

1. 9^{\(\frac{3}{2}\)} = (3²)^{\(\frac{3}{2}\)}
   
2. 32^{\(\frac{2}{2}\)} = (2⁵)^{\(\frac{2}{5}\)}
3. 16^{\(\frac{3}{4}\)} = (2⁴)^{\(\frac{3}{4}\)} = (2)^{4x\(\frac{3}{4}\)} = (2)³ = 8
4. 125^{\(\frac{1}{3}\)} = (5³)^{\(\frac{-1}{3}\)} = (5) ^{3x\(\frac{-1}{3}\)} = (5)^-1 = \(\frac{3}{2}\)

Question 3.
Simplify:

Solution:

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 1 Number Systems Ex 1.5

Question 1.
Classify the following numbers as rational or irrational –

1. 2 – √5
2. (3 + \(\sqrt{23}\) ) – \(\sqrt{23}\)
3. \(\frac { 2\sqrt { 7 } }{ 7\sqrt { 7 } } \)
4. \(\frac{1}{sqrt { 2 }}\)
5. 2π

Solution:

1. 2 – \(\sqrt{5}\) is an irrational number
2. (3 + \(\sqrt{23}\) ) – \(\sqrt{23}\) = 3 is a rational number
3. \(\frac { 2\sqrt { 7 } }{ 7\sqrt { 7 } } \) = \(\frac{2}{7}\) is a rational number
4. \(\frac{1}{\sqrt { 2 }}\) = is an irrational number
5. 2π is an irrational number.

Question 2.
Simplify each of the following expressions –

1. (3 + \(\sqrt{3}\) ) (2 +\(\sqrt{2}\))
2. (3 + \(\sqrt{3}\) ) (3 – \(\sqrt{3}\))
3. (\(\sqrt{5}\) + \(\sqrt{2}\))²
4. (\(\sqrt{5}\) – \(\sqrt{2}\)) (\(\sqrt{5}\) + \(\sqrt{2}\))

Solution:
1. (3 + \(\sqrt{3}\) ) (2 +\(\sqrt{2}\))
= 6 + 3\(\sqrt{2}\) + 2\(\sqrt{3}\) + \(\sqrt{6}\)

2. (3 + \(\sqrt{3}\)) (3 – \(\sqrt{3}\))
= (3)² – (\(\sqrt{3}\) )² = 9 – 3 = 6

3. (\(\sqrt{5}\) + \(\sqrt{2}\) )²
= (\(\sqrt{5}\))² + 2\(\sqrt{5}\) x \(\sqrt{2}\) + (\(\sqrt{2}\))²
= 5 + 2\(\sqrt{10}\) + 2 = 7 + 2\(\sqrt{10}\)

4. (\(\sqrt{5}\) – \(\sqrt{2}\)) (\(\sqrt{5}\) + \(\sqrt{2}\))
= (\(\sqrt{5}\))² – (\(\sqrt{2}\))² = 5 – 2 = 3

Question 3.
Recall, 7t is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = \(\frac{c}{d}\). This seems to contradiet the fact n is irrational. How will you resolve this contradiction?
Solution:
No contradiction will be there. Whenever we measure any length with any device, we only get an approx, rational value and so cannot realise that either c or d is irrational.

Question 4.
Represent \(\sqrt{9.3}\) on the number line.
Solution:
Steps:

1. Draw a line segment AC = 9.3 cm and extend it to B such that CB – 1 cm.
2. Draw the perpendicular bisector of AS and mark the mid point O of AB.
3. With O as centre and OB as radius, draw a semicircle.
4. At point C, draw perpendicular CD which intersect the semicircle at D such that CD = \(\sqrt{9.3}\).
5. With C as centre and CD as radius, draw an arc which intersect AB produced at E.

Proof:
OA = OB = \(\frac{10.3}{2}\) = 5.15cm
OD = 5.15 cm
OC = OB – BC = 5.15 – 1 = 4.15 cm
CD = \(\sqrt { OD^{ 2 }-OC^{ 2 } } \)
= \(\sqrt{9.3}\) cm

Question 5.
Rationalise the denominators of the following:

1. \(\frac{1}{\sqrt { 7 }}\)
2. \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
3. \(\frac{1}{\sqrt{5}+\sqrt{2}}\)
4. \(\frac{1}{\sqrt{7}-2}\)

Solution:
1. \(\frac{1}{\sqrt { 7 }}\)

2. \(\frac{1}{\sqrt{7}-\sqrt{6}}\)

3. \(\frac{1}{\sqrt{5}+\sqrt{2}}\)

4. \(\frac{1}{\sqrt{7}-2}\)

Laws of Exponents for real numbers:
If a, n and m are natural numbers, then

1. a^m x aⁿ = a^m+n
2. (a^m)ⁿ = a^mn
3. \(\frac { a^{ m } }{ b^{ n } } \) where m > n
4. a^m x b^m – (ab)^m

Example 1.

1. 2³ x 2⁴ = 2³⁺⁴ = 2⁷
2. (3²)³ = 3^2×3 = 3⁶
3. \(\frac { 5^{ 5 } }{ 5^{ 2 } } \) = 5^5-2 = 5³
4. 2³ x 3³ = (2 x 3)³ = 6³

Example 2.
1. 6^{\(\frac{2}{5}\)} x 6^{\(\frac{3}{5}\)}
Solution:
6^{\(\frac{2}{5}\)} x 6^{\(\frac{3}{5}\)} = 6^{\(\frac{5}{5}\)} = 6

2. 6^{\(\frac{1}{2}\)} x 7^{\(\frac{1}{2}\)}
Solution:
6^{\(\frac{1}{2}\)} x 7^{\(\frac{1}{2}\)} = (6 x 7)^{\(\frac{1}{2}\)} = 42^{\(\frac{1}{2}\)}

3. \(\frac{5^{6 / 7}}{5^{2 / 3}}\)
Solution:

Example 3.
Simplify

Solution:

MP Board Class 9th Maths Solutions