Frank ICSE

Frank ICSE Solutions for Class 9 Biology – Understanding Ecosystems

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Frank ICSE Solutions for Class 9 Biology – Understanding Ecosystems

PAGE NO: 94

Solution 1:
Environment is the sum of all external conditions and influences that affect organisms. The environment may be divided into biotic i.e. living and abiotic i.e. non-living components.

Solution 2:
Ecosystems are units consisting of living things and their specific habitats in the biosphere where living things interact with each other and their environment.

Solution 3:
Man made ecosystem is formed by major human modification or alteration in the natural environment. Examples of man-made ecosystems are aquarium, park, grassland, garden etc.

Solution 4:
Two major types of ecosystems are natural ecosystem and artificial ecosystem.

Solution 5:
Two features of forests are:

  1. It includes a complex assemblage of different kinds of biotic and abiotic factors.
  2. The temperature and rainfall conditions of a place determine the nature and characteristics of forests.

Solution 6:
The trees in coniferous forests are found in single strands with no undergrowth. Plants predominantly found here include firs, pines, spruces and hemlock while the ground is covered with mosses, grasses, sedges and herbs that are adapted to cold.

Solution 7:
Tropical rainforests are found in regions that experience high temperature, high humidity, heavy and well-distributed rainfall all year round. These regions lack seasonal changes and there is little difference between the length of days and nights.
Tropical rainforests mainly occur inside the World’s equatorial regions. They are restricted to the small land area between the Tropic of Capricorn and the Tropic of Cancer.

Solution 8:
Forest biome refers to the naturally occurring community of flora and fauna occupying the forest.
Features of forest biomes are:

  1. In tropical rainforest biomes, there is an amazing biodiversity of plants and animals. Trees are tall, with buttressed trunks and shallow roots, mostly evergreen, with large dark green leaves. Plants such as orchids, bromeliads, vines, lianas, ferns, mosses, and palms are present in tropical forests. Fauna include numerous birds, bats, amphibians, reptiles, small mammals, and a huge diversity of insects.
  2. In deciduous rainforest biomes, many types of tall and short trees, shrubs, herbs, mosses and lichens are found occupying five different layers. Also a huge variety of fauna like panda, brown bear, hedgehog etc are found.
  3.  Not much biodiversity occurs in coniferous forest biomes. The trees here are found in single strands with no undergrowth. Plants predominantly found here include firs, pines, spruces and hemlock while the ground is covered with mosses, grasses, sedges and herbs that are adapted to cold. Fur bearing animals are found abundantly like brown bear, fox, mink, beavers, deer and large birds of prey like red tailed hawks. Many animals migrate or hibernate during the extremely cold winters.

Solution 9:
Frank ICSE Solutions for Class 9 Biology - Understanding Ecosystems 1

Solution 10:
(i) (d) all plants and animal species along with environment.
(ii) (c) an artificial ecosystem
(iii) (c) communities of organisms interacting with one another
(iv) (c) Tansley
(v) (c) 100-150 cm
(vi) (b) Decomposers

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Frank ICSE Solutions for Class 9 Biology – Respiration in Plants

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Frank ICSE Solutions for Class 9 Biology – Respiration in Plants

PAGE NO: 87

Solution 1:
Frank ICSE Solutions for Class 9 Biology - Respiration in Plants 1

Solution 2:
Frank ICSE Solutions for Class 9 Biology - Respiration in Plants 2

Solution 3:
Frank ICSE Solutions for Class 9 Biology - Respiration in Plants 3

Solution 4:
Yes, respiration is the reverse of photosynthesis.

Solution 5:

  1. – (b) nutrients are oxidized without using molecular oxygen by the process of fermentation.
  2. – (c) is the best organic substrate for respiration.
  3.  – (a) partial breakdown of food substance.
  4.  – (e) the series of change from glucose to pyruvic acid in respiration.
  5.  – (d) the intermediate substance in the breakdown of glucose.

PAGE NO: 88

Solution 6:
Frank ICSE Solutions for Class 9 Biology - Respiration in Plants 4

Solution 7:
(a) Anaerobic
(b) Glycolysis
(c) Pyruvic acid
(d) Oxygen
(e) Cytoplasm

Solution 8:
(a) False
(b) False
(c) False
(d) False
(e) False

Solution 9:
(a) To show that heat is evolved during respiration
(b) In flask A, moist seeds respire and produce heat that increases the temperature.
(c) If formalin was not used, bacteria will grow on the dry seeds and respire anaerobically to produce a little heat.

PAGE NO: 89

Solution 10:
glucose ATP

Solution 11:
Frank ICSE Solutions for Class 9 Biology - Respiration in Plants 5

Solution 12:
(a) To absorb carbon dioxide produced during respiration
(b) If these are not soaked in disinfectant, the bacterial growth may be there in the tube Y and accurate result may not be obtained due to bacterial respiration.
(c) The germinating peas respire and oxygen is used which create a vacuum in the tube. So coloured water has risen in tube1.
(d) Respiration
(e) It is defined as the stepwise oxidation of glucose in the living cells to release energy.

Solution 13:
(a) mitochondria
(b) Fermentation

Solution 14:
Frank ICSE Solutions for Class 9 Biology - Respiration in Plants 6

Solution 15:
Frank ICSE Solutions for Class 9 Biology - Respiration in Plants 7

Solution 16:
(a) Respires
(b) Day and night
(c) Aerobic respiration
(d) Rises
(e) Controlled manner

PAGE NO: 90

Solution 17:
(a) Respiration
(b) ATP
(c) Aerobic respiration
(d) Anaerobic respiration
(e) Caustic potash and KOH
(f) Lime water
(g) Carbon dioxide and water
(h) Ethyl alcohol and carbon dioxide

Solution 18:

(i) (c) Glucose is converted to carbon dioxide and water, releasing energy.
(ii) (d) Glycolysis, Kreb’s cycle, electron transfer
(iii) (c) Energy is left in alcohol.
(iv) (d) To accept hydrogen and form water.
(v) (b) CO2 and alcohol
(vi) (c) mitochondria
(vii) (b) Hens Krebs
(viii) (d) fermentation
(ix) (b) ATP
(x) (c) ATP
(xi) (b) Two
(xii) (a) In cytoplasm
(xiii) (c) 673 Kcal
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Frank ICSE Solutions for Class 10 Chemistry – Analytical Chemistry

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Frank ICSE Solutions for Class 10 Chemistry – Analytical Chemistry

PAGE NO : 75
Solution 1:

  1. Cuprous salts = Colourless
  2. Cupric salts = Blue
  3. Aluminium salts = Colourless
  4. Ferrous salts= Light green
  5. Ferric salts = Yellow
  6. Calcium salts = Colourless

Solution 2:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 1

Solution 3:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 2

Solution 4:
K2SO4.

Solution 5:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 3

Solution 6:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 4

Solution 7:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 5

Solution 8:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 6

Solution 9:
Examples of amphoteric hydroxides are: Zn(OH)2, Al(OH)3.

Solution 10:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 7

Solution 11:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 8

PAGE NO : 76
Solution 12:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 9

Solution 13:
The chloride of a metal which is soluble in excess of ammonium hydroxide is zinc chloride i.e. ZnCl2.

Solution 14:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 10

Solution 15:

  1. PbO
  2. Al2O3
  3. Na2ZnO2

Solution 16:

  1. transition, Cr3+, Fe2+, MnO44-.
  2. Zn(OH)2
  3. NH4Cl
  4. Al2O3, Al
  5. NH4OH

Solution 1992-1:

  1. Addition of KCN
  2. Addition of excess of NaOH.
  3. Addition of excess ofNH4OH

Solution 1993-1:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 11

PAGE NO : 77
Solution 1995-1:

  1. The metal ion present in solution A is Pb2+
    .
  2. The cation present in solution B is Cu2+. The probable colour of solution B is blue.

Solution 1996-1:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 12

Solution 1996-2:
 The solutions for the tests will be prepared by dissolving the given powders separately in water.

  1. Solution of Calcium carbonate:
    Calcium carbonate is CaCO3 and contains Ca2+ ions. Sodium hydroxide solution NaOH can be used to identify Ca2+ since its addition to calcium carbonate solution will give white precipitates of Ca(OH)2 which are sparingly soluble in excess of NaOH.
  1. Solution of Lead carbonate:
    Lead carbonate is PbCO3and contains Pb2+ ions. Ammonium hydroxide solution NH4OH can be used to identify Pb2+ since its addition to lead carbonate solution will give white precipitates of Pb(OH)2 which are insoluble in excess of NH4OH.
  2. Solution of Zinc carbonate:
    Zinc carbonate is ZnCO3 and contains Zn2+ ions. Sodium hydroxide solution NaOH can be used to identify Zn2+ since its addition to zinc carbonate solution will give white gelatinous precipitates of Zn(OH)3 which are soluble in excess of NaOH.

Solution 1996-3:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 13

Solution 1997-1:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 14

Solution 1998-1:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 15

Solution 1999-1:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 16

Solution 2000-1:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 17

PAGE NO : 78
Solution 2001-1:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 18

Solution 2003-1:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 19

Solution 2003-2:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 20

Solution 2004-1:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 21

PAGE NO : 79
Solution 2005-1:

  1. B and E (Iron (II) sulphate and Magnesium sulphate)
  2. C and F (Iron (III) chloride and Zinc chloride)
  3. D (Lead nitrate)
  4. A (Copper nitrate)
  5. F (Zinc chloride)

Solution 2006-1:
Frank ICSE Solutions for Class 10 Chemistry - Analytical Chemistry 22

Solution 2009-1:
C ( Aluminium oxide)

Solution 2009-2:

  1. P is Ferric chloride
  2. Q is an ammonium salt
  3. R is ferrous sulphate

Solution 2009-3:

  1. When BaCl2
  2.  solution is added to the given solution ZnSO4
  3. gives a white precipitate while no precipitate is obtained with ZnCl2 solution.
  4. When NaOH solution is added to the given solution, iron (II) chloride gives dirty green precipitate while reddish brown precipitate is obtained with iron(III) chloride.

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Frank ICSE Solutions for Class 9 Chemistry – Study of the First Element – Hydrogen

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Frank ICSE Solutions for Class 9 Chemistry – Study of the First Element – Hydrogen

PAGE NO :146
Solution 1:

  1. (a) Symbol of hydrogen is ‘H’ and its molecular formula is H2.
  2. (b) 2 H means 2 atoms of hydrogen while H2 means one molecule of hydrogen.
  3. (c) Lavoisier gave “Hydrogen” its name because it means water product.

Solution 2:
Hydrogen is found in free state as well as in combined  state.

  1. In free state, it is found in traces in the earth’s crust and atmosphere. On the sun and bright stars, hydrogen is very abundant.It is suggested that the source of sun’s energy is fusion of hydrogen to helium.
  2. In combined state it is found in the compounds like water, acids, organic materials and minerals. Natural gas coming out from oil wells, coal mines and volcanoes invariably contains this gas.

Solution 3:
Hydrogen resembles the alkali in following manner-

  1. Electronic configuration – Hydrogen as well as alkali metals have one electron in their valence shell.
    H (1) = 1      ; Li(3)= 2, 1   ; Na(11) =2, 8, 1
  2. Ion formation – Hydrogen loses one electron to form H+ ion like the alkali metals which form Li+, Na+, K+ etc.
  3. Valency electrons – Like alkali metals, hydrogen exhibit valency of one in its compounds , as it has only one electron in its outermost shell.
  4. Combination with non-metals – Like alkali metals, hydrogen combines with non-metals such as oxygen, chlorine and sulphur forming their oxides, chlorides and sulphides respectively.
  5. Reducing action – Like alkali metals, hydrogen is also a very good reducing agent.

Solution 4:
Hydrogen resembles the alkali in following manner-

  1. Electronic configuration – All the halogens have seven electrons in their outermost shell and need just one more electron to attain stable inert gas configuration. Similarly, hydrogen with one electron in its outermost shell requires one electron to attain a stable inert gas(Helium) configuration.
  2. Valency – Hydrogen and halogen both show a valency of one.
  3. Non-metals – Hydrogen is non-metallic like halogen.
  4. Atomicity – Hydrogen as well as halogens are diatomic gases, e.g.H2,Cl2.Br2 etc.
  5. Ion formation – Halogens have a strong tendency to gain an electron to form halide ions.In similar way, hydrogen shows tendency to gain one electron to form hydride ion (H–).

Solution 5:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 1

Solution 6:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 2

Solution 7:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 3

Solution 8:
Hydrogen gas  obtained in the laboratory by dilute H2SO4 and granulated zinc have following impurities-

  1. Hydrogen sulphide (H2S),
  2. Sulphur dioxide (SO2),
  3. Oxides of nitrogen,
  4. Phosphine (PH3),
  5. Arsine (AsH3),
  6. Carbon dioxide, Nitrogen oxides,
  7. Water vapour
    They are removed by passing through-

    1. Lead nitrate solution-It absorbs H2S.
    2. Silver nitrate solution (AgNO3)-It absorbs PH3, AsH3
    3. Caustic potash (solid KOH)-It absorbs CO2, SO2, oxides of nitrogen
    4. Anhydrous calcium chloride or P2O5-It absorbs moisture.

Solution 9:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 4

Solution 10:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 5
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 6

PAGE NO :147
Solution 11:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 7

Solution 12:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 8

Solution 13:

  1. (a) Nickel, hydrogen
  2. (b) above, dilute mineral
  3. (c) covalent, electronegative
  4. (d) CuO, hydrogen, water
  5. (e) CO, H2
  6. (f) alkali
  7. (g) nascent hydrogen
  8. (h) water

Solution 14:

  1. (a) False
  2. (b) False
  3. (c) True
  4. (d) True
  5. (e) True
  6. (f) True
  7. (g) True
  8. (h) False

Solution 15:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 9
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 10

PAGE NO :148
Solution 16:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 11

Solution 17:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 12

Solution 18:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 14

Solution 19:
Some pieces of zinc granules are placed in Woulfe’s bottle and the apparatus is made in air tight. Now dilute sulphuric acid is poured through the thistle funnel.
Observation: The reaction takes place at room temperature even without heating. A brisk effervescence with the evolution of gas is seen inside the bottle.
If burning candle is brought near the bubbles of hydrogen gas, they explode.This proves that bubbles were full of hydrogen gas and they move up in the air because they are lighter than air.

Solution 20:

  1. (a) Meteorological balloons – The low density and high lifting power of hydrogen made it useful in meteorological balloons used for studying air currents and weather conditions. However, due to its highly inflammable nature it has been replaced by helium which has a lifting only slightly less than that of hydrogen.
  2. (b) In metallurgy – Hydrogen acts as a very good reducing agent. It is used to obtain metals by reducing their oxides.
  3. (c) In fuel – Hydrogen has very high heat of combustions, therefore it is used as fuel  in the form of coal gas, water gas and liquid hydrogen(for rocket propulsion).
  4. (d) In making fertilizers – A large quantity of hydrogen is used in the manufacture of ammonia by Haber process. Ammonia is used in manufacture of HCl and methyl alcohol.

Solution 21:
Frank ICSE Solutions for Class 9 Chemistry - Study of the First Element - Hydrogen 15

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Frank ICSE Solutions for Class 9 Biology – Seeds: Structure and Germination

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Frank ICSE Solutions for Class 9 Biology – Seeds: Structure and Germination

PAGE NO: 78

Solution 1:
(a) Seed is defined as a fertilized mature ovule which possesses an inactive embryo and reserve food for its further development.
(b) The process by which the dormant embryo of the seed resumes active growth and forms a seedling is known as germination.

Solution 2:
(a) Albuminous seed – In some dicotyledons and monocotyledons, the
food is stored mainly in the endosperm. Such seeds are called albuminous seeds. Example – Seeds of castor, cereals and grasses.
(b) Dormancy – Seed dormancy is a condition of plant seeds that prevents germination under optimal environmental conditions. Here the seed is in a state of apparent inactivity and will not grow even if favorable conditions are provided, until a definite time has elapsed.
(c) Hypogeal germination – In this germination, the seed remains inside the soil since epicotyl elongates faster than hypocotyl. Hence the cotyledons remain inside the soil. Example – Wheat, rice, pea, mango.
(d) Epigeal germination – It is a type of germination in which cotyledons are pushed above the soil into the air and light. This occurs due to rapid growth and elongation of the hypocotyl. Example – Bean, cotton, castor, papaya, onion, tamarind.

Solution 3:
This is because the seed is in a state of dormancy. In this case, even if all the favorable conditions are provided, the seed remains in a state of apparent inactivity and only germinates after a definite time has elapsed.

Solution 4:
(a) Seed coat is the outer covering of seed. It protects the inner contents of the seed.
(b) Micropyle allows entry of water into the embryo.
(c) Endosperm contains stored food mostly as starch.
(d) Cotyledons store food material for the embryo.

Solution 5:
(a) Plumule
(b) Coleorhiza
(c) Endosperm
(d) Micropyle
(e) Root and shoot
(f) Endosperm
(g) Epigeal germination
(h) Hypogeal germination
(i) Orchis seed
(j) Seed of Lodoicea moldivica

PAGE NO: 79

Solution 6:
Frank ICSE Solutions for Class 9 Biology - Seeds Structure and Germination 1

Solution 7:
(a) castor, papaya
(b) grasses, wheat
(c) pea, mango
(d) wheat, rice

Solution 8:
The factors necessary for germination are:

  1. Water – Water is essential for seed germination since protoplasm becomes active only when saturated with water. Water facilitates the necessary chemical changes in food material. Also enzymatic reaction occurs only in the water medium. Water when imbibed by the seed coat makes it soft and swollen. Then the seed coat bursts open, helping the embryo come out easily.
  2. Temperature – A suitable temperature is essential for seed germination since many physiological processes occur within the seed during germination. Seeds fail to germinate below 0?C or above 45?C. Optimum temperature for seed germination is 15-30?C.
  3. Oxygen – During germination, embryo resumes growth and for this energy is required. This energy comes from the oxidation of food material stored in the endosperm or cotyledons. This process requires oxygen.

Solution 9:
Apparatus required for three beans experiment are beaker, bean seeds and wooden piece.
The air-dried seeds are attached to a piece of wood, one at each end and one in the middle. This is then placed in a beaker and water is poured into it till the middle seed is half immersed in it. The beaker is then left in a warm place for a couple of days. From time to time, water is added to maintain the original level.
It is observed that after a couple of days that the bean in the middle germinates normally since it has sufficient water, oxygen and temperature. The bottom seed gets sufficient water and temperature but not oxygen hence it may develop a radicle but doesn’t grow further. The upper seed gets oxygen and temperature but not water and hence fails to germinate.
This experiment shows that water, temperature and oxygen are essential for seed germination and that germination will not occur if any one of these factors are absent.

Solution 10:
If the seeds are sown too deep in the soil, they may not get sufficient oxygen required for respiration and hence will fail to germinate.

Solution 11:
Frank ICSE Solutions for Class 9 Biology - Seeds Structure and Germination 2

Solution 12:
(i) (d) in endosperm
(ii) (d) castor bean
(iii) (d) all the above
(iv) (d) maize
(v) (a) double coconut
(vi) (b) germination
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Frank ICSE Solutions for Class 9 Biology – Pollination and Fertilization

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Frank ICSE Solutions for Class 9 Biology – Pollination and Fertilization

PAGE NO: 72

Solution 1:
Pollination is the transfer of pollen grains from the anther to the stigma of the same or another flower.
The male gametes are produced inside pollen grains located in the anthers of androecium whereas the female gametes are produced in the ovules located in the ovary of gynoecium. For forming zygote, the male gametes need to be transferred to the gynoecium for fusing with the female gametes. This is achieved through pollination. Pollination occurs through insects, wind or other agents.
There are two types of pollination – Self pollination and cross pollination.

Solution 2:
The two modes of pollination are:

(i) Self-pollination – It is the transfer of pollens produced within the anther of a flower to the stigma of the same flower or to the stigma of another flower of the same plant. In such flowers, pollination is ensured since the flowers bear similar genetic characters. Self pollination can occur in bisexual or monoecious flowers. Examples of plants showing self pollination are Mirabilis, Arachis etc.
(ii) Cross pollination – It is the transfer of pollen grains from the anthers of a flower of one plant to the stigma of a flower of another plant. Cross pollination occurs in unisexual or dioecious flowers such as papaya, maize, jasmine, rose etc.

Solution 3:
Frank ICSE Solutions for Class 9 Biology - Pollination and Fertilization 1

Solution 4:
Adaptations required by self pollinated plants are:

  • Bisexuality – Self pollination occurs only in bisexual flowers.
  • Homogamy – Both anther and stigma need to mature at the same time.
  • Cleistogamy – Flowers which are bisexual and never open are called cleistogamous flowers. They are small, colourless, odourless and without nectar. The pollen grains fall on the stigma inside the closed flower. Example – Arachis

Adaptations required by cross pollinated plants are:

  • Unisexuality – The stamens and carpels are found in different flowers. The male and female flowers may be borne on the same or different plants.
  •  Dichogamy – In bisexual flowers, stamens and carpels mature at different times.

It is of two kinds:

  1.  Protandry wherein stamens mature before carpels. E.g – jasmine
  2. Protogyny wherein carpels mature before stamens. E.g. – Rose
  •  Heterostyly – Here the style is either longer or shorter, thereby preventing self pollination.
  •  Herkogamy – Stigma and stamen mature at the same time, but some type of barrier prevents self pollination. E.g. – In caryophyllaceous flower, the stigma projects beyond the stamens so that pollens cannot fall on it.
  • Self-sterility – Pollen of one flower cannot fertilize the female gametes of the same flower.

Solution 5:
Frank ICSE Solutions for Class 9 Biology - Pollination and Fertilization 2
Frank ICSE Solutions for Class 9 Biology - Pollination and Fertilization 3

Solution 6:
Fertilisation is defined as the fusion of the male and female gametes.

Solution 7:
Frank ICSE Solutions for Class 9 Biology - Pollination and Fertilization 4

Solution 8:
In angiosperms, during fertilization, one male gamete fuses with the egg cell and forms diploid zygote in a process called syngamy. The other male gamete fuses with the two polar nuclei to form a triploid nucleus called primary endosperm nucleus. This process is called triple fusion. Since fertilization takes place twice here, so this process is called double fertilization.
Significance – Due to double fertilization, triploid nucleus develops into endosperm which serves as nutrition for embryo.

Solution 9:
Fruit is a ripened ovary containing one or more seeds.

Solution 10:
After fertilization, ovary undergoes two important changes:

  • The ovules develop into seeds
  • The ovary walls thicken and ripen into pericarp or fruit wall.

Solution 11:
Yes, fruits are important for the plant since the seeds mature inside it. Fruits are colourful and tasty and hence eaten by animals. This helps in far and wide dispersal of the seeds.

Solution 12:
(i) (c) entomophily
(ii) (a) bats
(iii) (a) ornithophily
(iv) (a) syngamy
(v) (c) pomology
(vi) (b) true fruits
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Frank ICSE Solutions for Class 9 Biology – Flowers

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Frank ICSE Solutions for Class 9 Biology – Flowers

PAGE NO: 64

Solution 1:
A flower is the reproductive unit in angiosperms. It is a modified shoot in which internodes are shortened and leaves are modified into floral structure. Flower is meant for sexual reproduction.
A typical flower has four different kinds of whorls arranged successively on the swollen parts of a flower stalk. Flower stalk consists of the stalk called pedicel and the swollen upper part called thalamus bearing the floral leaves.
The different floral whorls are calyx, corolla, androecium and gynoecium. Calyx and corolla are accessory whorls, while androecium and gynoecium are reproductive whorls.

  •  Calyx – The calyx is the outermost whorl of the flower and its members are called sepals. Generally, sepals are green, leaf like and protect the inner whorls of the flower in bud stage. They are also involved in producing food by photosynthesis. The calyx may be gamosepalous (sepals united) or polysepalous (sepals free).
  • Corolla – It is the second whorl composed of floral leaves called petals. Petals are usually brightly coloured to attract insects for pollination. Petals also protect the inner whorls. Like calyx, corolla may be also free (gamopetalous) or united (polypetalous). The shape and colour of corolla vary greatly in plants.
  •  Androecium – It is the third whorl and is the male reproductive whorl of a flower. Androecium is composed of one or more stamens. Each stamen consists of three parts:
  1. Filament – It is the lower stalk of the stamen.
  2. Anther – Filament bears a bilobed fertile structure called anther at its distal end. Each lobe contains two pollen sacs. The pollen grains are produced in pollen-sacs.
  3. Connective – Filament of the stamen is extended in between the two anther lobes called connective.
  • Gynoecium – It is the innermost whorl and the female reproductive part of the flower. Gynoecium is made up of one or more carpels. A carpel consists of three parts namely stigma, style and ovary.
    Ovary is the swollen basal part containing ovules. Each ovary bears one or more ovules attached to a flattened, cushion-like structure called placenta.
    Style is the elongated thread like structure attached to the apex of the ovary. It connects the ovary to the stigma.
    The stigma is situated at the tip of the style and is the receptive surface for pollen grains.
  • Frank ICSE Solutions for Class 9 Biology - Flowers 1

Solution 2:
(a) Inflorescence – The arrangement of flowers on the floral axis is called inflorescence.
Function – Inflorescence facilitates the best arrangement and display of flowers on a branch without any sort of overcrowding. It also facilitates pollination via a prominent visual display and more efficient pollen uptake and deposition.
(b) Gynoecium – It is the innermost whorl of the flower bearing the female reproductive parts.
Function – The ovary of gynoecium produces ovules which bear the female gamete.
(c) Placentation – The manner in which placenta and ovules are arranged inside the ovary wall is known as placentation.
Function – Placentation helps in the best arrangement of ovules within the ovary. Placentation also helps in plant classification.
(d) Incomplete flower – A flower lacking one whorl out of the four whorls is said to be incomplete flower.
Function – An incomplete flower contains either male or female reproductive organs.
(e) Perianth – When the calyx and corolla are not distinct in a flower (eg. – lily), the whorl is collectively called perianth.
Function – The members of perianth, called tepals are usually brightly coloured and bear scent. This attracts insects which aids in pollination. They also protect the flower in bud condition.

Solution 3:
Frank ICSE Solutions for Class 9 Biology - Flowers 2

Solution 4:
The flower is the reproductive unit in the angiosperms and is meant for sexual reproduction. Flowers produce seeds from which new plants grow in future. So the main function of flower is to perpetuate the species.
There are six different types of flowers. These are complete, incomplete, bisexual, unisexual, actinomorphic and zygomorphic.

Solution 5:
Frank ICSE Solutions for Class 9 Biology - Flowers 3
Frank ICSE Solutions for Class 9 Biology - Flowers 4

Solution 6:
In certain flowers like tomato and brinjal, the calyx remains attached even after the formation of the fruit and does not wither away. Such calyx is called persistent calyx.

Solution 7:
Calyx is the outermost whorl of a flower which is composed of sepals. Generally these sepals are green, leaf like and protect the inner whorls of the flower in bud condition. They are also involved in producing food by photosynthesis.

Solution 8:
Frank ICSE Solutions for Class 9 Biology - Flowers 5

Solution 9:
Corolla is the second whorl composed of floral leaves called petals. Petals are usually brightly coloured to attract insects for pollination. Petals also protect the inner whorls. The shape and colour of corolla vary greatly in plants.

Solution 10:
The androecium and gynoecium are the essential parts of a flower because they are involved in sexual reproduction.
Androecium is the male reproductive organ of a flower and is involved in producing male gametes.
Gynoecium is the female reproductive part of the flower and produces the female gametes.

The non-essential or accessory parts of flowers are the calyx and corolla since they do not directly participate in the process of sexual reproduction leading to the development of seed.
Sepals of calyx are green, leaf like and protect the inner whorls of the flower in bud stage. They are also involved in producing food by photosynthesis.
Petals of corolla are usually brightly coloured to attract insects for pollination; they also protect the inner whorls.

Solution 11:
Frank ICSE Solutions for Class 9 Biology - Flowers 6

Solution 12:
Frank ICSE Solutions for Class 9 Biology - Flowers 7

Solution 13:
(a) Androecium – It is the third whorl and is the male reproductive organ of a flower. Androecium is composed of one or more stamens. Each stamen consists of three parts: Filament, Anther and Connective. The pollen grains are produced in pollen-sacs on the anthers.
(b) Gynoecium – It is the innermost whorl and is the female reproductive part of the flower. Gynoecium is made up of one or more carpels. A carpel consists of three parts namely stigma, style and ovary. Ovary is the swollen basal part containing ovules.
(c) Calyx – The calyx is the outermost whorl of the flower and its members are called sepals. Generally, sepals are green, leaf like and protect the inner whorls of the flower in bud stage. They are also involved in producing food by photosynthesis. The calyx may be gamosepalous (sepals united) or polysepalous (sepals free).
(d) Corolla – It is the second whorl composed of floral leaves called petals. Petals are usually brightly coloured to attract insects for pollination. Petals also protect the inner whorls. Like calyx, corolla may be also free (gamopetalous) or united (polypetalous).

PAGE NO: 65

Solution 14:
(a) Datura
(b) Cotton
(c) Cotton
(d) Sunflower
(e) Tomato
(f) Mulberry

Solution 15:
(i) (b) condensed stem
(ii) (b) jointed calyx
(iii) (c) thalamus
(iv) (a) reniform
(v) (c) capitulum

BiologyChemistryPhysicsMaths

Frank ICSE Solutions for Class 9 Physics – Motion in One Dimension

by

Frank ICSE Solutions for Class 9 Physics – Motion in One Dimension

PAGE NO: 61
Solution 1:
A body is said to be in state of rest if it does not change its position with respect to its surrounding objects with time.

Solution 2:
A vector quantity is that physical quantity which is represented by both magnitude and direction.

Solution 3:
No mass is not a vector quantity.

Solution 4:
A vector is represented by an arrow. The length of the arrow represents the magnitude of vector quantity and arrow head gives the direction of vector quantity.

Solution 5:
If a book is lying in almirah then it is at rest.

Solution 6:
A body is said to be in motion when it change its position with respect to its surrounding objects with time.

Solution 7:
Yes rest and motion are relative to each other.

Solution 8:
Out of Force and Energy, Force is a vector quantity.

Solution 9:
Examples of scalars are distance and mass.

Solution 10:
Out of these positions, (i) and (ii) positions of body lie on same straight line as direction of these two are same.

Solution 11:
A vector quantity is represented only when its magnitude and direction are specified so this quantity is a vector quantity.

PAGE NO : 62
Solution 12:
Passengers sitting in a train are at rest with respect to each other.

Solution 13:
Yes we are at rest as well as motion because we are at rest with respect to a observer which is itself at rest and we are in motion with respect to a observer which is in motion.

Solution 14:
The platform is in motion with respect to train. As train is moving with respect to platform so platform would also look in motion with respect to train.

Solution 15:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 1

Solution 16:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 2

Solution 17:
The physical quantity representing the magnitude and its direction is a vector quantity.

Solution 18:

  • Yes we can add two scalars.
  • Yes we can add two vectors.
  • Yes we can multiply two scalars.
  • No we cannot add a scalar quantity to a vector quantity.
  • Yes we can subtract two scalars.
  • No we cannot subtract a scalar quantity from a vector quantity. Reverse is also not true.
  • Yes we can multiply vectors.

Solution 19:
The actual length of the path covered by a moving object irrespective of its direction of motion is called the distance travelled by the object.

Solution 20:
No the distance covered by a body cannot be less than the magnitude of its displacement.

Solution 21:
The displacement of a moving body is defined as the change in its position along a particular direction

Solution 22:
SI unit for measurement of distance and diplacement is metre denoted by m.

Solution 23:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 3

Solution 24:
Yes a body can have negative displacement.

Solution 25:
If a body is moving in a straight line then the displacement of a body is equal to the distance travelled by it.

Solution 26:

  • Distance is a scalar quantity whereas displacement is a vector quantity.
  • Distance is always positive but displacement can be negative,zero or positive.

Solution 27:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 4

Solution 28:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 5

Solution 29:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 6

Solution 30:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 7

Solution 31:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 8

PAGE NO: 79
Solution 1:
Speed of a body can be defined as distance covered by the body in unit time.

Solution 2:
Average speed of a body can be defined as ratio of total distance covered by a body In total time.

Solution 3:
Both speed and average speed have same unit and that is ms-1.

Solution 4:
No speed and average speed of a body have different meaning.

Solution 5:
60 km/hr can be converted into m/s to compare with 15m/s.
60 km/hr = (60 x 1000)/3600 = 16.66 m/s. so speed 60 km/hr is greater.

Solution 6:
20m/s can be converted inti km/hr as
20 m/s = (20 x 3600)/1000 = 72 km/hr.

Solution 7:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 9

Solution 8:
SI unit of velocity is ms-1.

Solution 9:
No as their direction is different they don’t have same velocity.

Solution 10:
we convert all the speeds in m/s to compare them.
36 km/hr = (36 x 1000)/3600 = 10m/s.
2 km/min = (2 x 1000)/60 = 33.3 m/s.
7 m/s = 7 m/s.
So increasing order of speed is 7m/s < 10m/s <33m/s.

Solution 11:
let total distance be S.
Boy covers distance S/2 with speed u then time taken by him to cover this distance would be T1 =S/2u.
Again boy covers rest of the distance S/2 with speed v then time taken by him to cover this distance would be T2 =S/2v.
So total time taken by boy to cover the distance S is T = T1 + T2.
Total time T= S/2 (1/u +1/v) = s(u+v)/2uv.
And average speed = S/T = 2uv/(u+v).

Solution 12:
Yes uniform speed and constant speed have same meaning.

Solution 13:
let S be the distance between P and Q.
Body covers forward journey distance S (P to Q) with speed u then time taken by him to cover this distance would be T1 =S/u.
Again body covers backward journey distance S (Q to P) with speed v then time taken by him to cover this distance would be T2 =S/v.
So total time taken by body to cover the distance S is T = T1 + T2.
Total time T= S (1/u +1/v) = s(u+v)/uv.
And average speed = 2S/T = 2uv/(u+v).

Solution 14:
As body goes from P to Q and then return back to P so the displacement of the body would be zero and hence average velocity would also be zero.

Solution 15:
let distance between P and Q is S.
Speed of car while travelling from P to Q is 20 m/s.
Let car take time T1 to travel from P to Q then T1= S/20.
Speed of car while travelling from Q to P is 30 m/s.
Let car take time T2 to travel from Q to P then T2= S/30.
Total time = T1 + T2 = S/20 +S/30 =S/12.
So average speed of journey = total distance/ total time = 2S/(S/12) = 24 m/s.
Average speed of journey is 24 m/s.

Solution 16:
Speed is a scalar quantity whereas velocity is a vector quantity. So speed doesn’t have its direction and velocity has a particular direction.

Solution 17:
Speed and velocity of a moving body become equal when the body moves in a straight line.

Solution 18:
When the velocity of a moving body doesn’t change with time then the velocity of the body is said to be constant or uniform.Yes uniform velocity and constant velocity are one and the same thing.

Solution 19:
Acceleration of a body is rate of change of its velocity with respect to the time.

Solution 20:
SI unit of acceleration is ms-2.

PAGE NO : 80
Solution 21:
If the acceleration of a moving body is constant in magnitude and direction then the path of the body must not be a straight line because in circular motion also acceleration of a body is constant in magnitude and always directed towards the centre.
So the path of the body may be a straight line and may be a circular one.

Solution 22:
No the relation S = v x t cannot used to find the total distance covered by a body moving with non-uniform speed.

Solution 23:
Yes area under a speed time graph in a given interval gives the total distance covered by a body.

Solution 24:
Yes the motion is uniform and the uniform speed is given by area under speed time graph divided by time interval.
So speed = 500/20 =25 m/s.

Solution 25:
Positive acceleration corresponds to situation when velocity is continuously increasing with respect to the time.

Solution 26:
Negative acceleration corresponds to situation when velocity is continuously decereasing with respect to the time.

Solution 27:
If a body falls towards earth then it would have a positive acceleration.

Solution 28:
If a body has acceleration of 8.5 ms-2 then it means its velocity is increasing at a rate of 8.5 ms-1 per second.

Solution 29:
SI unit of retardation is ms-2.

Solution 30:
first convert 60 km/h in m/s.
60 km/hr =(60 x 1000)/3600 = 16.7 m/s.
This is initial velocity of car i.e u = 16.7 m/s.
As car stops in 10 seconds so final velocity is =0 m/s.
So acceleration = (v-u)/t = (0-16.7)/10 = -1.67 ms-2.
Acceleration of car is = -1.67 ms-2.

Solution 31:
-30 m/s is speed.

Solution 32:
Velocity corresponds to the rate of change of displacement.

Solution 33:
No the speed of a body cannot be negative.

Solution 34:
A flying bird most likely to have a non uniform velocity.

Solution 35:
Let initial velocity be u.
Final velocity is v= 0 m/s.
Time taken by body to come to rest = 10 sec.
Retardation =2.5 ms-2.
We know v = u +at.
Then u = v – at.
u = 0 – (-2.5 x 10) = 25 m/s.
So initial velocity of the body is 25 m/s.

Solution 36:
Equation of motion gives us the picture of motion of moving body.

Solution 37:
First equation of motion is v = u + at.
Second equation of motion is s= ut + 1/2a t2.
Third eqution of motion is v2 – u2 =2as.

Solution 38:
Four variables are present in each equation of motion.

Solution 39:
Four variables are present in each equation of motion and if any of three is known to us then fourth can be easily find with the help of these equation of motion.

Solution 40:
SI unit of acceleration and retardation is ms-2.

Solution 41:
Distance is the physical quantity which is equal to the area under speed-time graph.

Solution 42:
A uniformly accelerated motion is one in which speed is constantly increasing or decreasing with time while a non uniform motion is one in which speed is not constantly changing with time.

Solution 43:
No we cannot use this relation for a body moving with uniform acceleration.

Solution 44:
Slope of a graph is given as rate of change of y coordinates to the x coordinate. In speed time graph speed is on the y axis and time is on the x axis. And we define acceleration as rate of change of speed with respect to time. So slope of a speed time graph gives acceleration.

Solution 45:

  • Motion of blades of an electric fan.
  • Motion of moon around earth.

Solution 46:
A straight line curve on speed time graph indicates that acceleration of the body is uniform and a zigzag or curved line indicates that acceleration of a body is not uniform.

Solution 47:
Two quantities are directly proportional to each other.

Solution 48:
As we distance = speed x time.
Speed = 42 km/hr.
Time = 10 m = 1/6 hr.
Distance = 42 x 1/6 =7 km.

Solution 49:
Initial velocity u =10 ms-1.
Acceleration a = 2 ms-2.
Time t = 10 s.
By using first equation of motion
V = u + at.
V = 10 + 2 x 10.
V (final velocity) = 30 ms-1.

Solution 50:
Initial velocity u = 10 km/hr. = (10 x 1000)/3600 = 8.33 ms-1.
Final velocity = 64 km/hr = (64 x 1000)/3600 = 17.77 ms-1.
Time = 10 s.
Acceleration = (v-u)/t = (17.77- 8.33)/10 = 9.44/10 = 0.94 ms-2.

PAGE NO : 81
Solution 51:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 10

Solution 52:
No a body cannot have a speed negative.

Solution 53:
No2 distance covered by body during nth second is not more than the distance covered in n seconds.

Solution 54:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 11

Solution 55:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 12

Solution 56:
If speed time graph is moving upward then the body is accelerating and if it is starting from origin then it means the body has initial velocity =0.

Solution 57:
Speed time graph is moving upward then the body is accelerating and if it is not starting from origin then it means the body has some initial velocity.

Solution 58:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 13
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 14

Solution 59:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 15

Solution 60:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 16

Solution 61:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 17

Solution 62:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 18

PAGE NO: 83
Solution 1:
Displacement and velocity are two examples of vectors.

Solution 2:
SI unit of retardation is ms-2.

Solution 3:
Velocity is the physical quantity associated with the rate of change of displacement with time.

Solution 4:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 19

Solution 5:
There are three types of rectilinear motion Translational , vibrational and rotational.

Solution 6:
A body is said to have a uniform velocity if it covers equal displacement in equal interval of time.

Solution 7:
Acceleration is a vector quantity.

Solution 8:
Slope of speed time graph represents acceleration.

Solution 9:
If a stone is dropped from a certain height then it undergoes non uniform velocity motion.

Solution 10:
This means the body has a positive acceleration.

Solution 11:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 20

Solution 12:

  • No a body with constant acceleration cannot have a zero velocity.
  • No a body with an acceleration in vertical direction cannot move horizontally.
  • No in an accelerated motion a body cannot have a constant velocity.

Solution 13:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 21

Solution 14:

  • In displacement-time graph a straight line parallel to time axis shows that body is at rest position.
  • In displacement-time graph a straight line inclined to the time axis with an acute angle means body is moving with a positive velocity.

Solution 15:
No a accelerating body cannot have constant speed.

Solution 16:

  • In displacement-time graph a straight line shows body is at rest if it is parallel to time axis and shows a body is moving with uniform velocity if it is inclined to x axis.
  • In velocity-time graph a straight line shows body is moving with uniform constant velocity if it is parallel to x axis and shows body is moving constant acceleration of it is inclined to x axis.

Solution 17:
Average speed during different time intervals for a uniform motion is same.

Solution 18:
Velocity of a stone thrown vertically upward at its maximum height is Zero.

Solution 19:
Velocity of a stone thrown vertically upwards decrease because acceleration due to gravity is acting on downward direction.

Solution 20:
Linear velocity would be equal to linear speed if body is moving in a straight line.

Solution 21:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 22

Solution 22:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 23

PAGE NO : 84
Solution 23:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 24

Solution 24:
During circular motion

  • Speed remains constant.
  • Velocity changes continuously.

Solution 25:
The statement is not correct , the correct statement is “the earth is moves round the sun with constant speed”.

Solution 26:
As in circular motion direction changes continuously with motion so after two complete revolutions we can say that direction has changed infinite times.

Solution 27:
As after completing 3 revolution in circular motion the displacement is = 0. so the ratio of distance covered to the displacement is infinite.

Solution 28:
The graph becomes straight line with positive slope with time axis and represents almost a constant acceleration.

Solution 29:
Retardation is negative of acceleration so retardation the body is +3.4 ms-2.

Solution 30:
Bus is moving with initial velocity of u = 60 km/hr.
60 km/hr = ( 60 x 1000)/3600 = u = 16.66 ms-1.
Reaction time = t =1/15 sec.
Distance would the bus had moved before pressing the bus would be = u x t.
S = 16.66 x 1/15 = 1.1 m.
Now if the driver is intoxiacated then reaction time would be t = 1/2 seconds.
So S becomes S = u x t = 16.66 x 1/2 = 8.33m.

Solution 31:
Time difference of 0.1 s denotes the time taken by sound to go from device to wall and back to wall. As the distance between wall and device is 15 m so total distance covered by sound is 2 x 15 m =30 m.
So speed of sound is = total distance covered/time taken = 30/0.1 =300 ms-1.
So speed of sound is 300 ms-1.

Solution 32:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 25

Solution 33:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 26

Solution 34:
slope of velocity time graph represents acceleration of the body.

Solution 35:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 27

PAGE NO : 85
Solution 36:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 28

Solution 37:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 29

Solution 38:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 30

Solution 39:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 31

Solution 40:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 32

Solution 41:
let total distance be S.
Boy covers distance S/2 with speed A then time taken by him to cover this distance would be T1 =S/2A.
Again boy covers rest of the distance S/2 with speed B then time taken by him to cover this distance would be T2 =S/2B.
So total time taken by boy to cover the distance S is T = T1 + T2.
Total time T= S/2 (1/A +1/B) = s(A+B)/2AB.
And average speed = S/T = 2AB/(A+B).

Solution 42:
Car travls 30 km distance with speed 60 km/hr
Time taken by car to travel this distance = 30/60 = 0.5 hr.
Car travels another distance of 30 km with speed of 20 km/hr.
Time taken by car to travel this distance = 30/20 = 1.5 hr.
Total time taken = 1.5 + 0.5 = 2 hr.
Total distance = 30+ 30 = 60 km.
Average speed of car = 60/2 = 30 km/hr.

Solution 43:
Train travels first 40 km at speed of 30 km/hr.
Time taken by train to cover this distance is = distance/speed = 40/30 = 4/3 hr.
Let speed of train to cover next 80 km is v .
Then time taken by train to cover these 80 km is 80/v.
Total time becomes T = 4/3 +80/v = ( 4v + 240)/3v.
Total distance= 120 km.
Average speed = 60 km/h (given)
However average is given by = total distance /total time.
So (120 x 3 x v)/(4v +240) = 60
360 v = 240v +14400
120v = 14400
v= 14400/120 =120 km/hr.
so train has to cover those 80 km at a speed of 120 km/hr.

Solution 44:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 33

Solution 45:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 34

Solution 46:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 35

Solution 47:
Initial velocity of car u = 18 km/hr.
Final velocity of car v= 36 km/hr.
Time taken by body = 15 min. = 1/4 hr
Acceleration of car a = ( v- u )/t = (36 – 18 ) x 4 = 72 kmh-2.

Solution 48:
Initial speed of car u = 50 km/h.
Final speed of car v = 55 km/h.
Time taken by car to attain this speed is = 1 sec. = 1/3600 hr.
Acceleration of the car is = (55 – 50 ) x 3600 = 18000 kmh-2.

Solution 49:
(a) 7200 km/h2 = ( 7200 x 1000)/(3600 x 3600) = 5/9 ms-2.
(b) 1/36 m/s2 = (1 x 3600 x 3600)/(36 x 1000) = 3600 kmh-2.

Solution 50:
initial velocity u = 20 m/s.
Acceleration = 5 m/s2.
T = 2 s.
We know v= u + at.
v= 20 + 5 x 2= 30 m/s.

Solution 51:
acceleration of the car = 10 ms-2.
Initial velocity u = 10 m/s.
Final velocity v = 30 m/s.
We v = u + at.
T = (v- u)/a
T = (30 – 10 )/10 = 2 sec.
Time taken by car is 2 sec.

PAGE NO : 86
Solution 52:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 36

Solution 53:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 37

Solution 54:
Let total distance be S.
Body covers distance S/2 with speed 40 ms-1 then time taken by him to cover this distance would be T1 =S/2 x 40.
Again body covers rest of the distance S/2 with speed 60 ms-1 then time taken by him to cover this distance would be T2 =S/2 x 60.
So total time taken by body to cover the distance S is T = T1 + T2.
Total time T= S/2 (1/40 +1/60) = s(40+60)/2 x 40 x 60 = s/48.
And average speed = S/T = 48 ms-1.
So average speed is 48 ms-1.

Solution 55:
As displacement for the motion from A to B and B to A is zero so the average velocity of the body would be zero.

Solution 56:
Initial velocity of body u = 0.5 ms-1.
Final velocity of the body v = 0 ms-1 as body comes to rest finally.
Retardation of body = 0.05 ms-2.
We know that v = u + at.
0 = 0.5 – 0.05t
T = 0.5/0.05 = 10 sec.

Solution 57:
Initial speed of train = 90 km/hr
Speed of train imn m/s = ( 90 x 1000 )/3600 = 25 m/s.
Retardation of the train = 2.5 ms-2.
Final speed of train at platform = 0 m/s.
We know that v2 – u2 = 2as.
0 – 25 x 25 = 2 x (-2.5) x S
S = 625/5 = 125 m.
So driver should apply the brakes 125 m before the platform.

Solution 58:
Train travels first 30 km at speed of 30 km/hr.
Time taken by train to cover this distance is = distance/speed = 30/30 = 1 hr.
Let speed of train to cover next 90 km is v .
Then time taken by train to cover these 90 km is 90/v.
Total time becomes T = 1 +90/v = ( v + 90)/v.
Total distance= 120 km.
Average speed = 60 km/h (given)
However average is given by = total distance /total time.
So (120 x v)/(v +90) = 60
120 v = 60v +5400
60v = 5400
v= 5400/60 =90 km/hr.
so train has to cover those 90 km at a speed of 90 km/hr.

Solution 59:
speed of train = 30 km/hr.
Speed in m/s = ( 30 x 1000 )/3600 = 50/6 m/s.
Lenth of train = 50 m.
Let lenth of bridge be s metre.
Train has to cover total distance of 50+s to cross that bridge.
Time taken by train to cover this distance = 36 sec.
So as time taken = total distance /total time taken.
36 = ( 50 +s ) x 6/ 50.
1800 = 300 + 6s
6s = 1500.
S = 1500/6 = 250m
Length of bridge is 250 m.

Solution 60:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 38

Solution 61:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 39

PAGE NO : 87
Solution 62:
Frank ICSE Solutions for Class 9 Physics - Motion in One Dimension 40

Solution 63:
(i) No vehicle is moving with uniform velocity.
(ii) Vehicle B is moving with constant acceleration.
(iii) At 6 seconds both vehicles would meet.
(iv) Velocity of both the vehicles is 60 m/s when they meet.
(v) Vehicle B is ahead at the end of 7th sec and by 70 m.

Solution 64:
The given question is wrong as distance can never DECREASE with progress of time.

PhysicsChemistryBiologyMaths

Frank ICSE Solutions for Class 9 Biology – Biotechnology Applications

by

Frank ICSE Solutions for Class 9 Biology – Biotechnology Applications

PAGE NO: 53

Solution 1:
Biotechnology is the study and use of techniques using living organisms or their products for the benefit of human race. It is the controlled use of biological agents such as microorganisms or cellular components for beneficial uses.

Solution 2:
Yes.

Solution 3:
Since ancient times, microbes have been exploited for fermentation, baking etc. even without any idea of the mechanism behind them.
The various ways in which microbes were utilized in ancient times are:

  1. In ancient times (before 2500 B.C.), the Aryans used to prepare ‘Soma’ as offering to God.
  2. Preparations of curds, cheese, paneer, butter etc. from milk are age-old techniques.
  3. In Roman times, cheese was processed from milk and its nutritive value was recognized.
  4. For baking, leftover dough was applied from a previous batch of bread on leavened bread.

Solution 4:
Some food items prepared using microbes are bread, idli and dosa (South India), soya sauce, koji (Japan), temph (Indonesia), curd, cheese, butter and gari (West Africa)

Solution 5:
Two pioneer institutes in the field of Biotechnology in India are:

  1. Bhabha Atomic Research Centre, Mumbai
  2.  Council of Scientific and Industrial Research, New Delhi

Solution 6:
Four industrial applications of biotechnology are:

  1. Biotechnology is used to produce various alcoholic beverages like wine, beer, whisky, brandy and rum. For this purpose, brewer’s yeast is used for fermenting malted cereals and fruit juices. Depending on the type of the raw material used for fermentation and the type of processing, different types of alcoholic drinks are obtained.
  2.  Biotechnology also helps in the commercial production of non-alcoholic beverages like tea and coffee. Here microbes are used in a fermentation process called curing.
  3. Vinegar i.e. acetic acid is obtained by the fermentation of fruit juices.
  4.  A number of organic acids like citric acid, lactic acid, butyric acid etc. are obtained by biotechnological methods by employing several acid producing microbes.
  5. Biotechnological techniques help in the production of certain enzymes for industrial use. For example proteases, lipases and amylases are obtained from cultures of yeasts like Saccharomyces and Torula and certain bacteria.
    (Write any 4)

Solution 7:

  1. Biofertilisers are organisms that enrich the nutrient quality of the soil. The main sources of biofertilisers are bacteria, fungi and cyanobacteria. Certain bacteria and blue-green algae fix atmospheric nitrogen and make it available to plants.
  2. Certain microorganisms control or prevent spread of other organisms. They are utilized as bio-control agents. Example – Certain fungi which kill agricultural weeds are utilized as bio-weedicides.

Solution 8:

(a) In sewage treatment plants, microbes are utilized as scavengers to remove organic matter from sewage. Bacteria, algae and fungi play important roles in sewage treatment.
(b) Gobar gas or biogas is obtained from cowdung, garbage etc. when it is placed in biogas plants where anaerobic bacteria produce methane etc. The waste slurry is used as manure.

Solution 9:
Five applications of biotechnology in the field of medicine are:

  1. Antibiotics – Antibiotics are chemical substances, which are produced by some microbes and can kill or retard the growth of other pathogenic microbes, without harming the host. Biotechnology has helped us to obtain pure and large amounts of antibiotics from microbes. Examples of antibiotics obtained from microbes are penicillin, tetracycline, streptomycin etc. The world’s first discovered antibiotic penicillin is produced by Penicillium species of fungi. It destroys bacteria causing tonsillitis, sore throat, gonorrhea, local infections and pneumonia.
  2. Vaccines – Vaccines are killed or weakened pathogens employed to provide immunity against the diseases caused by them. Using biotechnology, we have been able to produce cheaper, purer, safer and more potent vaccines.
  3. Antibodies – Antibodies against disease causing pathogens can be obtained through clone cultures. These help to provide immunity against diseases.
  4. Hormones – Using recombinant DNA technology, humans have succeeded in producing several important hormones. For example – The bacteria E.coli have been successfully used to produce large quantities of human insulin. This is a boon to the many diabetic patients who are deficient in insulin and hence are unable to control their blood sugar. Also this was an improvement over the earlier used animal insulin which was expensive and allergic to many people.
  5. Diagnostic kits – Today many diagnostic kits are available for detecting many diseases especially those caused by parasites. These kits give more reliable and quicker results easily.

Solution 10:
Koji – Japan
Gari – Africa
Temph – Indonesia
Vinegar – Fermentation
Baking – Yeast

Solution 11:
(a) Idli, dosa
(b) Wine, beer
(c) Insulin
(d) Criminals
(e) biological sciences, technology

Solution 12:

  1. (d) all of the above
  2. (b) acid fermentation
  3. (b) Soma
  4. (a) weakened pathogens
  5. (a) acetic acid

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Frank ICSE Solutions for Class 9 Chemistry – Atomic Structure

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Frank ICSE Solutions for Class 9 Chemistry – Atomic Structure

PAGE NO :113
Solution 1:
Frank ICSE Solutions for Class 9 Chemistry - Atomic Structure 1

Solution 2:
Frank ICSE Solutions for Class 9 Chemistry - Atomic Structure 2

Solution 3:

  1. (a) Mass of an atom – Protons and neutrons
  2. (b) Size of an atom – Electrons

Solution 4:

Three fundamental particles of an atom are-

  1. Proton
  2. Electron
  3. Neutron

Solution 5:

  1. (a) An atom – An atom is the smallest particle of an element which may or may not be capable of independent existence.
  2. (b) An element – An element is usually defined as a pure substance that contains only one kind of particles. These particles may be atoms or molecules.

Solution 6:
Atomic number – The number of protons present in the nucleus of an atom is called the atomic number of the atom.
It is denoted by ‘Z’.

Solution 7:
The protons and neutrons are collectively called as nucleons.

Solution 8:
Isotopes are atoms of the same element, having the same atomic number, same chemical properties but different mass number, i.e., the atoms differ in the number of neutrons. The three isotopes of hydrogen atoms are-

  1. Protium
  2. Deuterium
  3. Tritium

Solution 9:
Electrons take part in a chemical reaction. Therefore, the chemical properties of an element depend upon the electronic configuration. Since, isotopes of an element have the same atomic number and hence same electronic configuration. So, they exhibit the same chemical properties.

Solution 10:
Frank ICSE Solutions for Class 9 Chemistry - Atomic Structure 3

Solution 11:
The outermost orbit of an element is called valence shell. The electrons present in the outermost orbit (valence shell) of an element are called valence electrons.

Solution 12:
Atoms may have incomplete octet. During the formation of a molecule, an atom of a particular element gains, loses or shares electrons until it acquires a stable configuration of eight electrons in its valence shell.

Solution 13:

  1. (a) Atomic number = Number of protons = 20
  2. (b) Mass number = Number of protons +Number of electrons = 20 + 20 = 40
  3. (c) Electronic configuration = 2, 8, 8, 2
  4. (d) Valency = 2

Solution 14:

  1. (a) A=1
    B=3
    C=2
    D=1
    E=1
  2. (b) E contains the greatest number of neutrons.
  3. (c) A contains the least number of electrons.
  4. (d) B contains equal number of electrons and neutrons.
  5. (e) A is a metal.
  6. (f) Fluorine is the most reactive of the non-metals.
  7. (g) Both are electronegative and non-metals.
  8. (h) D and E belongs to a particular family.
  9. (i) A3B will be the formula if atoms of A combines with atoms of B.
  10. (j) ‘B’ will combine in  a trivalent  element while ‘C’ will combine as a bivalent element.

PAGE NO :114
Solution 15:
“During the formation of the molecule, an atom of a particular element gains, loses electrons or shares electrons until it acquires a stable configuration of eight electrons in its valence shell” i.e. until it acquires octet.

Solution 16:
K can accommodate maximum of 2 electrons.
L can accommodate maximum of 8 electrons.
M can accommodate maximum of 18 electrons.

Solution 17:
Cathode rays are formed at the negative electrode of the discharge tube experiment.

Solution 18:

  1. In K maximum number of 2 electrons can be accommodated.
  2. In L maximum number of 8 electrons can be accommodated.
  3. In M maximum number of 18 electrons can be accommodated.
  4. In N maximum number of 32 electrons can be accommodated.

Solution 19:
Frank ICSE Solutions for Class 9 Chemistry - Atomic Structure 4

Solution 20:
Electronic configuration of magnesium is -2, 8, 2. Since, it has 2 electrons in its valence shell, so its valency is 2.

Solution 21:
Number of electrons in Sodium = 11
Number of protons in sodium = 11
Number of neutrons in sodium = 12
Number of nucleons in sodium = 23

Solution 22:
Inert elements are the elements which have completely filled valence shell. Since, they are already stable and do not need more electrons, they do not combine with other atoms. So, they exist as monoatoms in molecule.

Solution 23:

  1. (a) Hydrogen
  2. (b) K shell.
  3. (c) Magnesium
  4. (d) Isotopes
  5. (e) Helium has zero valency.

Solution 24:

  1. (a) The number of protons = 9
  2. (b) The number of neutrons = 19 – 9 = 10
  3. (c) The number of electrons = 9

Solution 25:
Atomic number is the number of protons of an atom which is unique to an atom but mass number is the total of number of protons and number of neutrons which may or may not be same to other atoms since there is probability of combination of number of protons and number of neutrons be same for two atoms.

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Frank ICSE Solutions for Class 9 Chemistry – Water

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Frank ICSE Solutions for Class 9 Chemistry – Water

PAGE NO :96
Solution 1:
The molecular formula of water is H2O, this shows that water is not an element but it is a compound made up of two elements hydrogen and oxygen  combined in a fixed ratio, i.e., 1:8 by mass. This also shows that the properties of water is totally different from those of hydrogen and oxygen as compounds have completely different properties than the elements of which it is made up of.

Solution 2:
Two characteristics properties responsible for making water a good solvent are:

  1. Water molecule can form hydrogen bonds with the molecules of other compounds.
  2. Water molecule is polar in nature.

Solution 3:
Frank ICSE Solutions for Class 9 Chemistry - Water 1

Solution 4:

  1. (a) Physical properties of water:
    1. Pure water is colourless, transparent, odourless and tasteless liquid.
    2. Boiling and Freezing points: At normal atmospheric pressure water boils at 100oC and freezes at 0oC.
    3. Pure water being a covalent compound is a very poor conductor of heat and electricity.On addition of electrolytes it becomes a good conductor of heat and electricity.
    4. Solvent properties:Water is a remarkable solvent dissolves many substances forming aqueous solutions because of its high dielectric constant.
    5. Anomalous behavior: On cooling water contracts in volume, as do other liquids, but at 4o C, it starts expanding, and continues to do till the temperature reaches 0oC,the point at which it converts to ice.
  2. (b)
    1. Boiling point of water: It increases with the addition of salt in water.
    2. Freezing point of water: It decreases with the addition of salt in water.
    3. Density of water: It increases with the addition of salt in water.

Solution 5:
Due to its high specific heat capacity, water is used in cooling systems for e.g., as coolant in motor car radiators.

Solution 6:
Frank ICSE Solutions for Class 9 Chemistry - Water 2

Solution 7:

  1. (a) Water reacts with metals oxides: corresponding hydroxides are formed which act as alkalis.
  2. (b) Water reacts with non metals oxides: corresponding acids are formed.

Solution 8:
Frank ICSE Solutions for Class 9 Chemistry - Water 3

Solution 9:
We can test the odourless, colourless liquid by:

  1. Measuring its boiling point, if comes out to be 100 oC then it confirms the liquid to be water.
  2. Measuring its melting point, if comes out to be 0 oC then it confirms the liquid to be water.

Solution 10:
Frank ICSE Solutions for Class 9 Chemistry - Water 4

Solution 11:

  1. (a) Water being the universal solvent ,acts as an important mode of transport in plants,also water taken from natural source contains dissolved salts in them which is essential for the growth and development of plants
  2. (b) Water constitutes 3/4th of our human body and it is necessary for the survival of human beings and being the universal solvent helps in transportation and also regulates the body temperature.It also contains dissolved salts which supply essential minerals which are necessary for our body.

Solution 12:
Frank ICSE Solutions for Class 9 Chemistry - Water 5

Solution 13:
Frank ICSE Solutions for Class 9 Chemistry - Water 6

Solution 14:
Density of water is 0.997 gcm-3 at 4oC . The density of water decreases when the temperature increases above 4oC or decreases below 4oC.
At 40C, water has its maximum density and minimum volume. At any temperatur above or below 4 C, the density of water decreases. This property is called anomalous expansion of water. The significance of this unique property of water is that it enables marine life to exist in the colder regions of the world, because even when water freezes on the top, it is still liquid below the ice layer.

Solution 15:
At 40C, water has its maximum density and minimum volume. The property of anomalous expansion of water enables marine life to exist in the colder regions of the world because even when water freezes on the top, it is still liquid below the ice layer.

PAGE NO :97
Solution 16:
Frank ICSE Solutions for Class 9 Chemistry - Water 11

Solution 17:
Frank ICSE Solutions for Class 9 Chemistry - Water 7

Solution 18:

  1. (a) Liquids: alcohols, acids
  2. (b) Solids: Sugar, urea
  3. (c) Gases: Oxygen, carbon dioxide

Solution 19:
Rain water is the purest form of natural water.

Solution 20:

  1. (a) Solid in a liquid: Solubility decreases with rise in temperature in an exothermic process, for example, calcium sulphate(CaSO4), sodium sulphate(NaSO4), and in an endothermic reaction solubility increases with rise in temperature,for example potassium nitrte(KNO3) and sodium nitrate(NaNO3).
  2. (b) A gas in a liquid: Solubility decreases with rise in temperature.Thus, gases dissolves readily in cold water than in hot water.

Solution 21:
Frank ICSE Solutions for Class 9 Chemistry - Water 8

Solution 22:

  1. (a) Density of water is 0.997 gcm-3 at 4oC .The density of water decreases when the temperature decreases below 4oC. So ice has less density of 0.92 and is lighter  then that of water hence it floats on water.
  2. (b) A solution is said to be saturated if at a particular temperature if no more of the solute can be dissolved in it at that temperature, when the temperature is increased more solute particles can be dissolved in that saturated solution as the solubility of most of the substances generally increases with rise in temperature and thereby making it unsaturated.
  3. (c) Solubility of gases decreases with increase in temperature hence hot water contains less dissolved air than cold water.

Solution 23:

  1. (a) Solute: The substance which dissolves in a medium to produce a solution is called the solute.
  2. (b) Solvent: It is the medium in which the solute dissolves.
  3. (c) Solution: A homogeneous mixture that has a uniform composition throughout the volume of the mixture. It is the medium in which the solute dissolves.

Solution 24:

  1. (a) When we heat a saturated solution more solute can be dissolved in it, it as mostly solubility increases on raising the temperature and the solution then becomes an unsaturated solution.
  2. (b) When a hot saturated solution is cooled slowly and is kept undisturbed the excess salt does not separates out . Thus the solution becomes a supersaturated solution which contains more solute in it than it can hold at room temperature.
  3. (c) When the quantity of solvent increases,greater amount of solute can be dissolved.

Solution 25:
Frank ICSE Solutions for Class 9 Chemistry - Water 9

Solution 26:
Frank ICSE Solutions for Class 9 Chemistry - Water 10

Solution 27:
Solubility of a solid depends on:

  1. Size of particles: Smaller the size of particles of the solute, greater is the solubility.
  2. Contact between solute and the solvent: Increased contact between the solute and the solvent increases the solubility.
  3. Temperature: The variation of solubility with temperature depends on its nature i.e. whether the reaction is exothermic or endothermic.
    For exothermic reactions-The solubility decreases on increasing the temperature.
    For endothermic reaction-The solubility increases on increasing the temperature.

Solution 28:
Henry’s law states that:
At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

Solution 29:

  1. Z is least reactive so it will be found free in nature.
  2. X is more reactive so it will react with oxygen more readily and with ease.
  3. Z
  4.  Y is iron whose oxide is brown coloured Fe2O3.
  5. X
  6. The piece of Y i.e., iron is more reactive then copper so it will displace copper from copper nitrate and a brown layer of copper will be developed on the iron piece.

Solution 30:
Drop in, a crystal of the solute that is in the solution. If the crystal dissolves its unsaturated, if it stays the same size then its saturated and if it gets bigger than its supersaturated.

  1. This is because an unsaturated solution will be able to take in more solute, so it dissolves the crystal.
  2. A saturated solution will not take in more solute but will also not deposit any solute so the crystal stays the same size.
  3. A supersaturated solution wants to deposit its excess solute and become saturated, it just needs something to start it along.

The crystal starts it up and it will deposit its excess solute onto the crystal making the crystal bigger.

PAGE NO :98
Solution 31:

  1. (a) Increase in weight: when sodium chloride when exposed in air it gains moisture and its weight increases.
  2. (b) Decrease in weight: iron when exposed to air gets rusted due to the presence of moisture in the air it gets corroded, and hence its weight is reduced.
  3. (c) No effect on weight: when conc. Sulphuric acid when exposed to air  does not react and no change occurs in weight.

Solution 32:
Water pollution may be defined as the contamination of water by foreign substances which make it harmful for health of animals or plants or aquatic life, make it unfit for drinking and for domestic, industrial and agriculture use.

Solution 33:
Major pollutants of water are:
Ground water pollutants: Septic tanks, industry by products like pesticides, fertilizers, tanneries, mining wastes.
Surface water pollutants: No. of gases present in atmosphere like SO2, CO2, H2S, NOx, CO, etc., pollute surface water.
Lake water pollutant: organic wastes from hills, toxic effluents from urban areas, industrial effluents, dumping of huge amounts of sediments etc.,
River water pollutants: Industrial discharge, sewage discharges, detergents, discharges from drug, paper, and textile industries
Marine water pollutant: major pollutant is oil spills.

Solution 34:
Various sources of water pollution are:

  1. Natural process: Washing away of decomposed and animal wastes into main stream of water.
  2. Human activity: 
    1. (a)Discharge of Household detergents:  detergents used as a cleaning agent produce foam and pollute water. They do not undergo bio-degradation.
    2. (b) Discharge of industrial effluents: Industrial activities generate a variety of waste products which are generally discharged into water streams.The pollutants associated with the industrial effluents are  organic matter, inorganic dissolved salts, suspended solids. They inhibit oxidation or organic compounds; stabilize the colloidal impurities which do not aggregate to settle down.
    3. (c) Sewage:  Sewage is cloudy dilute aqueous solution containing minerals and organic matter.Sewage from homes and industries contains decomposable organic matter ,inorganic cations and anions,toxic metals etc.Pouring the drains and sewers in fresh water bodies causes water pollution.
      Water pollution due to sewage creates the following problems:

      1. Self purifying ability of water is lost and it becomes unfit for domestic purpose.
      2. Self regulatory capabilities of aquatic organism is retarded.
      3. Sewages produce pathogens which are diseases causing bacteria and result in water born gastro-intestinal diseases.

Solution 35:
Water treated for safe effluents involves sequential treatment:

  1. Primary treatment: The primary treatment involves physico-chemical processes to reduce settle able suspended solids of the wastes water and smoothened out individual effluent flow variations.It involves physic-chemical processes such as sedimentation,aeration,adsorption,oxidation etc.
  2. Secondary treatment: In this, the dissolved and colloidal organic matter present in waste is removed by biological processes involving bacteria and other micro organisms.
    These process may be aerobic or anaerobic.
    Aerobic treatment: Purification is carried out by aerobes in the presence of molecular oxygen.
    Anaerobic treatment: Purification of waste is achieved by anaerobes in complete absence of molecular oxygen.
  3. Tertiary treatment: It is the final treatment for polishing the effluents from secondary treatment. By this process suspended solids are removed, bacteria are removed, and organic and inorganic solids are removed.

Solution 36:
Two water born diseases are:

  1. Gastroenteritis
  2. Bacterial dysentery

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Frank ICSE Solutions for Class 9 Chemistry – Physical and chemical changes

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Frank ICSE Solutions for Class 9 Chemistry – Physical and chemical changes

PAGE NO :74
Solution 1:
A physical change is a temporary change in which no new substance is formed and the composition or identity of the substance is not altered although certain specific physical properties may be changed.

Solution 2:
A chemical change is a permanent change in which the original substance gives rise to one or more substances with different properties.

Solution 3:
The reactions in which heat is evolved are called exothermic reactions while the reactions in which heat is absorbed are called endothermic reactions.

Solution 4:

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

Solution 5:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 1

Solution 6:
Possible conditions for a chemical change are-

  1. One or more new substance is formed during reaction.
  2. The change occurring during the reaction is permanent.
  3. The mass of the substance undergoing a chemical change is generally altered.
  4. Chemical change involves  making and breaking of bond.

Solution 7:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 2

Solution 8:

(a) Redox reaction – The reaction in which both oxidation and reduction takes place simultaneously is known as rtedox reaction. Oxidation is a reaction that involves the addition of oxygen or the removal of the hydrogen. Reduction is a reaction that involves the addition of hydrogen or the removal of oxygen.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 3
(b) Oxidation – Oxidation is a reaction that involves the addition of oxygen or the removal of the hydrogen. In electronic concept, it is defined as the process in which an atom, molecule or ion loses one or more electrons. This results in increase in the positive charge or decrease in negative charge on the resulting species.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 4
(c) Reduction – Reduction is a reaction that involves the addition of hydrogen or the removal of oxygen. In the electronic concept, it is defined as the process in which an atom, molecule or ion gains one or more electrons. This results in increase in the negative  charge or decrease in positive charge on the resulting species.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 5

Solution 9:

  1. Exothermic reaction
  2. Endothermic reaction

Solution 10:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 6

Solution 11:
The chemical reactions which occur with the absorption of light energy are called photochemical reactions.
Examples-
Decomposition of silver nitrate takes place in the presence of light.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 7

Solution 12:
As the burning substance combines with oxygen ,the total mass of the products should be greater than that of the burning substance. For example,when,magnesium is burnt,a new substance magnesium oxide is formed,whose weight is greater than that of the original magnesium.
Experiment – A crucible is weighed containing about 0.5 gm of magnesium.Now crucible is heated.When magnesium begins to burn,the lid is put back on the crucible and the lid is occasionally raised to allow air to enter and burn the magnesium such that no product is lost.When,all the magnesium has been burnt up, the crucible is allowed to cool and then on weighing it  we observe that there is gain in weight.

Solution 13:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 8

Solution 14:
Three conditions necessary for burning are-

  1. The substance to be burnt must be combustible.
  2. A supporter of combustion such as air or oxygen must be present.
  3. A combustible substance must be heated to its ignition temperature.

PAGE NO :75
Solution 15:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 9

Solution 16:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 10

Solution 17:
When oxidation occurs there is a loss of electrons but simultaneously there is a gain of electrons by other species which is called reduction. These both process occur simultaneously so we can say that both oxidation and reduction go hand in hand and such reactions are known as redox reaction.

Solution 18:

  1. Copper is oxidized to copper sulphate while sulphur in sulphuric acid is reduced to sulphur dioxide.
  2. Silver in silver oxide is reduced to silver while oxygen in hydrogen peroxide is oxidised to molecular oxygen.

Solution 19:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 11
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 12

Solution 20:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 13

Solution 21:
A candle is a stick of paraffin wax with cotton wick.As a candle burns wax melts and trickles down. It gets solidified shortly. This is physical change. Paraffin wax is a mixture of hydrocarbons. When wick catches fire, paraffin wax melts, evaporates and burns in air like any hydrocarbon to give carbon dioxide and water. This is a chemical change.

Solution 22:

  1. physical
  2. chemical
  3. chemical
  4. physical

Solution 23:
Two examples are:-

  1. Burning of wood-carbon get oxidized and oxygen gets reduced.
  2. Rusting-In it iron is oxidized.

Solution 24:

  1. Chromium(VI) .
  2. Hydrogen peroxide
  3. Barium carbonate.
  4. Silver nitrate.
  5. Manganese dioxide.

Solution 25:
Ignition temperature – Ignition temperature is the lowest temperature up to which temperature of a substance must be raised so that it catches fire.
A combustible substance must be heated to its ignition temperature for burning.

Solution 26:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 14

Solution 27:
On heating few crystals of iodine in a test tube, the grey crystals sublimes and dense violet fumes are seen. On cooling, the vapours again form the crystals. So, a physical change can be reversed.

Solution 28:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 20

Solution 29:
When water is freezed and evaporated, these both are physical changes because-

  1. The change is temporary and reversible.
  2. No new substance is formed and the chemical composition of the original substance remains the same.
  3. Mass of the substance remains unchanged
  4. The amount of energy required to bring about a physical change is generally equal to the amount of energy required to reverse the change. Hence,there is no net energy change involved.

Solution 30:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 15

Solution 31:
Air is necessary for burning. Incorrect amount of air in fuel combustion accounts for the largest losses in combustion system. If the fuel does not get enough air for combustion it will generate smoke and a potentially unhealthy mixture of gas products.

Solution 32:

  1. (a) Combustible substances -The substances that catch fire and burn easily. Ex-Wood, Charcoal, petrol, kerosene etc.
    Non-combustible substances-Substance which cannot burn in air or oxygen are called as non- combustible substances. Ex-Nitrogen gas, carbon dioxide etc.
  2. (b) Two substances other than oxygen that support combustion are-
    1. Hydrogen
    2. Nitrogen

Solution 33:

  1. (a)
    1. Burning of coal in air releases CO2 in air.
    2. Respiration releases carbon dioxide and water vapours.
  2. (b)
    1. Photosynthesis removes CO2 from the atmosphere. Plants take carbon dioxide from the atmosphere in the presence of sunlight and use it to synthesise glucose with the liberation of oxygen.
    2. Some man made chemical activities such as setting of mortar also use atmospheric carbon dioxide and helps in removing carbon dioxide.

Solution 34:
Nitrogen is inert in nature and does not support combustion while oxygen supports combustion.If proportions of nitrogen and oxygen in the air were reversed then the rate of combustion of substances will increase.

Solution 35:
Heating of sulphur – If some powdered sulphur is heated gently in a glass test tube, it melts to a pale yellow liquid. Flame is removed to stop heating, it is quickly changed back to solid sulphur.

PAGE NO :76
Solution 36:
Activity series – The arrangement of the metals in the decreasing order of their chemical reactivity is called the activity series.
In displacement reactions, a more reactive element (metal or non-metal) displaces a lesser reactive element from its compound. With the help of the activity series, it is possible to predict which metals will displace other metals from their solutions.

Solution 37:
Balance of oxygen and carbon dioxide is maintained in nature because there is a natural oxygen cycle and a natural carbon cycle operating all the time by which the desired proportions of the two gases in the air are maintained. This is also known as carbon dioxide-oxygen cycle.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 16

Solution 38:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 17

Solution 39:
Carbon dioxide from the atmosphere enters the plant through photosynthesis, where carbohydrates are produced. From green plants, the carbon in the form of carbohydrates, etc. enter the animal and human bodies. The atmospheric carbon dioxide gets dissolved in oceans by diffusion. Marine algae and photosynthetic bacteria obtain carbon dioxide from water.
Carbon dioxide returns to the atmosphere by respiration, combustion of fossil fuels like coal, wood, petroleum etc., weathering of rocks, volcanic eruptions etc.
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 18

Solution 40:
Frank ICSE Solutions for Class 9 Chemistry - Physical and chemical changes 18

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