Veerendra

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:

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

PAGE NO: 87

Solution 1:

Solution 2:

Solution 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:

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:

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:

Solution 15:

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

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:

Solution 3:

Solution 4:
K₂SO_4.

Solution 5:

Solution 6:

Solution 7:

Solution 8:

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

Solution 10:

Solution 11:

PAGE NO : 76
Solution 12:

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

Solution 14:

Solution 15:

1. PbO
2. Al2O3
3. Na2ZnO2

Solution 16:

1. transition, Cr³⁺, Fe²⁺, MnO₄^4-.
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:

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:

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:

Solution 1997-1:

Solution 1998-1:

Solution 1999-1:

Solution 2000-1:

PAGE NO : 78
Solution 2001-1:

Solution 2003-1:

Solution 2003-2:

Solution 2004-1:

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:

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 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:

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:

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

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:

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:

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

Solution 7:

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

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.
• 

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:

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:

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:

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:

Solution 12:

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

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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:

Solution 16:

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:

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:

Solution 28:

Solution 29:

Solution 30:

Solution 31:

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:

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:

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:

Solution 55:

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:

Solution 59:

Solution 60:

Solution 61:

Solution 62:

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:

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:

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:

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:

Solution 22:

PAGE NO : 84
Solution 23:

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:

Solution 33:

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

Solution 35:

PAGE NO : 85
Solution 36:

Solution 37:

Solution 38:

Solution 39:

Solution 40:

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:

Solution 45:

Solution 46:

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:

Solution 53:

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:

Solution 61:

PAGE NO : 87
Solution 62:

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

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

BiologyChemistryPhysicsMaths

Frank ICSE Solutions for Class 10 Chemistry – Study Of Acids, Bases and Salts

PAGE NO : 62
Solution 1:

Solution 2:

1. (i) Hydrogen chloride HCl
   (ii) Nitric acid HNO₃
2. (i) Carbonic acid H₂CO₃
   (ii) Oxalic acid (COOH)₂
3. (i) Sulphuric acid H₂SO₄
   (ii) Hydrogen chloride HCl
4. (i) Carbonic acid H₂CO₃
   (ii) Acetic acid

Solution 3:

Solution 4:

1. The pH of a solution is defined as the negative logarithm (base 10) of the hydronium ion concentration present in the solution.
   pH =-log₁₀ [H₃O⁺]
2. The three applications of pH scale are:
   • It is used to determine the acidic or basic nature of the solution.
   • It is used to determine hydronium ion concentration present in the solution.
   • It is used to find out neutrality of the solution.

Solution 5:

Solution 6:

Solution 7:

1. Base in solution furnishes the ions:
   Hydroxide ion/ oxide ion and a metallic ion.
2. A weak alkali furnishes the ions:
   Hydroxide ion and metallic ion and molecules of weak alkali./
3. An acid in a solution furnishes the ions:
   Hydronium / Hydrogen ion and a negative ion.

Solution 8:

PAGE NO : 63
Solution 9:

1. CaO
2. NaOH
3. CuO
4. Cu[(OH)₂]
5. H₂CO₃
6. Ferric hydroxide [Fe (OH)₃].
7. CuO
8. NH₃

Solution 10:
Anhydrous hydrogen chloride is not an acid but its aqueous solution is a strong acid because anhydrous means without water and we know that the property of acidity is shown by a substance only when it is dissolved in water or its aqueous solution is prepared.

Solution 11:

Solution 12:
Strength of an acid measures the ease with which the acid can ionize to produce hydrogen or hydronium ions when dissolved in water. Those acids which can easily ionize to form hydrogen ions are called strong acids while those which can partially ionize to form hydrogen ions are called weak acids.
Strength of an acid depends upon many factors such as:

1. Molecular structure of the acid
2. The temperature
3. Properties of the solvent

Solution 13:

Solution 14:
Solution B with pH value 9 will give pink colour with phenolphthalein.
Concept Insight: Bases give pink colour with phenolphthalein because a base will abstract two protons from phenolphthalein and the resulting phenolphthalein ion provides pink colour to the solution.

Solution 15:
Two indicators for identification of acid are methyl red and Thymol blue.

Solution 16:

Solution 17:

Solution 18:

Solution 19:

Solution 20:

1. Efflorescence: It is the phenomenon by which hydrated salts on exposure to dry air, lose their water of crystallization and crumble to powder.
2. Hygroscopy: It is the phenomenon by which substances absorb moisture from air, but only sufficiently so as to become wet.
3. Water of crystallization: It is the fixed amount of water that is present in a crystal as an integral part of its constitution. Hydrated salts are salts having water of crystallisation.

Solution 21:
Deliquescence is the phenomenon by which certain salts absorb moisture from air, lose their water of crystallization and dissolve in it to form a saturated solution.
The substances which exhibit deliquescence are called deliquescent.
For example Caustic soda NaOH, Caustic potash KOH.

Solution 22:

Solution 23:

Solution 24:

1. Common salt gets wet during rainy season because the commercially available salt contains impurities, like magnesium chloride, which are deliquescent substances. These absorb moisture from atmosphere and make the table salt wet.
2. (i)  Na₂CO₄.10H₂O = Washing soda
   (ii) MgSO₄.7H₂O = Epsom salt
   (iii)CuSO₄.5H₂O = Blue vitriol
   (iv) ZnSO₄.7H₂O = White vitriol

PAGE NO : 64
Solution 1996-1:

1. pH of a solution having pH 7 can be increased by adding a base to it such as NaOH.
2. pH can be decreased by adding an acid such as HCl to it.
   If a solution changes colour of litmus from red to blue, it shows that its pH is above 7.

Solution 1996-2:

Solution 1996-3:

1. Zinc sulphate = Zinc and dilute sulphuric acid
2. Copper sulphate = Copper oxide and dilute sulphuric acid
3. Sodium sulphate = Sodium carbonate solution and dilute sulphuric acid
4. Lead sulphate = Lead carbonate and dilute sulphuric acid

Solution 1997-1:
The term acid salt means the salt formed by partial replacement of the hydrogens present in the acid by metallic or ammonium ions.
For example: NaHCO3

Solution 1997-2

1. pH scale is used to express the acidic or basic nature of solution.
2. pH of pure water is 7 since it does not have any impurities.
3. (a) A soluble oxide of A will have pH less than the pH of pure water i.e. below 7.
   (b) A solution of ‘B’ will have more pH than pure water i.e. above 7.

Solution 1997-3:

1. Water of crystallization: It is the fixed amount of water that is present in a crystal as an integral pat of its constitution. Compounds having water of crystallization are called hydrous salts.
   For example: Sodium carbonate Na2CO3 has 10 molecules of water present as water of crystallization Na₂CO₃.10H₂O
2. Anhydrous: Hydrous salt on heating lose their water of crystallization, such salts are then called anhydrous.
   For example:Na₂CO₃.10H2O on losing 10 molecules of water forms Na₂CO₃

Solution 1997-4:

Solution 1998-1:

1. Water of cystallization.
2. White.
3. Efflorescence.
4. Sodium chloride.

Solution 1998-2:
Those acids which ionize partially in aqueous solution and thus they contain ions as well as molecules of the acid. Organic acid such as CH₃COOH, is a weak acid.

Solution 1998-3:

PAGE NO : 65
Solution 1998-4:
The name and formula of the acid salt which gives sodium ions and sulphate ions in solution is Sodium hydrogen sulphate NaHSO₄

Solution 1999-1:

Solution 2000-1:

Solution 2001-1:

PAGE NO : 66
Solution 2002-1:

Solution 2003-1:

1. Hydronium, positive.
2. Acid, metal.

Solution 2003-2:

Solution 2004-1:
methods for preparation:

1. Preparation of copper(II) chloride.
   Action of an acid on an oxide or carbonate
2. Preparation of iron(III) chloride.
   Direct combination
3. Preparation of iron (II) chloride.
   Action of an acid on a metal
4. Preparation of lead (ii) chloride
   Precipitation (double decomposition)
5. Preparation of sodium chloride
   Neutralization of an alkali by an acid.

Solution 2005-1:

PAGE NO : 67
Solution 2005-2:
Positive, hydroxyl, Salt, Neutralization.

Solution 2005-3:
When neutral litmus solution is added to sodium hydrogen carbonate solution, litmus solution turns red

Solution 2006-1:

1. From pink to colourless.
2. From orange to pink.
3. From colourless to red.

Solution 2007-1:

1. Hydronium
2. Hydroxide
3. Salt
4. Water
5. Hydrogen

Solution 2007-2:

Solution 2008-1:

1. Complex salt.
2. Alkali.

Solution 2009-1:
Acidified potassium dichromate paper

PAGE NO : 68
Solution 2009-2:

1. Solution B.
2. Solution A.
3. Solution B
4. Solution of ammonium hydroxide NH₄OH is a weak alkali.

Solution 2009-3:

ChemistryBiologyPhysicsMaths

Frank ICSE Solutions for Class 9 Physics – Measurement

PAGE NO: 15
Solution 1:
Measurement is an act or the result of comparison of a quantity whose magnitude is unknown with a predefined standard.

Solution 2:
The physical quantities like mass, length and time which do not depend on each other are known as fundamental quantities.

Solution 3:
Length, mass, time are the three fundamental quantities.

Solution 4:
Unit is a standard quantity of the same kind with which a physical quantity is compared for measuring it.

Solution 5:
A standard metreis equal to 1650763.73 wavelengths in vacuum, of the radiation from krypton isotope of mass 86.

Solution 6:
Three systems of unit are

• C.G.S system – fundamental unit of length is centimetre(cm), of mass is gram(gm), of time is second(s).
• F.P.S system– fundamental unit of length is foot(ft), of mass is pound(lb), of time is second(s).
• M.K.S system– fundamental unit of length is metre(m), of mass is kilogram(kg), of time is second(s).

Solution 7:
The SI unit of mass is Kilogram. One standard kilogram is equal to the mass of a cylinder of nearly same height and diameter and made up of platinum and iridium alloy.

Solution 8:
Three units of length greater than a metre are

• Decameter = 10 metre
• Hectometer = 100 metre
• Kilometer = 1000 metre

Solution 9:

Solution 10:
Light year is defined as the distance travelled by light in vacuum in one year.

Solution 11:
Two units of length smaller than a metre are

• Decimeter = 0.1 metre
• Centimeter = 0.01 metre

Solution 12:
Leap year because it is a unit of time.

Solution 13:
Order of magnitude of a physical quantity is defined as its magnitude in powers of ten when that physical quantity is expressed in powers of ten with one digit towards the left decimal.
For example, volume= 52.37 m³ then the order of magnitude is 10²m³.

Solution 14:
No, micron is not same as millimeter because micron is equal to 10^-6metre while a millimeter is equal to 10^-3metre.

Solution 15:

Solution 16:

Solution 17:
A leap year refers to a year in which February has 29 days and the total days in the year are 366 days.

PAGE NO: 16
Solution 18:

Solution 19:

Solution 20:

PAGE NO: 28
Solution 1:
When one complete rotation is given to the screw hand, it moves forward or backward by a distance is called pitch of the screw. It is distance between two consecutive threads of the screw.
Pitch of the screw = distance travelled by screw in n rotations/n rotations

Solution 2:
No, least count is not same as pitch because least count is found by dividing pitch by number of divisions on the circular scale.

Solution 3:
Two uses of vernier caliper are

• Measuring the internal diameter of a tube or a cylinder.
• Measuring the length of an object.

Solution 4:
Two limitations of metre rule

• There comes an error of parallax due to thickness of the metre rule.
• We cannot use metre rule for measuring small thickness.

Solution 5:
When one complete rotation is given to the screw hand, it moves forward or backward by a distance called pitch of the screw. It is distance between two consecutive threads of the screw.
Pitch of the screw = distance travelled by screw in n rotations/n rotations
Least count refers to the smallest reading that can be accurately measured while using an instrument. The least count is the value of one division on its scale.

Solution 6:
Initial level of water in cylinder = 30 ml
Level of water in cylinder after immersing piece of copper = 50 ml
Volume of copper piece = 50-30 = 20 ml

PAGE NO: 29
Solution 7:

Solution 8:
The ratchet is used in a screw gauge to hold the object under measurement gently between the studs.

Solution 9:
If the zero of the circular scale does not coincide with the zero of the main scale (pitch scale), this is known as zero error. There are two types of zero error –

1. If the zero of the circular scale remains below the line of graduation then it is called positive zero error
2. If the zero of the circular scale lies above the line of graduation then it is called negative zero error
   For positive zero error correction, the zero error should always be subtracted from the observed reading
   For negative zero error correction, the zero error must be added to the observed reading.

Solution 10:
Two scales in a screw gauge are

• A linear scale called the main scale graduated in half millimeters
• A circular scale divided into 50 or 100 equal parts.

Solution 11:
Due to constant use, there is space for the play of screw gauge but gradually this space increases with the use or wear and tear, so that when the screw is moved by rotating it in some direction, it slips in the nut and does not cover any linear distance for some rotation of the screw head. The error due to this is known as backlash error.
It is avoided by turning the screw always in the same direction.

Solution 12:
Following procedure is used to measure the diameter of a wire

• Calculate the least count and zero error of the screw gauge.
• Place the wire in between the studs. Turn the ratchet clockwise so as to hold the wire gently in between the studs. Record the main scale reading.
• Now record the division of circular scale that coincides with the base line of main scale. This circular scale division multiplied by least count will give circular scale reading.
• The observed diameter is obtained by adding the circular scale reading to the main scale reading. Subtract the zero error if any, with its proper sign, from the observed diameter to get the true diameter.

Solution 13:

Solution 14:
Screw gauge measures a small length to a high accuracy because it has the lowest least count among the given three instruments. And low least count means high accuracy

Solution 15:

Solution 16:

Solution 17:

Solution 18:
If the zero of the circular scale remains below the line of graduation then it is called positive zero error. When there is positive zero error, then the instrument reads more than the actual reading. Therefore in order to get the correct reading, the zero error should always be subtracted from the observed reading.

Solution 19:
Pitch of the screw gauge = 0.5mm = 0.05 cm
Circular scale divisions = 100
Least Count of screw gauge = pitch of the gauge/circular scale divisions
= 0.05/100
= 0.0005cm

Solution 20:
If the zero of the circular scale lies above the line of graduation then it is called negative zero error. When there is negative zero error, then the instrument reads less than the actual reading. Therefore in order to get the correct reading, the zero error should always be added to the observed reading.

Solution 21:

• False, because the accuracy is higher in case of screw gauge due to lower least count value of 0.01mm
• True
• False, because its least count is limited to 0.1 cm. thus this length can be measured with an instrument of least count of 0.001 cm i.e. screw gauge
• False, the ratchet is used to hold the object under measurement gently between the studs.
• True

Solution 22:
The space occupied by a body is known as its volume. SI unit of volume is cubic metre (m³)

Solution 23:
The space occupied by a body is known as its volume. SI unit of volume is cubic metre (m³)

Solution 24:
1 m³ = 1000 litre
1 litre = 1/1000 m³
= 0.001 m³

PAGE NO: 30
Solution 25:

Solution 26:
SI unit of volume is cubic metre or metre³ (m³).
The relation between liter and metre³
1 metre³ = 1000 liter

Solution 27:
Pitch of the screw = 0.5 mm
Least count = 0.001 mm
Number of divisions = pitch/least count
= 0.5/0.001
= 500

Solution 28:

Solution 29:
Precautions to be taken while measuring volume of a solid lighter than water using displacement method

• The sinker should be insoluble in water
• The sinker should have a high density than water.
• Lower meniscus should be read to note down the readings and error due to parallax should be avoided.

Solution 30:
Measurement of volume of an irregular solid soluble in water using a graduated cylinder.

• In this case, kerosene or any liquid whose density is lighter than water and in which the solid is not soluble is used.
• Fill the graduated cylinder with the liquid.
• Record the lower meniscus of liquid and let the value be V₁.
• Tie the solid whose volume is to be measured to a strong string and lower it into the water gently.
• Note the reading carefully and let the value be V₂
• Volume of the solid, V = V₂ – V₁

PAGE NO: 38
Solution 1:

Solution 2:
A seconds pendulum is a pendulum which takes 2 seconds to complete one oscillation. The length of seconds pendulum, where g = 9.8ms^-2, is nearly 1 m.

Solution 3:
A stopwatch is used to measure short intervals of time.

Solution 4:
SI unit of frequency is hertz (Hz).

Solution 5:
When a pendulum completes one oscillation in one second, then the frequency is one hertz.

Solution 6:
The time period, T and frequency of oscillation, f are related as,
T = 1/f or f = 1/T

Solution 7:
One complete to and fro motion of a pendulum about its mean position is known as oscillation. Amplitude is the magnitude of the maximum displacement of the bob from the mean position on either side when an oscillation takes place.

Solution 8:
SI unit of amplitude is metre (m).

Solution 9:
A seconds pendulum is a pendulum which takes 2 seconds to complete one oscillation. The length of seconds pendulum, where g = 9.8ms^-2, is nearly 1 m.

Solution 10:

Solution 11:

Solution 12:
When a pendulum is taken from earth to moon surface, its time period will increase because the acceleration due to gravity on moon is less than that on earth and the time period depends inversely on square root of acceleration due to gravity.

Solution 13:
If time period of a pendulum becomes infinite, the pendulum will not oscillate at all as pendulum will take infinite time to complete one oscillation.

Solution 14:
Effective length of a simple pendulum is the distance of the point of oscillation (i.e. the centre of the gravity of bob) from the point of suspension.

Solution 15:

Solution 16:

Solution 17:

Solution 18:
The time period of a pendulum is independent of mass of the bob.

Solution 19:

Solution 20:
The quantity of matter contained Mass of a body can be measured by using a beam balance. in a body is called its mass. Mass is always constant for a given body.

Solution 21:
A beam balance works on the principle of moments. According to the principle of moments, under equilibrium condition, the clockwise moment due to the body on one side of beam equals the anti clockwise moment due to standard weights on the other side of beam.

Solution 22:
Precautions to be taken to measure the mass of a body using beam balance are

• The beam must be gently lowered before adding or removing weights from the pan.
• The weights should not be carried with bare hands to avoid the change in weights due to moisture and dust particles from the surrounding.
• The lever should be turned gently, in order to prevent knife edges from chipping.
• Never keep the wet or hot objects on the pan.
• The weights should be placed into weight box after use.
• Whenever you are near the actual weight, you should carefully try the weights in the descending order.

Solution 23:
SI units of time and mass are second (s) and kilogram (kg) respectively.

Solution 24:
Conditions for a beam balance to be true are

• Both the pans must be of equal weights.
• Both the arms must be of equal lengths.

PAGE NO: 44
Solution 1:
Least count of an instrument refers to the smallest reading that can be accurately measured while using the instrument. For an instrument provided with a scale the least count is the value of one division on its scale.

Solution 2:
Maximum possible error is 0.1 cm.

Solution 3:
Slope of a graph indentifies the proportional relationship between the quantities plotted.

Solution 4:

Solution 5:

Solution 6:
Accuracy is the extent to which a reported measurement approaches the true value of the quantity measured. This extent is usually described by the least count of the instrument and since the least count for a given instrument is limited hence, the accuracy of the instrument is limited.

Solution 7:
Two types of error in a measurement are

• Random errors-these errors are due to various factors. In a number of observations we get different readings every time.
  These errors can be minimized by taking observations a large number of times and taking the arithmetic mean of the readings.
• Gross error– these errors are due to carelessness of the observer like parallax, improper setting of the instrument.
  These errors can be minimized only when the observer is careful in setting up of instrument and taking readings.

Solution 8:
3000g is the most accurate measurement because it has maximum number of significant figures = 4.

Solution 9:
Basically there is no difference between the quantity being measured but there is a difference of significant figures in the measurement.

1. Number of significant figures = 3
2. Number of significant figures = 4
3. Number of significant figures = 5
   Since (3) part has maximum number of significant figures = 5, therefore it is most accurate among the given three.

PAGE NO: 46
Solution 1:
Unit is a standard quantity of the same kind with which a physical quantity is compared for measuring it.

Solution 2:
The units which can neither be derived from one another, nor can they be further resolved into other units are known as fundamental units.

Solution 3:
The units which can be expressed in terms of fundamental units of mass, length and time are known as derived units.

Solution 4:
A standard metre is equal to 1650763.31 wavelengths in vacuum, of the radiation from krypton isotope of mass 86.

Solution 5:
One standard kilogram is equal to the mass of a cylinder of nearly same height and diameter and made up of platinum and iridium alloy.

Solution 6:
SI unit of electric current is Ampere (A).

PAGE NO: 47
Solution 7:
Light year is defined as the distance travelled by light in vacuum in one year.

Solution 8:
1 Parsec is bigger because 1 Parsec is 3.26 times a light year.

Solution 9:
1 Fermi is smaller because 1 Fermi is 10^-15 m while 1 micron is 10^-6 m.

Solution 10:
Parsec refers to the distance at which an arc of length equal to 1 astronomical unit subtends an angle of one second at a point.
No, parsec is not same as astronomical unit (A.U.).
1 Parsec = 2 x 10⁵ A.U.

Solution 11:
Least count of a vernier caliper used in laboratory is 0.1mm = 0.01cm

Solution 12:
Vernier caliper is an instrument used for measuring small lengths of solid objects where an ordinary scale cannot be applied. We can measure the length accurately up to the order of 10^-2 cm, 10^-3 cm depending upon the vernier used. Therefore a vernier caliper is important to measure the fraction of a smallest division of a measuring scale which otherwise could not be done by the judgment of the eye.

Solution 13:
Least count of an instrument refers to the smallest reading that can be accurately measured while using the instrument. For an instrument provided with a scale the least count is the value of one division on its scale.

Solution 14:
No, we cannot measure the thickness of a paper with vernier caliper as its least count is only 0.1mm. We should use screw gauge instead as its least count is 0.01 mm as the thickness of the paper is in the range of 10^-2 mm.

Solution 15:
If the zero of the one scale (vernier scale or circular scale of screw gauge) does not coincide with the zero of the main scale, this is known as zero scale, zero error arises. There are two types of zero error –

• If the zero of the scale remains below the line of graduation of the main scale then it is called positive zero error
• If the zero of the scale lies above the line of graduation of the main scale then it is called negative zero error

Solution 16:
Screw gauge consists essentially of a screw with a uniform pitch which moves in a nut, thus it is named as screw gauge because the major working part is a screw.

Solution 17:

Solution 18:

Solution 19:
Material used for making screw gauge is stainless steel to avoid expansion and contraction due to change in weather as stainless steel absorbs a little heat.

Solution 20:
When one complete rotation is given to the screw hand, it moves forward or backward by a distance is called pitch of the screw. It is distance between two consecutive threads of the screw.
Pitch of the screw = distance traveled by screw in n rotations/n rotations

Solution 21:
If the zero of the circular scale lies above the line of graduation then it is called negative zero error. When there is negative zero error, then the instrument reads less than the actual reading. Therefore in order to get the correct reading, the zero error should always be added from the observed reading.

Solution 22:
Due to constant use, there is space for the play of screw gauge but gradually this space increases with the use or wear and tear, so that when the screw is moved by rotating it in some direction, it slips in the nut and does not cover any linear distance for some rotation of the screw head. The error due to this is known as backlash error.
It is avoided by turning the screw always in the same direction.

Solution 23:
A screw are threaded to twist in, when turned with a screw driver while nails are smooth to slide in straight when pounded with hammer.

Solution 24:
Screw has two types of motions: linear and circular motions.

Solution 25:
Unit of Least count of an instrument is cm.

Solution 26:
1 micron = 10^-6 m.

Solution 27:
A physical balance works on the principle of moments. According to the principle of moments, under equilibrium condition, the clockwise moment due to the body on one side of beam equals the anti clockwise moment due to standard weights on the other side of beam.

Solution 28:
1 light year = 9.46 x 10¹⁵ m

Solution 29:

Solution 30:

Solution 31:
Yes, the vibration is same as the oscillation.

Solution 32:
The time period, T and frequency of oscillation, f are related as,
T = 1/ for f = 1/T

Solution 33:
An ideal pendulum is a simple pendulum consists a heavy mass (called the bob) considered as a point mass suspended by a thread which is considered to be mass less and inextensible or non-elastic, from a fixed point or rigid support and in which there is no friction between the support and the string.

Solution 34:
Wall clock with a pendulum will run at a faster rate in winter as it pendulum rod get shorter and the pendulum will swing at a faster rate thus the clock would run faster in winters.

Solution 35:
Measurement is needed for precise description of any phenomenon happening in the world. For example, if a body is freely falling down to the ground, to understand this phenomenon we must know its velocity, time it will take to reach the ground , etc and to get answer to all our questions we need measurement.

Solution 36:

Solution 37:

Solution 38:
The maintenance of standard units is essential because any variation in these standards would lead to wrong measurements, misleading results and confusing generalizations. The standards are preserved in such a way that they do not undergo any change with the change in temperature, pressure, humidity and other environmental changes.

Solution 39:
Main characteristics of a standard unit are as follows

• It must be well defined.
• It must be of proper size. Very small or large size may cause inconvenience.
• It should be easily accessible
• It must be reproducible at all places without any difficulty.
• It must be accurately defined and must not change with time, place and physical conditions such as pressure, humidity, etc.
• It must be widely acceptable all over the world.

Solution 40:
The units which can neither be derived from one another, nor can they be further resolved into other units are known as fundamental units. Some of the fundamental units are metre (length), kilogram (mass), second (time), Kelvin (temperature), ampere (current), etc.

Solution 41:

Solution 42:

Solution 43:

PAGE NO: 48
Solution 44:
If the zero of the circular scale does not coincide with the zero of the main scale (pitch scale) when the end of the movable screw is brought in contact with the fixed end then the screw gauge is said to have a zero error.

Solution 45:
In this case, the zero error is positive
Least count of screw gauge = 0.01 mm
Thus, zero error = 0 + 4 x L.C. = 0.04 mm

Solution 46:
In this case, the zero error is negative
Least count of screw gauge = 0.01 mm
Thus, zero error = (50-47) x L.C.
= 3 x 0.01
= 0.03 mm

Solution 47:
No, we cannot measure the diameter of a wire by wrapping it around a pencil because it is not very accurate. We can use screw gauge for this purpose as it can measure the diameter correct up to 1/100 of millimeter or even less.

Solution 48:

Solution 49:

Solution 50:
Number of threads =20
Distance covered in 20 threads = 10 mm
Pitch of the screw gauge = 10/20 =0.5 mm
No of divisions on circular scale = 50
Least count = pitch/no of divisions = 0.01 mm

Solution 51:

• Oscillation – One complete to and fro motion of a pendulum about its mean position is known as oscillation.
• Time period – The time taken by a simple pendulum for an oscillation is known as the time period of a simple pendulum.
• Frequency -the number of oscillation made by the pendulum in one second is called frequency. Its SI unit is Hertz (Hz).
• Amplitude – Amplitude is the magnitude of the maximum deviation of the bob from the mean position on either side when an oscillation takes place.

Solution 52:

Solution 53:
Mass of the metal = 540g
Volume = 200cm³
Density = mass of metal/ volume
= 540 /200 = 2.70 g/cm³

Solution 54:
Mass of copper = 540 g
Density of copper = 9 g/cm³
Volume of copper used in the alloy = mass of copper / density
= 540/9 = 60 cm³
Mass of iron = 240 g
Density of iron = 8 g/cm³
Volume of iron used in the alloy = mass of iron / density
= 240/8 = 30 cm³
Total mass of the alloy = 540 + 240 = 780 g
Total volume of the alloy = 60 + 30 = 90
Density of the alloy = mass of the alloy / density of the alloy
= 780 / 90 = 8.67 g/cm³

Solution 55:

PAGE NO : 49
Solution 56:

Solution 57:
For measuring the length of an object using a vernier calipers, these steps are followed:

• First of all we find the least count and zero error of the vernier calipers.
• Place the object whose length is to be measured below the lower jaws and move the jaw till it touches the object. Record the main reading.
• Note the division on the vernier scale that coincides with some division of the main scale. Multiply this number of vernier division with least count. This is vernier scale reading.
• Record the observed length by adding the main scale reading and the vernier scale reading. Also, subtract zero error with its proper sign, if any, from the observed length to find the true length of the object.

Solution 58:

Solution 59:
Following procedure is used to measure the diameter of a wire

• Calculate the least count and zero error of the screw gauge.
• Place the wire in between the studs. Turn the ratchet clockwise so as to hold the wire gently in between the studs. Record the main scale reading.
• Now record the division of circular scale that coincides with the base line of main scale. This circular scale division multiplied by least count will give circular scale reading.
• The observed diameter is obtained by adding the circular scale reading to the main scale reading. Subtract the zero error if any, with its proper sign, from the observed diameter to get the true diameter.

Solution 60:
In order to measure the length of an object using a metre rule, the metre rule must be placed with its marking close to the object, such that the zero marking on the scale coincides with one end of the object. Then the reading on the scale corresponding to the other end of the object will give the length of the object.
Precautions to be taken for measuring the length of the object, the eye must be kept vertically above the end of the object to avoid parallax and the corresponding marking along the line should be carefully read.
The meter scale can measure up to an accuracy of 1mm or 0.1 cm

Solution 61:

Solution 62:

Solution 63:

Solution 64:

Solution 65:

Solution 66:

• Oscillation – One complete to and fro motion of a pendulum about its mean position is known as oscillation.
• Amplitude – Amplitude is the magnitude of the maximum deviation of the bob from the mean position on either side when an oscillation takes place.
• Frequency – the number of oscillation made by the pendulum in one second is called frequency. Its SI unit is Hertz (Hz).
• Time period – The time taken by a simple pendulum for an oscillation is known as the time period of a simple pendulum.

Solution 67:

PAGE NO: 50
Solution 68:

Solution 69:
To measure mass of a body using a physical balance

1. Before starting, bring the plumb line just above the pointed projection by adjusting the leveling screws at the base. The beam is then gently raised using the lever. And it should be ensured that the pointer swings equally on both sides of the zero mark of the scale.
2. Now lower the beam gently and given body is kept on left pan.
3. Next, place some weight on the right pan form the weight box using the forceps.
4. Now the lever is turned towards right so that the beam rises and the power begins to swing to pointer swing on either side. It must be carefully noted that the side to which the pointer moves more, denotes lesser mass on that side.
5. Go on adjusting the standard weights till the pointer swings equally on both sides of the zero mark.
6. At this stage, the total mass of weights on the right pan gives the mass of the body.
   Three precautions to be taken to measure the mass of a body using beam balance are
   • The beam must be gently lowered before adding or removing weights from the pan.
   • The weights should not be carried with bare hands to avoid the change in weights due to moisture and dust particles from the surrounding
   • Whenever you are near the actual weight, you should carefully try the weights in the descending order.
     Conditions for a beam balance to be true are
7. Both the pans must be of equal weights.
8. Both the arms must be of equal lengths.

Solution 70:

PhysicsChemistryBiologyMaths

Frank ICSE Solutions for Class 10 Chemistry – Chemical Bonding

PAGE N0 : 39
Solution 1:
Chemical bond: A chemical bond may be defined as the linkage that stands for the force which actually holds the atoms together within the molecule.
Chemical bonding: The phenomenon during which a chemical bond is formed is called chemical bonding.

Solution 2:
Atoms combine to attain the electronic configuration of nearest inert gases as the atoms of inert gases are very stable having 8 electrons or duplet (or 2 electrons in case of helium atom) in their outermost shell.

Solution 3:
Electrovalent compounds: The chemical compounds containing electrovalent bonds are called electrovalent or ionic compounds.
For example: Sodium chloride (NaCl).
Covalent compounds: The chemical compound, formed as a result of mutual sharing of electrons or electron pairs thereby establishing a covalent bond is called a covalent or molecular compound.
For example: Hydrogen molecule (H₂)

Solution 4:
The conditions for the formation of an electrovalent bond are:

1. Low ionization energy of electropositive atom
2. High electron affinity of the electronegative atom.
3. Large electronegativity difference.
4. High lattice energy.
   Concept Insight:
   • Lower is the ionization energy of atom, higher is its tendency to lose electron to form a cation and form ionic bond.
   • Higher the value of electron affinity of an atom, greater will be its tendency to form anion and form ionic bond.
   • If the electronegativity difference of two elements is higher, more easy will be the transfer of electrons and hence more chances of ionic bond formation.
   • Lattice energy is the energy released when positive and negatively charged atoms called ions come closer to form a crystal because the attractive forces among the oppositely charged ions tend to decrease the energy of the system. Higher is the lattice energy, greater will be the ease of formation of the compound.

Solution 5:

Solution 6:
Hydrogen chloride has a polar covalent bond because in hydrogen chloride the higher electronegativity of chlorine atom attracts the shared electron pair towards itself. As a result, the chlorine atom gets a partial negative charge while the hydrogen atom gets a partial positive charge. Hence such a covalent bond with charge separation is called polar covalent bond.
While methane has a non polar covalent bond because in case of methane molecule the shared electron pairs are at equal distance from the carbon and hydrogen atoms, because neither the carbon atom nor the hydrogen atom has enough electronegativity difference between each other to attract the shared pairs of electrons towards itself. Hence no charge separation occurs in the covalent bond due to which it is called non polar covalent bond.
Concept Insight: When a covalent bond is formed between the atoms of the same elements of equal electronegativity then the electron pairs are shared equally between the atoms and the bond so formed is called non polar covalent bond. On the other hand, if the covalent bond is formed between atoms of different elements, with difference in electro negativity, the electrons are not shared equally between the atoms. The more electronegative atom pulls the bonded pair of electrons towards itself and acquires negative charge while the other less electro negative atom acquires positive charge and the bond becomes polar covalent bond.

Solution 7:
In terms of electron transfer, oxidation is defined as the phenomenon in which an atom loses electron to form a positively charged cation while reduction is defined as the phenomenon in which an atom gains electron to form a negatively charged ion called anion.
During formation of ionic bond one atom undergoes oxidation while another atom undergoes reduction.

PAGE NO : 48
Solution 8:

Solution 9:

Solution 10:

1. Sodium chloride dissolves in water because it is an ionic compound and water is also a polar covalent compound. Water decreases the electrostatic forces of attraction among the sodium and chloride ions due to which these ions become free in water, hence sodium chloride dissolves.
   On the other hand, carbon tetra chloride has non polar covalent bond and water has polar covalent bond. Hence, water is unable to break the non polar covalent bond of carbon tetra chloride. So it is insoluble in water
2. Helium does not form He2 molecule as it has its outermost shell complete i.e. two electrons in its valence shell. Due to this complete valence shell helium atom is very stable hence does not participate in chemical bonding to form  He2 molecule.
3. Pure water does not conduct electricity because it has a polar covalent molecule hence does not have ions in it which can conduct electricity.
   On adding sodium chloride to pure water, sodium chloride breaks apart into sodium and chloride ions because water being polar decreases the strong forces of attraction among sodium and chloride ions. Now, pure water has ions present in it which can conduct electricity.
4. Cl2 is a non polar molecule because the bond is between same atoms that is chlorine with zero electronegativity difference among them. So the shared electron pair is attracted equally by the two chlorine atoms hence there is no separation of charges in the bond formed so the chlorine molecule is non polar.
5. In case of HCl the bond is formed between two different atoms that is hydrogen and chlorine with enough electro negativity difference so that the shared electron pair is attracted towards more electronegative chlorine atom which acquires partial negative charge while the hydrogen atom acquires partial positive charge hence HCl is a polar molecule.
6. Metals have low ionization energy due to which they can lose their outermost electrons easily to form positive metallic ions hence metals are electropositive.
   For example Sodium metal always form Na⁺ions, Potassium forms K⁺ ions etc.

Solution 11:

1. (i) when the electro negativity difference between the two atoms is high then the bond formed will be purely ionic.
   (ii) When the electro negativity difference between the two atoms is low then the bond formed will be polar covalent bond.
   (iii) When the electro negativity difference between the two atoms is zero then the bond formed will be purely covalent.
2. Ionic compounds = NO, NH₄Cl, NH₄NO₃
   Covalent compounds = N2, NH2 , NO
   Polar compounds = NCl3

Concept Insight: Electro negativity difference between the bonded atoms determines the ease of transfer of electrons between the atoms. On the basis of extent of transfer of electrons between the two atoms the bond will be ionic, covalent or polar.

Solution 12:

Solution 13:

1. MgCl₂, CaCl₂.
2. Urea, Glucose.
3. CH₄, benzene.
4. SO₂, H₂S
5. H₂, N₂

Solution 14:
The necessary conditions for the formation of covalent molecule are:

1. Number of valence electrons: Both the participating atoms should have four or more valence electrons in their valence shell.
2. Equal electro negativities: The combining atoms should have equal electro negativities so that no transfer of electrons takes place.
3. Equal electron affinities: The combining atoms should also have equal electron affinities i.e. equal attraction for electrons.
4. Ionization energy: It should be high for both the atoms so that there is no chance of removal of electrons.
5. High nuclear charge and small inter nuclear distance: Both these conditions favor the formation of covalent bond because during the formation of a covalent bond the electron density gets concentrated between the nuclei of the combining atoms and this electronic charge is responsible for holding the two nuclei together.
   The properties of covalent compounds are:
6. Nature: They are generally volatile liquids or gases. Some may be gases like urea, sugar etc.
7. Low melting and boiling points: Since the intermolecular forces of attraction are weak, very small amount of heat energy is required to overcome these forces hence their melting and boiling points are low.
8. Electrical conductivity: Since covalent compounds are made up of molecules and not ions, so they do not conduct electricity.
9. Solubility: These are insoluble in water but soluble in organic solvents.
10. Ionization in solution: These do not ionize when dissolved in water except some polar covalent compounds like HCl.
11. Molecular reactions: These participate in reactions as a molecule so the reactions are called molecular reactions. These are slow reactions.

Solution 15:
Coordinate bond: The bond formed between two atoms by a pair of electrons, provided entirely by one of the combining atoms, is called a coordinate bond or dative bond.
Conditions for the formation of coordinate bond:

1. One of the two atoms must have at least one lone pair of electrons.
2. Another atom should be short of at least a lone pair of electrons.

Solution 16:
Lone pair: A pair of electrons which is not shared with any other atom is known as the lone pair of electrons.
For example in NH₃, Nitrogen has a lone pair of electrons which is not shared with any hydrogen atom.
Shared pair: A pair of electrons which is shared with other atoms to form a bond is known as shared pair of electrons.
For example in HCl the pair of electrons responsible for bond formation between H and Cl is called shared pair.

Solution 17:

Solution 18:

1. The forces of attraction between the molecules of covalent compounds are weak because the molecules are neutral. So, they are generally gases or liquids or soft solids.
2. Covalent compounds have low melting and boiling point because the intermolecular forces of attraction among the molecules of covalent compounds are weak. Hence very small amount of heat energy is required to overcome the attraction between the molecules.
3. On the basis of principle like dissolves like we can interpret the insolubility of non polar covalent compounds. Since water is a polar covalent compound that is it has positively and negatively charged ends but the non-polar covalent compounds do not have any kind of charge separation. So water molecules are unable to interact with the molecules of non polar compound and break apart the intermolecular forces of attraction among non-polar molecules making them soluble in water.
4. Polar covalent compounds are good conductors of electricity because when these are dissolved in water, they ionize and act as electrolyte to produce ions which are responsible for conduction of electricity.
   
   For example polar covalent compound HCl in water behaves as:
   These hydronium and chloride ions produced on dissolution of HCl in water are responsible for conduction of electricity.

PAGE NO : 41
Solution 2002-1:

Solution 2004-1:

Solution 2005-1:

PAGE NO : 42
Solution 2006-1:

Solution 2006-2:

1. (b)
2. (a)

Solution 2007-1:

Solution 2008-1:
ionises when dissolved in water

Solution 2008-2:

1. Covalent bond.
2. Co-ordinate bond

Solution 2009-1:

1. solid
2. low

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Frank ICSE Solutions for Class 10 Chemistry – Periodic Properties and Variation of Properties

PAGE NO : 16
Solution 1:
Modern periodic law states that the physical and chemical properties of elements are a periodic function of their atomic numbers i.e., if the elements are arranged in the order of their atomic numbers, the elements with similar properties are repeated after definite regular intervals.
Concept Insight: The elements are characterized by their atomic number as well as atomic weight. Modern periodic law uses atomic number which is number of protons or number of electrons present in an atom of an element.

Solution 2:
Modern periodic table consists of eighteen groups and seven periods.
Concept Insight: Classification of elements on the basis of increasing atomic number is known as Modern Periodic Table. The vertical columns are called groups and the horizontal rows are called periods.

Solution 3:
The recurrence of similar properties of elements after certain regular intervals when they are arranged in the order of increasing atomic numbers is called periodicity.
Concept Insight: Periodicity in properties is due to the repetition of similar outer electronic configuration of elements at certain regular intervals.

Solution 4:
In general, the elements belonging to a group have the same number of valence electrons .For example, all the group 1 elements have valency one since they have only one electron in their outermost shell.
In general, the elements belonging to a period do not have same valency but their valence shell remains the same. For example, second period has 8 elements with atomic number 3 to 10 but in all of them the valence electrons are present in shell number two.
Concept Insight: For elements in a group the number of electrons present in the outermost shell is the same and therefore the elements have same valency and or elements in a period number of electrons present in the outermost shell of elements in a period increase from left to right but the shell does remains the same.

Solution 5:
Fluorine has lower electron affinity than chlorine because of the small size of fluorine which results in stronger repulsion between the electron and the electrons already present in the atom of fluorine. Hence the energy released in accepting an electron is lesser in fluorine than that of chlorine.
Concept Insight: For answering this question you should recall that electron affinity is the energy released when an electron is added to an isolated gaseous atom to form an anion. Due to small size of fluorine atom, the valence shell is already crowded, hence when an electron is added to a fluorine atom in gaseous state there occurs strong repulsion between the added electron and those already present in the atom hence less amount of energy is released.

Solution 6:

1. The element with highest first ionization energy: Neon (Ne)
2. The element with highest electro negativity: Fluorine (F)
3. The element with largest atomic size: Lithium (Li)
4. The most reactive non-metal: Fluorine
5. The most reactive metal: Lithium
6. Concept Insight:
   • Neon has highest ionization energy since it is a noble gas and has its octet complete which makes it very stable.
   • Fluorine has highest electronegativity as we know that electro negativity increases along a period due to decreasing atomic size and increasing nuclear charge.
   • Lithium has the largest atomic size since it is an alkali metal i.e., belongs to group 1 and we know that as we move from left to right in a period atomic size decreases. So lithium has largest size while fluorine has smallest size in second period.
   • Fluorine is most reactive non metal as it requires only one electron to complete its octet and become stable.
   • Lithium is the most reactive metal as it can complete its octet by losing its single electron present in its outermost shell.

PAGE NO : 17
Solution 7:

1. The most metallic element will be found at C.
2. The most non-metallic element will be found at D.

Concept Insight: For answering this question you should recall metallic character increases down the group and also increases with the increasing size of the atom. Since elements of group 1 has largest atomic sizes among all the elements of periodic table so the most metallic element belongs to group 1.
Similarly, non-metallic character decreases down the group and increases with the decreasing size of atom.

Solution 8:

Solution 9:

1. 18, 7.
2. First.
3. Seventeen.
4. Electron affinity.
5. Decrease, increase
6. Fluorine
7. Zero.

Solution 10:

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

Concept Insight: (i) There is inverse relation between atomic size and electron affinity. More is the size, less is the electron affinity and vice versa because more is the size of atom, more is the distance between the nucleus and last shell to which electron enters. This results in decrease in force of attraction between the nucleus and incoming electron and hence the electron affinity decreases. Fluorine has smaller size than chlorine so it must have less electron affinity than chlorine.

Solution 11:

1. Li < Be < B
2. I < Br < F < Cl
3. SiO₂ < P₂O₅ < SO₃ < Cl₂O₇
4. I⁺ < I < I^–

Concept Insight:

1. Li, Be, B belongs to second period and ionization energy increase as we move left to right in a period due to increased nuclear charge and decrease of atomic size.
2. Electron affinity decreases down the group due to increase in atomic size as it results in more distance between nucleus and last shell to which incoming electron enters. Hence, incoming electron feels less attraction from the nucleus.
3. In a period, acidic nature of oxide increases.
4. Size of a cation is always smaller than the corresponding atom due to decrease in number of electrons and increase in effective nuclear charge i.e., greater force of attraction by the nucleus on the electrons.
   Size of an anion is always more than the corresponding atom due to decrease in effective nuclear charge i.e., lesser force of attraction by the nucleus o the electrons.

Solution 12:
The statement that in each period, the atomic size gradually decreases with increase in atomic number means that as move from left to right in a period, nuclear charge increases by one unit in each succeeding element while the number of shells remains the same. Due to this increased nuclear charge, the electrons of all the shells are pulled closer to the nucleus thereby bringing the outer most shell closer to the nucleus. With the result, the atomic size decreases across a period.
For example, in the second period from lithium to fluorine, lithium has the largest size while fluorine has the smallest size.

Solution 13:
In the case of noble gases or inert gases there are exceptions and the atomic radius or size of the elements are greater than the other elements of the period to which these elements belong.

PAGE NO : 18
Solution 14:
As we move down a group, the atomic radii increase because a new shell is added at each succeeding element though the number of electrons in the outer most shell remains the same. Thus, the atomic size of elements increases in size downward.
Although nuclear charge also increases in going down the group but the effect of nuclear charge on atomic size is much less than the increase due to addition of a new shell.
In group 17, the atomic size follows the trend:
F < Cl < Br < I

Solution 15:
The elements of third period are:
Na, Mg, Al, Si, P, S, Cl, Ar
The most metallic element is sodium i.e., Na and the most non-metallic element is chlorine i.e., Cl.
Concept Insight:
In a period, metallic character decreases on moving from left to right because of decrease in size of atom due to which elements cannot lose electron easily.

Solution 16:

Solution 17:

Solution 18:
Electron affinity is the energy released when an electron is added to an isolated gaseous atom to form the negative ion (anion).
Unit: Its units are electron volt (eV).
Its SI units are Kilojoules per mole(KJmol)^-1

Solution 19:
Out of A and B, A will ionize more easily to form a negative anion because of the high value of electron affinity, energy released during addition of electron will be high hence the resulting anion formed will be more stable than the corresponding atom.

Solution 20:

1. Larger the atomic size, farther is the valence electron from the nucleus and lesser is the pull exerted on it. As a result, electron can be easily removed from the valence shell and hence more metallic is the element.
2. Halogens need only one electron to complete their octet and become stable their atomic size is very less hence the distance between their last shell and nucleus is very less, as a result the force of attraction between the nucleus and the incoming electron is less and hence the electron affinity is high for halogens.
3. When an atom loses or gain electron to form ion, the number of electrons present in the outermost shell also changes. Corresponding to that effective nuclear charge on the changed number of electrons also change which further changes the size of an atom as there is inverse relation between effective nuclear charge and size of atom.
4. K and Li belongs to group 1 i.e., metals and we know that for metals chemical reactivity of elements increases down the group because chemical reactivity increases as electropositive or metallic character increases.
5. The electronegativity of chlorine is higher than sulphur because both of them belong to third group and chlorine follows sulphur. We know that, within a period electronegativity increases as we move from left to right because of decrease in atomic size and increase in nuclear charge.
6. Group 17 elements are non metals because they have 7 electrons in their valence shell and ionize by accepting 1 electron to form an anion.
   For example group 17 elements F, Cl, Br and I all have 7 electrons in their valence shell and ionize by accepting 1 electron to form F^–, Cl^–, Br^– and I^–.
   Group 1 elements are metals because they have tendency to lose the one electron present in their valence shell and form positive ion.
   For example, group 1 elements Li, Na, K, Rb, Cs have tendency to lose the one electron present in their valence shell and form positive ions Li^+., Na^+., K^+., Rb^+.and Cs^+.

Solution 21:

1. (C) i.e. 2, 8, 2 because it has only 2 electrons in its valence shell which can be lost to form a di positive cation.
2. (C) i.e. 0.72, 0.72
3. (d) i.e. element forms basic oxide because the element is a metal as it has valency 1.
4. (a) i.e. F because it belongs to group 17 whose elements have valency 7 and thus requires only 1 electron to complete their octet.

Solution 2000-1:

1. Number of elements in period 1 = 2
   In period 2 = 8
   In period 3 = 8.
2. Elements in period 1 are Hydrogen (H) and Helium (He).
3. Atomic size of elements decreases on moving from left to right in a period.

Solution 2000-2:

1. The elements at the end of period 2 and period 3 both have their outermost shell complete and belong to noble gases.
2. An element in group 7 is likely to be non metallic in character since group 7 element will have 7 electrons in its valence shell.
3. Metallic.

PAGE NO : 19
Solution 2001-1:

1. Atomic number.
2. Period, non-metallic.
3. More.
4. Number of outer electrons.

Solution 2002-1:
A group is a vertical column of elements having the same number of valence electrons and same valency in the periodic table. There are 18 groups in the periodic table.

Solution 2002-2:
Within a group the element with the greatest metallic character and largest size is expected to be present at the bottom of the group.

Solution 2002-3:
Ionization potential decreases down the group because atomic size increases down the group which decreases the effective nuclear charge over the valence electron which further can now be removed easily.

Solution 2002-4:
There are 8 elements in period 2.

Solution 2003-1:

1. Al₂(SO₄)₃
2. Covalent.
3. The elements of group VIIA all have same number of electrons in their valence shell and same valency.
4. Neon
5. 8 electrons are present in the valence shell of the element with atomic number 18.
6. Electron affinity.
7. Electronic configuration of element in the third period which gains one electron to become an anion is 2, 8, 7.
8. Decreases, number of valence shell electrons/outermost shell electrons, valence shell/ outermost shell.

PAGE NO : 20
Solution 2004-1:

1. Na Mg Al Si P S Cl.
2. (a) lower, higher.
3. remains the same.

Solution 2005-1:

1. b
2. d
3. c
4. a
5. c

Solution 2006-1:

1. Second period.
2. Nitrogen. It should be placed between oxygen andcarbon.
3. Beryllium < nitrogen < fluorine
4. Fluorine (F)

Solution 2007-1:

1. Thallium
2. Boron
3. 3
4. BCl₃
5. The elements in the group to the right of this boron group will be less metallic in character because on moving to the right of the periodic table metallic character decreases as ionization energy deceases and tendency to lose electron also decreases.

Solution 2008-1:

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

Solution 2008-2:

1. (i) First element in period 2 is Lithium and last element is Neon.
   (ii)Atomic size increases on moving from top to bottom of a group.
   (iii)Chlorine among halogens has the greatest electron affinity.
   (iv) All elements in group 7 have same number of valence shell electrons.
2. (i) metallic
   (ii) smallest
3. (i) Ba i.e. Barium will form ion most readily since it is at the bottom of a group its ionization energy is low because its atomic size is more. Due to this effective atomic charge of nucleus over the valence shell electron is least and it can be removed easily.
   (ii) Electro negativity of an element measures the capacity of an element to attract the shared pair of electrons in a bond towards itself.

Solution 2009-1:
(d) Fluorine

PAGE N0 : 22
Solution 2009-2:

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